{"id":30234,"date":"2018-08-01T11:00:23","date_gmt":"2018-08-01T11:00:23","guid":{"rendered":"https:\/\/cbselibrary.com\/?p=30234"},"modified":"2020-10-21T12:35:17","modified_gmt":"2020-10-21T07:05:17","slug":"a-new-approach-to-icse-physics-part-2-class-10-solutions-electric-circuits-resistance-ohms-law","status":"publish","type":"post","link":"https:\/\/cbselibrary.com\/a-new-approach-to-icse-physics-part-2-class-10-solutions-electric-circuits-resistance-ohms-law\/","title":{"rendered":"A New Approach to ICSE Physics Part 2 Class 10 Solutions Electric Circuits, Resistance & Ohm\u2019s Law"},"content":{"rendered":"

A New Approach to ICSE Physics Part 2 Class 10 Solutions Electric Circuits, Resistance & Ohm\u2019s Law<\/span><\/h2>\n

Exercise – 1<\/strong><\/span><\/p>\n

Question 1.<\/strong><\/span>
\nIn which direction conventional current and electronic current flow from a source of electricity ?
\nAnswer:<\/strong><\/span>
\nElectronic current is always in opposite direction to conventional current.<\/p>\n

    \n
  1. When both the bodies are positively charged and are in contact the body having more +ve charge is at higher potential conventional current from A to B + 100 > + 70 from higher potential to lover potential.
    \n\u2234 Electronic current from B to A.
    \n\"A<\/li>\n
  2. When both A and b negative charge conventional current: from higher potential (\u201470) to lower potential (\u2014100) i.e. from B to A electronic current: from A to B.
    \n\"A<\/li>\n
  3. When A has +ve charge of 100 units and B has \u2014ve charge of 70 units Conventional Current: from A to B Higher potential to lover potential a electronic current from b to A.
    \n\"A<\/li>\n<\/ol>\n

    Question 2.<\/strong><\/span>
    \nDefine electric potential. State its practical unit and define it.
    \nAnswer:<\/strong><\/span>
    \nElectric Potential<\/strong> : \u201cIs the electrical state of a conductor which determines the direction of flow of charrge when two conductors are either in contact or joined by a metallic wire.\u201d
    \nOr<\/strong>
    \nELECTRIC POTENTIAL \u201cat a point is the work done in moving unit positive charge from infinity to that point in an electric field.
    \nS.I. unit is volt.
    \nVOLT :<\/strong> \u201cIf work done in moving 1 coulomb of charge from one point to other is 1 joule, the potential-difference between two points is said to be 1 volt.\u201d<\/p>\n

    Question 3.<\/strong><\/span>
    \nDefine quantity of charge. States its practical unit and define it
    \nAnswer:<\/strong><\/span>
    \nQuality of charge is \u201cThe number of charges (electrons) which drift from a higher to a lower potential is called quantity of charge.\u201d
    \nParticle unit of charge is coulomb.
    \nCoulomb:<\/strong> \u201cFlow of 6.25 \u00d7 1018<\/sup> electrons through a conductor constitute 1 coulomb.\u201d
    \nOr<\/strong>
    \n\u201cCharge carried by\u00a0 6.25 \u00d7 1018<\/sup> electrons is called 1 coulomb.\u201d<\/p>\n

    Question 4.<\/strong><\/span>
    \nDefine electric current State its practical unit and define it
    \nAnswer:<\/strong><\/span>
    \nElectric current: \u201cRate of flow of charge\u201d. I = Q\/t = ne\/t
    \nS.I. Unit \u2192 Ampere (A)
    \nAmpere :<\/strong> \u201cWhen a charge of 1 coulomb passes in 1 second current flowing is one ampere.\u201d<\/p>\n

    Question 5.<\/strong><\/span>
    \nState two multiples and two submultiples of the unit of electric potential and electric current
    \nAnswer:<\/strong><\/span>
    \nMultiple units of :<\/strong>
    \nElectric potential is<\/p>\n

      \n
    1. ( Kv ) kilovolt = 103<\/sup> V<\/li>\n
    2. Megavolt (Mv) = 106\u00a0<\/sup>V<\/li>\n<\/ol>\n

      Electric current :
      \n<\/strong><\/p>\n

        \n
      1. KA (kilo-ampere) = 1000 A<\/li>\n
      2. MA (Mega-ampere) = 106<\/sup><\/li>\n<\/ol>\n

        Sub multiple units of:<\/strong>
        \nElectric potential:<\/strong><\/p>\n

          \n
        1. mV milli volt = 10-3<\/sup> V<\/li>\n
        2. \u00b5v = micro volt = lO-6\u00a0<\/sup>V<\/li>\n<\/ol>\n

          Electric current <\/strong>:<\/p>\n

            \n
          1. mA = milli ampere = 10-3<\/sup> A<\/li>\n
          2. \u00b5A = micro-ampere = 10-6<\/sup> A<\/li>\n<\/ol>\n

            Question 6.<\/strong><\/span>
            \nWhat do you understand by the terms potential difference? State its practical unit
            \nAnswer:<\/strong><\/span>
            \nPotential difference: \u201cIs the amount of work done in moving a unit positive charge from one point to other.\u201d Practical unit – volt.<\/p>\n

            Question 7.<\/strong><\/span>
            \nDefine
            \n(a)<\/strong> open electric circuit
            \n(b)<\/strong> closed electric circuit.
            \nAnswer:<\/strong><\/span>
            \n(a) Open electric circuit :<\/strong> \u201cAn electric circuit in which low of current stops, because of an open switch is called an open_electric circuit.
            \n(b) Closed electric circuit :<\/strong> \u201cAn electric circuit in which a current flows continuously, because the switch is closed is called a closed electric .
            \n\"A<\/p>\n

            Question 8.<\/strong><\/span>
            \nWhat do you understand by the term electric resistance? state its practical unit.
            \nAnswer:<\/strong><\/span>
            \n\u201cThe obstruction offered to the flow of current by a conductor (wire) is called its RESISTANCE.\u201d
            \nS.l. UNIT \u2014 is 1\/OHM or (Ohm)-1<\/sup><\/p>\n

            Question 9.<\/strong><\/span>
            \nWhat do you understand by the term electric conductance? State its practical unil
            \nAnswer:<\/strong><\/span>
            \n\u201cThe reciprocal of resistance is called electric CONDUCTANCE\u201d. i.e.
            \nConductance = 1\/Resistance
            \nS.l. Unit of Conductance \u03a9-1<\/sup> or OHM-1<\/sup><\/p>\n

            Question 10.<\/strong><\/span>
            \nWhat is a superconductor ? Name two materials and the temperature at which they become superconductors.
            \nAnswer:<\/strong><\/span>
            \nSUPER CONDUCTOR : \u201cThe substances which lose resistance when they are cooled to very low temperature (nearly absolute zero) are called super conductors, e.g. mercury at 4.12 k, LEAD. TIN, VANADIUM etc. and this phenomenon is called SUPER CONDUCTIVITY. The temperature at which they become super-conductors is called CRITICAL TEMPERATURE.<\/p>\n

            Question 11.<\/strong><\/span>
            \nState the laws of resistance.
            \nAnswer:<\/strong><\/span>
            \nLaws of resistance :<\/strong><\/p>\n

              \n
            1. Resistance of a conductor is directly proportional to its length R\u00a0\u03b1 l<\/li>\n
            2. Resistance of a conductor is Inversely proportional to its area of cross-section R \u03b1 1\/a<\/li>\n
            3. Resistance of a conductor depend upon its nature i.e. copper has less resistance than iron.<\/li>\n
            4. Resistance of a conductor increases with increase in temperature i.e. resistance of filament of a bulb is more when lighted as compared to when it is not lighted.<\/li>\n<\/ol>\n

              Question 12.<\/strong><\/span>
              \nDefine specific resistance and state its unit in CGS and SI system.
              \nAnswer:<\/strong><\/span>
              \nSpecific resistance:<\/strong> \u201cIs the resistance of a wire of that material of unit length and unit area of cross-section.\u201d
              \nUnit:<\/strong>
              \nIn C.G.S. system \u2192 [\u03a9 – cm] ohm – cm
              \nIn S.I. system \u2192 [\u03a9 – m] ohm – metre.<\/p>\n

              Question 13.<\/strong><\/span>
              \nName two materials in each case whose resistance
              \n(a)<\/strong> increases,
              \n(b)<\/strong> remains the same and
              \n(c)<\/strong> decreases with the rise in temperature.
              \nAnswer:<\/strong><\/span>
              \nTwo materials:<\/strong><\/p>\n

                \n
              1. Cu, iron, sp. resistance increases Tungston with increase in temp.<\/li>\n
              2. Metallic alloys like eureka, Manganin and German silver The sp. resistance remains the same with rise in temp.<\/li>\n
              3. Carbon and Rubber. Resistance decrease with increase in temperature.<\/li>\n<\/ol>\n

                Question 14.<\/strong><\/span>
                \nGive two differences between the electric resistance and electric specific resistance of a material
                \nAnswer:<\/strong><\/span>
                \nTwo differences between resistance and sp. resistance.
                \nResistance :<\/strong><\/p>\n

                  \n
                1. S.I. unit is ohm (\u03a9)<\/li>\n
                2. It is measured as ratio of pot. difference at the ends of a conductor to the current flowing through the conductor.<\/li>\n<\/ol>\n

                  Resistivity or sp. Resistance :<\/strong>
                  \nS.I. unit [\u03a9-m] ohm. metre. It is measured as the resistance offered by a conductor of unit length and unit area of cross-section.<\/p>\n

                  Multiple choice questions<\/strong><\/span><\/p>\n

                  Tick ( \u2713) the most appropriate option.<\/strong><\/p>\n

                  Question 1.<\/strong><\/span>
                  \nThe graph between V\/I for a conductor is a straight line. The slope of the graph represents :
                  \n(a)<\/strong> resistivity
                  \n(b)<\/strong> resistance
                  \n(c)<\/strong> electric potential
                  \n(d)<\/strong> none of these
                  \nAnswer:<\/strong><\/span>
                  \n(b)<\/strong> resistance<\/p>\n

                  Question 2.<\/strong><\/span>
                  \nTwo conductors A and B have 500 and 100 units of . negative charge when the conductors are connected by
                  \nan electric wire the conventional current flows from :
                  \n(a)<\/strong> A to B
                  \n(b)<\/strong> B to A
                  \n(c)<\/strong> Current does not flow
                  \n(d)<\/strong> none of these
                  \nAnswer:<\/strong><\/span>
                  \n(b)<\/strong> B to A<\/p>\n

                  Question 3.<\/strong><\/span>
                  \nA conductor at 4.2 K is found to offer no resistance. Such a conductor is called
                  \n(a)<\/strong> zero conductor
                  \n(b)<\/strong> superconductor
                  \n(c)<\/strong> absolute conductor
                  \n(d)<\/strong> none of these
                  \nAnswer:<\/strong><\/span>
                  \n(b)<\/strong> superconductor<\/p>\n

                  Question 4.<\/strong><\/span>
                  \nWhich of the following is non-ohmic resistance ?
                  \n(a)<\/strong> Copper wire
                  \n(b)<\/strong> Brass wire
                  \n(c)<\/strong> Copper wire wound on an electromagnet
                  \n(d)<\/strong> Constantan wire
                  \nAnswer:<\/strong><\/span>
                  \n(b)<\/strong> Brass wire<\/p>\n

                  Question 5.<\/strong><\/span>
                  \nWhich of the following an ohmic resistance ?
                  \n(a)<\/strong> Diode valve
                  \n(b)<\/strong> Filament of a bulb
                  \n(c)<\/strong> Carbon are light
                  \n(d)<\/strong> Manganin wire
                  \nAnswer:<\/strong><\/span>
                  \n(d)<\/strong> Manganin wire<\/p>\n

                  Question 6.<\/strong><\/span>
                  \nA conductor has a resistivity of 2.63 \u00d7 10-8<\/sup> \u03a9 m at 20\u00b0 C. If the temperature of conductor is raised to 200\u00b0C, its resistivity will :
                  \n(a)<\/strong> increase
                  \n(b)<\/strong> decrease
                  \n(c)<\/strong> remain unaffected
                  \n(d)<\/strong> none of these
                  \nAnswer:<\/strong><\/span>
                  \n(a)<\/strong> increase<\/p>\n

                  Question 7.<\/strong><\/span>
                  \nAmongst the following substance, the resistance will decrease with the increase in temperature in case of:
                  \n(a)\u00a0<\/strong>copper
                  \n(b)<\/strong> carbon
                  \n(c)\u00a0<\/strong>brass
                  \n(d)<\/strong> nichrome
                  \nAnswer:<\/strong><\/span>
                  \n(b)<\/strong> carbon<\/p>\n

                  Numericals on Specific Resistance<\/strong><\/span><\/p>\n

                  Practice Problems : 1<\/strong>
                  \nQuestion 1.<\/strong><\/span>
                  \nA wire of resistance 4.5 \u03a9 and length 150 cm, has an area of cross-section of 0.04 cm-2<\/sup>. Calculate sp. resistance of the wire.
                  \nAnswer:<\/strong><\/span>
                  \n\"A<\/p>\n

                  Question 2.<\/strong><\/span>
                  \nA wire of length 40 cm and area of cross-section 0.1 mm2<\/sup> has a resistance of 0.8 fl Calculate sp. resistance of the wire.
                  \nAnswer:<\/strong><\/span>
                  \nI = 40 cm, area of cross-section a = 0.1 m m2<\/sup> = 0.1\/100 c m2<\/sup> R = 0.8 \u03a9
                  \n\"A<\/p>\n

                  Practice Problems : 2<\/strong><\/p>\n

                  Question 1.<\/strong><\/span>
                  \nResistance of a conductor of length 75 cm is 3.25 \u03a9. Calculate the length of a similar conductor, whose resistance is 13.25\u03a9.
                  \nAnswer:<\/strong><\/span>
                  \n\"A<\/p>\n

                  Question 2.<\/strong><\/span>
                  \nA conductor of length 85 cm has a resistance of 3.750. Calculate the resistance of a similar conductor of length 540 cm.
                  \nAnswer:<\/strong><\/span>
                  \n\"A<\/p>\n

                  Practice Problems : 3<\/p>\n

                  Question 1.<\/strong><\/span>
                  \nA resistance wire made from German silver has a resistance of 4.250. Calculate the resistance of another wire, made from same material, such that its length increases by 4 times and area of cross-section decreases by three times.
                  \nAnswer:<\/strong><\/span>
                  \n\"A<\/p>\n

                  \"A<\/p>\n

                  Question 2.<\/strong><\/span>
                  \nA nichrome wire of length l and area of cross-section a\/ 4 has a resistance R. Another nichrome wire of length 31 and area of cross-section a\/2 has a resistance of R1<\/sub> Find the ratio of R, : R.
                  \nAnswer:
                  \n<\/strong><\/span>As both the wire are made of same material, have same sp. resistance
                  \n\"A<\/p>\n

                  Exercise – 2<\/strong><\/span><\/p>\n

                  Question 1.<\/strong><\/span>
                  \n(a)<\/strong> Define series circuit.
                  \nAnswer:<\/strong><\/span>
                  \n(a) Series circuit:<\/strong> \u201cResistances are said to be connected in series if same current flows through them and the resistance are connected end to end.
                  \n(b)<\/strong> State three characteristics of a series circuit Ans. Characteristics of series circuit:
                  \nAnswer:<\/strong><\/span><\/p>\n

                    \n
                  1. Same current flows through each resistance.<\/li>\n
                  2. V = V1<\/sub> + V2<\/sub> + V3<\/sub> …. i.e. total potential drop is the sum of individual resistances.<\/li>\n
                  3. When we want higher resistance, connect them in series. (Resistance is more than individual resistances).<\/li>\n<\/ol>\n

                    Question 2.<\/strong><\/span>
                    \n(a)<\/strong> Define parallel circuit.
                    \nAnswer:<\/strong><\/span>
                    \nParallel circuit :<\/strong> \u201cResistances are said to be connected in parallel if one end of each is connected at a common terminal and other end of each at other common terminal and they have a common pot. difference.\u201d
                    \n(b)<\/strong> State three characteristics of a parallel circuit.
                    \nAnswer:<\/strong><\/span>
                    \nCharacteristics of parallel circuit:<\/strong><\/p>\n

                      \n
                    1. Pot. difference of each resistance is same.<\/li>\n
                    2. Current divides [I = I1<\/sub> + I2<\/sub> + I3<\/sub>…]<\/li>\n
                    3. I\/R = I\/R1<\/sub> + I\/R2<\/sub> + I\/R3<\/sub> reciprocal of total resistance is the sum of reciprocals of individual resistances.<\/li>\n
                    4. Total resistance is less than the least of individual resistances.<\/li>\n<\/ol>\n

                      Question 3.<\/strong><\/span>
                      \n(a)<\/strong> Stale Ohm\u2019s law.
                      \nAnswer:<\/strong><\/span>
                      \nOhm\u2019s Law :<\/strong> \u201cPhysical conditions like temp. remaining the same potential across the ends of a conductor is directly proportional to the current flowing\u201d.<\/p>\n

                      (b)<\/strong> What are the limitations of 0hm\u2018s law?
                      \nAnswer:<\/strong><\/span>
                      \nOhms\u2019Law is obeyed only when temperature remains constant.<\/p>\n

                      Question 4.<\/strong><\/span>
                      \nHow will you verify Ohm\u2019s law by voltmeter, ammeter method?
                      \nAnswer:<\/strong><\/span>
                      \n\"A<\/p>\n

                      Verification of Ohm\u2019s Law:<\/strong> Use the circuit as shown taking case the +ve of voltmeter and +ve of Ammeter should be connected to the +ve of battery and voltmeter in parallel key is closed and Rheostat is set to get the minimum reading in Ammeter and voltmeter. The rheostat is then gradually moved
                      \nand each time value of A and V are noted. The ratio of V\/I is always found constant. This verified ohm\u2019s law.<\/p>\n

                      Question 5.<\/strong><\/span>
                      \nHow will you verify Ohm\u2019s law by potentiometer method?
                      \nAnswer:<\/strong><\/span>
                      \nPotentiometer method to verify<\/p>\n

                      \"A<\/p>\n

                      Connect the potentiometer as shown, close the key and record the potential difference by pressing the jockey at 10 cm intervals of length of the potentiometer wire. Repeat the experiment for six different lengths of potentiometer wire and record the corresponding pot. differences.
                      \n\"A<\/p>\n

                      Find the ratio between potential difference and length. It is found.
                      \n\"A
                      \nV\/l = constant (I) (where I is current and the current in series circuit is constant quantity)
                      \nBut l \u03b1 R {By law of resistance}
                      \nV\/R = I or I = V\/R
                      \nHence ohm\u2019s law is verified.<\/p>\n

                      Question 6.<\/strong><\/span>
                      \nWhat are ohmic resistances ? Given two examples.
                      \nAnswer:<\/strong><\/span>
                      \nOhmic resistances :<\/strong> \u201cConductors which obey ohm\u2019s law are called ohmic resistances.\u201d
                      \nTwo examples :<\/strong> Vanadium, all pure metals like Cu, Al, etc.<\/p>\n

                      Question 7.<\/strong><\/span>
                      \nWhat are non-ohmic resistances ? Give two examples.
                      \nAnswer:<\/strong><\/span>
                      \nNon-ohmic resistances:<\/strong> \u201cThe resistances which do not obey ohm\u2019s law are called non-ohmic resistances.\u201d
                      \nTwo examples :\u00a0<\/strong>Diode valve, triode valve, transistors, filament of a bulb.<\/p>\n

                      Question 8.<\/strong><\/span>
                      \nDerive an expression for three resistances connected in series.
                      \nAnswer:<\/strong><\/span>
                      \nEQUIVALENT RESISTANCE OF RESISTORS
                      \n(i) in PARALLEL :
                      \n\"A
                      \n<\/strong><\/p>\n

                      Let three resistors are connect in such away that one end of each R1<\/sub>, R2<\/sub>, R3<\/sub> is connected at a common terminal (X) and the other end of each at common terminal (Y) through a battery. So that potential difference of each resistor is V and current I at X divides itself and I1<\/sub>, I2<\/sub> and I3<\/sub> flows through R,, R, and R3<\/sub> respectively and again combine at y and current I flows further.
                      \nI = I1<\/sub> + I2<\/sub> + I3<\/sub> …(i) we know that
                      \n\"A
                      \nEQUIVALENT or RESULTANT resistance of parallel conductors
                      \nThe reciprocal of EQUIVALENT RESISTANCE is equal to the sum of the reciprocals of individual resistors.
                      \nEQUIVALENT RESISTANCE OF RESISTORS
                      \nIn (ii)
                      \n\"A<\/p>\n

                      Let three resistors R1<\/sub>, R2<\/sub>, and R3<\/sub>, be connected in series i.e. resistors are joined one after the other as shown and same current I passes through each and each resistors has potential difference say V,, V, and V3<\/sub> so that total p.d. between A and D terminals is V
                      \nV = V1<\/sub> + V2<\/sub> + V3<\/sub> …(i) V = IR
                      \nIR = I R1<\/sub> + IR2<\/sub> + IR3<\/sub> V1<\/sub> = IR1<\/sub>
                      \nIR = I[R1<\/sub> + R2<\/sub> + R3<\/sub>] V2<\/sub> = IR2<\/sub>
                      \nV3<\/sub> = IR3<\/sub> put in (i)
                      \nR = R1<\/sub> + R2<\/sub> + R3<\/sub>
                      \ni. e. EQUIVALENT RESISTANCE of resistors in series is the sum of their individual resistance.<\/p>\n

                      Question 9.<\/strong><\/span>
                      \nDerive an expression for three resistances connected in parallel.
                      \nAnswer:<\/strong><\/span>
                      \nTo derive relation for resultant resistance of three resistances in parallel, consider three resistances
                      \n\"A<\/p>\n

                        \n
                      1. r1<\/sub>, r2<\/sub>, r3<\/sub>\u00a0One end of each is connected to common terminal X and other end of each at common terminaly.<\/li>\n
                      2. Here current I divides into I1<\/sub>, I2<\/sub>, I3<\/sub>, flowing through r1, r2, r3 and potential difference is v for all resistances.
                        \n\"A
                        \n\"A<\/li>\n<\/ol>\n

                        Question 10.<\/strong><\/span>
                        \nWhat do you understand by the term internal resistance of a cell ?
                        \nAnswer:<\/strong><\/span>
                        \nINTERNAL RESISTANCE OF A CELL :<\/strong>
                        \n\u201cThe resistance offered by the electrolyte inside the cell to the flow of current is called the internal resistance of cell.
                        \n\"A<\/p>\n

                        Question 11.<\/strong><\/span>
                        \nState the factors on which internal resistance of a cell depends.
                        \nAnswer:<\/strong><\/span>
                        \nFactors effecting the internal resistance of a cell :<\/strong><\/p>\n

                          \n
                        1. Surface area of electrodes larger surface area, lesser is the internal resistance.<\/li>\n
                        2. Distance between electrodes : more the distance, more is the internal resistance.<\/li>\n
                        3. Temp, of electrolyte r \u03b1 1\/T<\/li>\n
                        4. Higher the concentration of electrolyte greater is internal resistance.<\/li>\n<\/ol>\n

                          Question 12.<\/strong><\/span>
                          \nWhat is the difference between emf and terminal voltage of a cell ?
                          \nAnswer:<\/strong><\/span>
                          \nDifference between e.m.f. and terminal voltage :<\/strong>
                          \nTerminal voltage :<\/strong> When current is drawn from a cell i.e. the cell is in a closed circuit, the potential differences between the electrodes (terminals) of a cell is called terminal voltage,
                          \ne.m.f. : \u201cwhen no current is drawn from a cell i.e. when the cell is in open circuit, the pot. difference between the terminals of the cell is called electromotive force, (e.m.f.).<\/p>\n

                          Multiple choice questions<\/strong><\/span><\/p>\n

                          Tick (\u00a0\u2713 ) the most appropriate option.<\/strong><\/p>\n

                          Question 1.<\/strong><\/span>
                          \nIn a series circuit:<\/strong>
                          \n(a)<\/strong> p.d. across all resistors is same
                          \n(b)<\/strong> current flowing through all resistors is same
                          \n(c)<\/strong> The combined resistance of all resistors is less than individual resistors.
                          \n(d)<\/strong> none of the above
                          \nAnswer:<\/strong><\/span>
                          \n(b)<\/strong> current flowing through all resistors is same<\/p>\n

                          Question 2.<\/strong><\/span>
                          \nIn a parallel circuit:<\/strong>
                          \n(a)<\/strong> p.d. across all resistors is same
                          \n(b)<\/strong> current flowing through all resistors is same
                          \n(c)<\/strong> the equivalent resistance of all resistors is more than any of the individual resistors
                          \n(d)<\/strong> none of the above
                          \nAnswer:<\/strong><\/span>
                          \n(a)<\/strong> p.d. across all resistors is same<\/p>\n

                          Question 3.<\/strong><\/span>
                          \nTwo resistors of 2 \u2126. each are connected in a parallel. The equivalent resistance is :<\/strong>
                          \n(a)<\/strong> less than 2 \u2126 but more than 1\u2126
                          \n(b)<\/strong> one ohm
                          \n(c)<\/strong> four ohm
                          \n(d)<\/strong> between 4 \u2126 and 2 \u2126
                          \nAnswer:<\/strong><\/span>
                          \n(b)<\/strong> one ohm<\/p>\n

                          Question 4.<\/strong><\/span>
                          \nA new cell is marked 1.5 V. When connected to an external resistance, the voltmeter connected to its terminals reads 1.2 V. The drop in potential across the terminals of the cell is due to the :
                          \n(a)<\/strong> internal resistance of cell
                          \n(b)<\/strong> external resistance .
                          \n(c)<\/strong> both (a) and (b)
                          \n(d)<\/strong>none of these
                          \nAnswer:<\/strong><\/span>
                          \n(a)<\/strong> internal resistance of cell<\/p>\n

                          Question 5.<\/strong><\/span>
                          \nA potentiometer is connected to a cell through switch in series. To one end of the potentiometer is attached a voltmeter with the help of connecting wire and a jockey. When the jockey is moved over the potentiometer wire from zero end to 100 cm the reading shown by voltmeter is likely to :
                          \n(a)<\/strong> decrease
                          \n(b)<\/strong> increase
                          \n(c)<\/strong> does not change
                          \n(d)<\/strong> none of these
                          \nAnswer:<\/strong><\/span>
                          \n(b)<\/strong> increase<\/p>\n

                          Question 6.<\/strong><\/span>
                          \nWhen the current is drawn from a cell in a closed circuit, the potential difference between the terminals of cell is called :
                          \n(a)<\/strong>\u00a0e.m.f.
                          \n(b)<\/strong> p.d.
                          \n(c)<\/strong> terminal voltage
                          \n(d)<\/strong> both (a) and (b)
                          \nAnswer:<\/strong><\/span>
                          \n(c)<\/strong> terminal voltage<\/p>\n

                          Numerical Problems on Resistance<\/strong><\/span><\/p>\n

                          Practice Problems : 1<\/strong><\/p>\n

                          Question 1.<\/strong><\/span>
                          \n\"A
                          \nCalculate the equivalent resistance.<\/p>\n

                            \n
                          1. between points A and B<\/li>\n
                          2. between points A and C.<\/li>\n<\/ol>\n

                            Answer:<\/strong><\/span>
                            \nResistance 6\u2126, 3\u2126 and 2\u2126 are in parallel between A and B<\/p>\n

                              \n
                            1. Equivalent resistance between AB is R1\u00a0\u00a0\u2234\u00a0<\/sub>R1<\/sub> = 1\u2126
                              \nNow combination R1<\/sub> and 1\u2126 of BC are in series.<\/li>\n
                            2. Now between points A and C R = R1<\/sub> + 1 = 1 + 1 = 2\u2126<\/li>\n<\/ol>\n

                              Question 2.<\/strong><\/span>
                              \nIn figure, calculate equivalent resistance between points<\/p>\n

                              \"A<\/p>\n

                              (i) A and B
                              \n(ii) B and C
                              \n(iii) A and C.<\/p>\n

                              Answer:<\/strong><\/span>
                              \n(i) Equivalent resistance between AB 4\u2126 and 12 \u2126 are in parallel
                              \n\"A
                              \n(iii) Between A and CD
                              \nNow R1<\/sub> and R2\u00a0<\/sub>and in series
                              \nEquivalent Resistance between A and C
                              \n1 R = R1<\/sub> + R2<\/sub> = 3 + 4 = 7\u2126<\/p>\n

                              Practice Problems : 2<\/strong><\/p>\n

                              Question 1.<\/strong><\/span>
                              \nCalculate the equivalent resistance between points<\/p>\n

                              (i) B and E
                              \n(ii) A and F.<\/p>\n

                              \"A
                              \nAnswer:<\/strong><\/span>
                              \n(i) Equivalent Resistance between BE, (1\u2126, 2\u2126, 3\u2126 are in series) isR1
                              \nR1<\/sub>\u00a0= 1 + 2 + 3 = 6\u2126
                              \nR1<\/sub>\u00a0is in parallel to R2<\/sub> = 3\u2126
                              \ntheir resultant R.3
                              \n1\/r3<\/sub> = 1\/r1<\/sub> +1\/r2<\/sub> = 1\/6 + 1\/3 = 1+2\/6 = 1\/6 Between BE
                              \n:.R3\u00a0<\/sub>= 2\u2126
                              \n(ii) Equivalent resistance between A and F
                              \ni.e. R4<\/sub>, R2<\/sub>, Rs are in series
                              \n3\u2126, 2\u2126 and 3\u2126 are in series
                              \nR= 3 + 2 + 3 = 812<\/sub><\/p>\n

                              Question 2.<\/strong><\/span>
                              \nCalculate the equivalent resistance of circuit diagram shown in Fig. below.
                              \n\"A
                              \nResistances A and B are in series
                              \nR1\u00a0<\/sub>= 4 + 8 = 12\u2126
                              \nResistance C and D are in series
                              \nR2<\/sub> = 1.5 + 4.5 = 6\u2126
                              \nNow R1<\/sub>, E and R2<\/sub> are in parallel
                              \n:. Equivalent resistance between F and G
                              \n\"A<\/p>\n

                              Practice Problems : 3<\/strong><\/p>\n

                              Question 1.<\/strong><\/span>
                              \nEquivalent resistance of circuit diagram is 6\u2126. Calculate the value of x.
                              \n\"A<\/p>\n

                              Question 2.<\/strong><\/span>
                              \nEquivalent resistance of circuit diagram is 5\u2126. Calculate the value of x.
                              \n\"A
                              \nAnswer:<\/strong><\/span>
                              \nSince Equivalent resistance of 4\u2126 and parallel combination is 5\u2126 and 4\u2126 and parallel combination are in series.
                              \nResistance of parallel combination is = 5 – 4= l\u2126
                              \n\"A<\/p>\n

                              Numerical Problems on Ohm\u2019s Law<\/strong><\/span><\/p>\n

                              Practice Problems : 1<\/strong><\/p>\n

                              Question 1.<\/strong><\/span>
                              \nA current of 0.2 A flows through a conductor of resistance 4.50. Calculate p.d. at the ends of conductor.
                              \nAnswer:<\/strong><\/span>
                              \nHere I = 0.2 A, R = 4.5\u2126
                              \np.d. at the ends of conductor V = IR
                              \nV = 0.2 \u00d7 4.5 = 0.9 V<\/p>\n

                              Question 2.<\/strong><\/span>
                              \nA bulb of resistance 4000 is connected to 200 V mains. Calculate the magnitude of current.
                              \nAnswer:<\/strong><\/span>
                              \nR = 400\u2126, V = 200 V
                              \nI = V\/R 200\/400 = 0.5 A<\/p>\n

                              Question 3.<\/strong><\/span>
                              \nAn electric heater draws a current of 5 A, when connected to 220 V mains. Calculate the resistance of its filament.
                              \nAnswer:<\/strong><\/span>
                              \nI = 5 A, V = 220 V, R = ?
                              \nR = v\/I = 220\/5 = 44\u2126<\/p>\n

                              Practice Problems : 2<\/strong><\/p>\n

                              Question 1.<\/strong><\/span>
                              \nFour resistors of resistance 0.5 \u2126, 1.5\u2126, 4\u2126 and 6\u2126 are connected in series to a battery of e.m.f. 6 V and negligible internal resistance. Calculate :<\/p>\n

                                \n
                              1. current drawn from the cell<\/li>\n
                              2. p.d. at the ends of each resistor.<\/li>\n<\/ol>\n

                                Answer:<\/strong><\/span>
                                \n\"A<\/p>\n

                                Question 2.<\/strong><\/span>
                                \nFigure shows a circuit diagram having a battery of 24 V and negligible internal resistance. Calculate :<\/p>\n

                                  \n
                                1. reading of the ammeter,<\/li>\n
                                2. reading of V1<\/sub>, V2<\/sub> and V3<\/sub>.
                                  \n\"A<\/li>\n<\/ol>\n

                                  Answer:<\/strong><\/span>
                                  \nAs 6\u2126 and 3\u2126 are in parallel
                                  \n\"A<\/p>\n

                                  Practice Problems : 3<\/strong><\/p>\n

                                  Question 1.<\/strong><\/span>
                                  \nThree resistors of 6\u2126, 2\u2126 and x are connected in series to a cell of e.m.f 3\/2 V, when the current registered in circuit is 1\/6 A. Draw the circuit diagram and calculate value of x
                                  \nAnswer:<\/strong><\/span>
                                  \nCircuit diagram:
                                  \n\"A<\/p>\n

                                  Question 2.<\/strong><\/span>
                                  \nCarefully study the circuit diagram in figure and calculate the value of resistor x.
                                  \n\"A
                                  \nAnswer:<\/strong><\/span>
                                  \n\"A\"A<\/strong><\/span><\/p>\n

                                  Practice Problems : 4<\/strong><\/p>\n

                                  Question 1.<\/strong><\/span>
                                  \nThree resistors of 4\u2126, 6\u2126. and 12\u2126 are connected in parallel The combination of these resistors is connected in series to a resistance of 2\u2126 and then to a battery of e.m.f 6 V and negligible internal resistance.
                                  \n(a) Draw the circuit diagram
                                  \n(b) Calculate the current in main circuit
                                  \n(c) Calculate the current in each of the resistors in parallel
                                  \nAnswer:<\/strong><\/span>
                                  \n(a) Circuit diagram
                                  \n\"A<\/p>\n

                                  \"A<\/p>\n

                                  Question 2.<\/strong><\/span>
                                  \nStudy the circuit diagram in figure carefully and calculate:
                                  \n(a) current in main circuit
                                  \n(b) current in each of the resistors in parallel circuit.
                                  \n\"A
                                  \n\"A
                                  \n\"A<\/p>\n

                                  Practice Problems : 5<\/strong><\/p>\n

                                  Question 1.<\/strong><\/span>
                                  \nFigure shows a circuit diagram containing 12 cells, each of e.m.f 1.5 V and intenal resistance 0.25\u2126 Calculate:
                                  \n\"A
                                  \n(a)<\/strong> Total internal resistance
                                  \n(b)<\/strong> Total e.m.f.
                                  \n(c)<\/strong> Total external resistance
                                  \n(d)<\/strong> Reading shown by the ammeter
                                  \n(e)<\/strong> Current in 12\u2126 and 8\u2126 resistors
                                  \n(f)<\/strong> p.d. across 2.2 resistor
                                  \n(g)<\/strong> Drop in potential across the terminals of the cell
                                  \nAnswer:<\/strong><\/span>
                                  \nNumber cells in series = n – 12
                                  \n(a)<\/strong> Total internal resistance of 12 cells = n r = 12 \u00d7 0.25 = 3\u2126
                                  \n(b)<\/strong> Total e.m.f. = 12> <1.5 = 18 v
                                  \n(c)<\/strong> Total external resistance : 4 + 8 = 12 \u2126 in series
                                  \n\"A
                                  \n\"A
                                  \n\"A<\/p>\n

                                  Question 2.<\/strong><\/span>
                                  \nFour cells, each of e.m.f 2 V and internal resistance 0.2 \u2126 each are connected in series to form a battery. This battery is connected to an ammeter, a resistance 1.2 and then to a set of resistance of 4 \u2126, 6 \u2126 and 12 \u2126 in parallel to complete the overall circuit in series.
                                  \n(a)<\/strong> Draw circuit diagram of arangment.
                                  \n(b)<\/strong> Calculate total internal resistance
                                  \n(c)<\/strong> Total e.m.f.
                                  \n(d)<\/strong> Current recorded by ammeter.
                                  \n(e)<\/strong> Current flowing through 6 wire in parallel.
                                  \n(f)<\/strong> Drop in potential across the terminals of the battery.
                                  \nAnswer:<\/strong><\/span>
                                  \n(a) Circuit diagram :
                                  \n\"A
                                  \n(b) Total internal resistance r = 4 \u00d7 0.2 = 0.8 \u2126
                                  \n(c) Total e.m.f. = 4 \u00d7 2 = 8 v
                                  \n(d) Current recorded by ammeter ?
                                  \nR1<\/sub> across CD 1\/R1<\/sub> = 1\/4 + 1\/6 + 1\/12 = R1<\/sub> = 2 \u2126
                                  \nTotal resistance (0.8 + 1.2 + 2) = 4 \u2126
                                  \nI = v\/R = 8\/4 = 2a
                                  \n(e) Vacross CD = I R1<\/sub>= 2 \u00d7 2 = 4 V
                                  \nCurrent through 6 \u2126 = v\/R = 4\/6 = 0.67 A
                                  \n(f) droop in potential across the terminals of the battery E – V = Ir = 2 \u00d7 0.8 = 1.6 V<\/p>\n

                                  Practice Problems : 6<\/strong><\/p>\n

                                  Question 1.<\/strong><\/span>
                                  \nTwo cells, each of e.m.f 1.5 V and internal resistance 1 \u2126 are connected in parallel, to form a battery. The battery is connected to an externari resistance of 0.5 \u2126 and two resistances of 3 \u2126 and 1.5 \u2126 in parallel.
                                  \n(a)<\/strong> Draw the circuit diagram.
                                  \n(b)<\/strong> Calculate the current in main circuit.
                                  \n(c)<\/strong> Calculate the current in 1.5 \u2126 resistor.
                                  \n(d)<\/strong> Calculate the drop in potential across the terminal of the battery.
                                  \nAnswer:<\/strong><\/span>
                                  \n(a)<\/strong> Circuit diagram
                                  \n\"A
                                  \n(b)<\/strong> Current in the main circuit
                                  \nTotal internal resistance of two parallel cells I\/R1<\/sub> = 1 + 1
                                  \nR = 1\/2 = O.5 \u2126
                                  \nEffective resistance between PQ = 1\/R2<\/sub> = 1\/3 + 2\/3
                                  \nR2<\/sub> = 1 \u2126
                                  \nTotal resistance of circuit R = R1<\/sub> + R2<\/sub>+ external resistance
                                  \nR= O.5 + I + O.5 = 2\u2126, V= 1.5 V
                                  \ncurrent in main circuit
                                  \nI = V\/R = 1.5\/2 = O.75 A
                                  \n(c)<\/strong> Current in 1 .5 \u2126 resistor
                                  \np.d. between PQ = IR2
                                  \nV1<\/sub> = 0.75 \u00d7 1
                                  \nV1<\/sub> = 0.75 V
                                  \nV1<\/sub> = 0.75
                                  \nI1<\/sub>= V1<\/sub>\/1.5 = 0.75\/1.5 = O.5 A
                                  \n(d)<\/strong> Drop in potential across the terminals of battery E – V = Ir = 0.75 \u00d7 0.5 = 0.375 V<\/p>\n

                                  Question 2.<\/strong><\/span>
                                  \nFour cells, each of e.m.f. 1.5 V and internal resistance 2 \u2126. each are connected in parallel to form a battery. The battery is connected to an external resistance of 0.5 \u2126. and three resistances of 12 \u2126, 6 \u2126 and 4 \u2126. in parallel.<\/p>\n

                                    \n
                                  1. Draw the circuit diagram.<\/li>\n
                                  2. Calculate current in main circuit<\/li>\n
                                  3. Calculate current in 4 D resistor.<\/li>\n
                                  4. Calculate drop in potential across the terminals of\u00a0 battery.<\/li>\n<\/ol>\n

                                    Answer:<\/strong><\/span>
                                    \n1.circuit diagram is drawn
                                    \n\"A
                                    \n4. Drop in potential across the terminals of battery
                                    \nE – V = 1r<\/sub>
                                    \n= 0.5 \u00d7 0.5 = 0.25 V<\/p>\n

                                    Practice Problems : 7<\/strong><\/p>\n

                                    Question 1.<\/strong><\/span>
                                    \nA cell of e.m.f 1.5 V, records a p.d. of 1.35 V, when connected lo an external resistance R, such that current flowing through circuit is 0.75 A. Calculate the value of R and internal resistance of cell
                                    \nAnswer:<\/strong><\/span>
                                    \n\"A<\/p>\n

                                    Question 2.<\/strong><\/span>
                                    \nIn figure a current of 1 A flows through the circuit, when p.d. recorded at the ends of parallel resistors is 1 volt. Calculate the value of R and r.
                                    \n\"A
                                    \nAnswer:<\/strong><\/span>\u00a0For parallel resistors
                                    \n\"A
                                    \n\"A<\/p>\n

                                    Practice Problems : 8<\/strong><\/p>\n

                                    Question 1.<\/strong><\/span>
                                    \nA cell of e.m.f 1.8 V is connected to an external resistance of 2 \u2126, when p.d. recorded at the ends of resistance is 1.6 V. Calculate the internal resistance of the cell.
                                    \nAnswer:<\/strong><\/span>
                                    \nI = V\/R
                                    \n\"A<\/p>\n

                                    Question 2.<\/strong><\/span>
                                    \nStudy ttitel circuit diagram in Fig. 8.44, and hence, calculate the internal resistance of cel\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \"A<\/p>\n

                                    Practice Problems 9<\/strong><\/p>\n

                                    Question 1.<\/strong><\/span>
                                    \nA cell, when connected to an external resistance of 4.5 \u2126 shows a p.d of 1.35 V. If 4.5 \u2126 resistance is replaced by 2.5\u2126 resistance the p.d drops to 1.25 V. Calculate:
                                    \n(a)<\/strong> em.f.,
                                    \n(b)<\/strong> internal resistance of the cell
                                    \nAnswer:<\/strong><\/span>
                                    \nLet \u2018E\u2019 be the e.m.f and \u2018r\u2019 the internal resistance
                                    \nCase(l) r = R [e-v]\/v
                                    \n\"A<\/p>\n

                                    Question 2.<\/strong><\/span>
                                    \nStudy the figures carefully and hence calculate the value of E and r.
                                    \n\"A
                                    \nAnswer:
                                    \n\"A
                                    \n<\/strong><\/span><\/p>\n

                                    Questions from ICSE Examination Papers<\/strong><\/span><\/p>\n

                                    2003<\/strong><\/span><\/p>\n

                                    Question 1.<\/strong><\/span>
                                    \nStudy the diagram carefully and calculate :
                                    \n(a)<\/strong> the equivalent resistance between P and Q.
                                    \n\"A
                                    \n(b)<\/strong> the reading of the ammeter.
                                    \n(e)<\/strong> the electrical power between P and Q.
                                    \n\"A<\/p>\n

                                    2004<\/strong><\/span><\/p>\n

                                    Question 2.<\/strong><\/span>
                                    \nMention two factors which determine the internal resistance of a cell.
                                    \nAnswer:<\/strong><\/span>
                                    \nThe internal resistance of a cell depends on<\/p>\n

                                      \n
                                    1. Surface area of electrodes and<\/li>\n
                                    2. distance between the electrodes.<\/li>\n<\/ol>\n

                                      2005<\/strong><\/span><\/p>\n

                                      Question 3.<\/strong><\/span>
                                      \nFour resistances of 2.0\u2126 each are joined end to end to form a square A B C D. Calculate the equivalent resistance of the combination between any two adjacent corners.
                                      \nAnswer:<\/strong><\/span>
                                      \nResistors R1<\/sub>, R2<\/sub> and R3<\/sub> are in series, therefore their equivalent resistance is Rs = R1<\/sub> + R2<\/sub> + R3<\/sub>= 2 + 2 + 2 = 6 ohm.
                                      \n\"A
                                      \nNow Rs<\/sub> and R4<\/sub> are in parallel, therefore equivalent resistance of the combination between two adjacent corners is
                                      \n1\/Rp<\/sub> = 1\/Rs<\/sub> + 1\/R4<\/sub> = 1\/6 + 1\/2 = 2\/3
                                      \nTherefore Rp = 1.5 ohm.<\/p>\n

                                      Question 4.<\/strong><\/span>
                                      \nThe figure shows three ammeters A, B and C. The ammeter B reads O. 5 A. If all the ammeters have negligible resistance calculate: Calculate:
                                      \n\"A<\/p>\n

                                        \n
                                      1. the readings in the ammeters A and C<\/li>\n
                                      2. the total resistance of the circuit
                                        \n\"A<\/li>\n<\/ol>\n

                                        2006<\/strong><\/span><\/p>\n

                                        Question 5.<\/strong><\/span>
                                        \nA wire of uniform thickness with a resistance of 27 \u2126 is cut into three equal pieces and they are joined in parallel. Find the resistance of Ike parallel combination.
                                        \nAnswer:<\/strong><\/span>
                                        \nA wire is cut into 3 peices
                                        \nAlso Resistance of wire is proportional to length
                                        \nResistance \u221d l
                                        \nLet length of wire =3l = 27\u2126
                                        \nResistance of each piece, l = 27\/3 = 9 \u2126
                                        \nWhen connected in parallel Resultant resistance of combination
                                        \nR is 1\/R = 1\/9 + 1\/9 + 1\/9 = 3\/9 = 1\/3
                                        \n\u2234R= 3\u2126<\/p>\n

                                        Question 6.<\/strong><\/span>
                                        \nMention two factors on which the resistance of a wire depends.
                                        \nAnswer:<\/strong><\/span>
                                        \nTwo factors are<\/p>\n

                                          \n
                                        1. length of wire R \u221d \u00a1<\/li>\n
                                        2. Area of cross-section of wire<\/li>\n<\/ol>\n

                                          R \u221d 1\/a<\/p>\n

                                          Question 7.<\/strong><\/span>
                                          \nIn the figure below, the ammeter A reads 0.3 A. Calculate :-
                                          \n(a) the total resistance of the circuit.
                                          \n(b) the value of R.
                                          \n(c) the current flowing through R.
                                          \n\"A<\/p>\n

                                          2007<\/strong><\/span><\/p>\n

                                          Question 8.<\/strong><\/span>
                                          \nThe V-I graph for a series combination and for a parallel combination of two resistors is as shown in the figure below:
                                          \n\"A
                                          \nWhich of the two, A or B, represents the parallel combination? Give a reason for you answer.
                                          \nAnswer:<\/strong><\/span>
                                          \nSlop of\u00a0 V-I graph gives us the resistance of the resistor. Slope of A is less, therefore, combination A has less resistance as compared to B. Further, since the net resistance decreases in parallel combination, therefore, A represent the parallel combination.<\/p>\n

                                          Question 9.<\/strong><\/span>
                                          \nCalculate the value of the resistance which must be connected to a 15 \u2126 resistance to provide an effective resistance of 6 \u2126.
                                          \nAnswer:<\/strong><\/span>
                                          \nAs the effective resistance is 6 \u2126 less than 15 \u2126
                                          \n\u2234 Resistance R must be connected in parallel with 15 \u2126
                                          \n1\/15 + 1\/R = 1\/6 = 1\/R = 1\/6 – 1\/15 = 5\u20142\/30 = 1\/10
                                          \n\u2234R=10\u2126<\/p>\n

                                          Question 10.<\/strong><\/span>
                                          \nA cell of e.m.f. 1.5 V and internal resistance 1.0\u2126 is connected to two resistors of 4.0\u2126 and 20.0 \u2126 in series as shown in the figure:
                                          \n\"A<\/p>\n

                                            \n
                                          1. current in the circuit.<\/li>\n
                                          2. potential difference across the 4.0 ohm resistor.<\/li>\n
                                          3. voltage drop when the current is flowing.<\/li>\n
                                          4. potential difference across the cell.<\/li>\n<\/ol>\n

                                            Answer:<\/strong><\/span>
                                            \nHere e.m.f., E = 1.5 V
                                            \nInternal resistance 1.0 \u2126
                                            \nAll the resistances are connected in series
                                            \nThe total circuit resistance, R= 1 + 4 + 20 = 25\u2126
                                            \n\"A<\/p>\n

                                            The current,i = E\/R =1.5\/25 = O.06 A
                                            \nPotential difference across 4 \u2126 resistance = r \u00d7 i = 4 \u00d7 0.06 = 0.24 V
                                            \nVoltage drop across the cell = 0.06 \u00d7 1
                                            \n= 0.06 V
                                            \nPotential difference across the cell = 1.5\u2014 0.06
                                            \n= 1.44 V<\/p>\n

                                            2008<\/strong><\/span><\/p>\n

                                            Question 11.<\/strong><\/span><\/p>\n

                                              \n
                                            1. Sketch a graph to show the change in potential difference across the ends of an ohmic resistor and the current flowing in it Label the axis of your graph.<\/li>\n
                                            2. What does the slope of the graph represent?<\/li>\n<\/ol>\n

                                              Answer:<\/strong><\/span>
                                              \n(i)
                                              \n<\/strong>\"A<\/strong><\/p>\n

                                              Question 12.<\/strong><\/span>
                                              \nThree resistors of 6.0 \u2126 2.0 \u2126 and 4.0 \u2126 respectively are joined together as shown in the figure. The resistors are connected to an ammeter and to a cell of e.m.f. 6.0 V.
                                              \nCalculate:
                                              \n(i) tile effective resistance of the circuit.
                                              \n(ii) the current drawn from the cell
                                              \nAnswer:<\/strong><\/span><\/p>\n

                                              \"A<\/p>\n

                                              2009<\/strong><\/span><\/p>\n

                                              Question 13.<\/strong><\/span>
                                              \nThe equivalent resistance of the following circuit diagram is 4\u2126 Calculate the value of x.
                                              \nAnswer:<\/strong><\/span>
                                              \n(i) Equivalent resistance = 4\u2126 , 5\u2126 , \u00d7 \u2126 are in series
                                              \n\u2234 R1<\/sub> (5 + x)\u2126
                                              \n8\u2126 and 4\u2126 are in series
                                              \nR2<\/sub> = 8 + 4 = 12\u2126
                                              \nR1<\/sub> and R2<\/sub> are in parallel
                                              \n\"A
                                              \n3 (5 +x) 5 + x + 12
                                              \n15 + 3 x = x+ 17
                                              \n3 x – x = 17-15 = 2
                                              \n2 x = 2
                                              \nx = 2\/2 = 1\u2126<\/p>\n

                                              Question 14.<\/strong><\/span><\/p>\n

                                                \n
                                              1. Stale Ohm\u2019s Law.<\/li>\n
                                              2. Diagrammatically illustrate how you would connect a key, a battery, a voltmeter, an ammeter, an unknown resistance R and a rheostat so that it can be used to verify the above law ?<\/li>\n<\/ol>\n

                                                Answer:<\/strong><\/span><\/p>\n

                                                  \n
                                                1. Ohm\u2019s law states that current flowing in a conductor is directly proportional to the potential difference across its ends provided the physical conditions remains constant.
                                                  \nI \u221d V or V = IR
                                                  \nVerification of Ohm\u2019s Law: Use the circuit as shown taking case the +ve of voltmeter and +ve of Ammeter should be connected to the +ve of battery and voltmeter in parallel key is closed and Rheostat is set to get the minimum reading in Ammeter and voltmeter. The rheostat is then gradually moved
                                                  \nand each time value of A and V are noted. The ratio of v\/I is always found constant. This verified ohm\u2019s law.<\/li>\n<\/ol>\n

                                                  2010<\/strong><\/span><\/p>\n

                                                  Question 15.<\/strong><\/span>
                                                  \nSix resistances are connected together as shown in the figure. Calculate the equivalent resistance between the points A and B.
                                                  \n\"A
                                                  \nAnswer:<\/strong><\/span>
                                                  \nClearly, the resistance of 2\u2126, 3\u2126 and 5\u2126 are in series. Total resistance is 2\u2126 + 3\u2126 + 5\u2126 = 10\u2126 and the circuit reduces to as shown. Now, 10\u2126 and 10\u2126 are in parallel, there combined resistance R’ is:
                                                  \n\"A<\/p>\n

                                                  Question 16.<\/strong><\/span>
                                                  \n(a)<\/strong><\/p>\n

                                                    \n
                                                  1. A substance has nearly zero resistance at a temperature of 1 K. What is such a substance called ?<\/li>\n
                                                  2. State any two factors which affect the resistance of a metallic wire.<\/li>\n<\/ol>\n

                                                    (b)<\/strong> Five resistors of different resistances are connected together as shown in the figure. A 12 V battery is connected to the arrangement. Calculate :<\/p>\n

                                                      \n
                                                    1. the total resistance in the circuit.<\/li>\n
                                                    2. the total current flowing in the circuit.
                                                      \n\"A<\/li>\n<\/ol>\n

                                                      Answer:
                                                      \n<\/strong><\/span>(a) <\/strong><\/p>\n

                                                        \n
                                                      1. Superconductor<\/li>\n
                                                      2. The resistance of a metallic wire is affected by
                                                        \n(a) Its area of cross-section. Ra%
                                                        \n(b) Length of conductor R \u221d 1\/a<\/li>\n<\/ol>\n

                                                        (b)<\/strong> To solve the above Question, we have to first find the total resistance of the circuit.<\/p>\n

                                                          \n
                                                        1. Here we find that R1<\/sub> and R2<\/sub> are in parallel and their
                                                          \ncombined resistance R’ is given by R’= 10 \u00d7 40\/10+40 = 8\u2126
                                                          \nAlso, the resistances of 30\u2126, 20\u2126 and 60\u2126 are in parallel and their combined resistance R\u201d is given by
                                                          \n1\/R” = 1\/30 + 1\/20 + 1\/60
                                                          \n= 2 + 3 + 1\/60 = 6\/60
                                                          \nR” = 60\/6 = 10\u2126
                                                          \nNow, R’ and R\u201d are in series and their combined resistance R is given by .
                                                          \nR = R’ + R” = 8\u2126 + 10\u2126 = 18\u2126
                                                          \n.\u2019. Total resistance in the circuit = 18\u2126
                                                          \n\"A<\/li>\n<\/ol>\n

                                                          Question 17.<\/strong><\/span>
                                                          \n(a)<\/strong> Calculate the equivalent resistance between the points A and B from as shown in fig.
                                                          \n\"A
                                                          \n(b) <\/strong><\/p>\n

                                                            \n
                                                          1. Draw a graph of Potential difference (V) versus Current (I) for an ohmic resistor.<\/li>\n
                                                          2. How can you find the resistance of the resistor from this graph ?<\/li>\n
                                                          3. What is a non-ohmic resistance ?<\/li>\n<\/ol>\n

                                                            (c)<\/strong> Three resistors are connected to a 12 V battery as shown in the figure given below :
                                                            \n\"A<\/p>\n

                                                              \n
                                                            1. What is the current through the 8 Q resistor ?<\/li>\n
                                                            2. What is the potential difference across parallel combination of 6 Q and 12 Q ?<\/li>\n
                                                            3. What is the current through the 6 Q resistor ?<\/li>\n<\/ol>\n

                                                              Answer:<\/strong><\/span>
                                                              \n(a)<\/strong><\/p>\n

                                                                \n
                                                              1. 2 + 3 = 5\u2126 are in series
                                                                \n6 +- 4 = 10\u2126 are in series
                                                                \n5, 30, 10\u2126 are connected in parallel
                                                                \n\"A<\/li>\n<\/ol>\n

                                                                (b) <\/strong><\/p>\n

                                                                  \n
                                                                1. V – 1 for ohmic resistor
                                                                  \n\"A<\/li>\n
                                                                2. Resistance can be found by finding the reading of V corresponding to I from graph.
                                                                  \nR = V\/I<\/li>\n
                                                                3. Non-Ohmic Resistance : \u201cThe resistors which do not obey the Ohm\u2019s Law are called non-ohmic resistors.\u201d
                                                                  \n\"A
                                                                  \nr2<\/sub>, r3<\/sub> are in parallel<\/li>\n<\/ol>\n

                                                                  1\/Rp<\/sub> = 1\/12 +1\/6 = 3\/12 = 1\/4
                                                                  \nRp<\/sub> = 4
                                                                  \nr1<\/sub> and Rp<\/sub> are in series
                                                                  \n:. R = r1<\/sub> + r = 8 + 4 = 12\u2126<\/p>\n

                                                                    \n
                                                                  1. I = v\/R = 12\/12 = I A<\/li>\n
                                                                  2. P.D across parallel combination AB
                                                                    \nV = I Rp<\/sub>
                                                                    \nI \u00d7 4 = 4 V<\/li>\n
                                                                  3. Current through 6\u2126
                                                                    \nI = v\/r3<\/sub> = 4\/6 = 0.666
                                                                    \nI = 0.67 A<\/li>\n<\/ol>\n

                                                                    2012<\/strong><\/span><\/p>\n

                                                                    Question 18.<\/strong><\/span>
                                                                    \n(a)<\/strong> Calculate the equivalent resistance between P and Q in the following diagram:
                                                                    \n\"A
                                                                    \n(b)<\/strong> A cell is sending current in an external circuit. How does the terminal voltage compare with the e.m.f of the cell?
                                                                    \n(c)<\/strong> Three resistors are connected to a 6 V battery as shown in the figure in 8.59.
                                                                    \n\"A
                                                                    \nCalculate:<\/p>\n

                                                                      \n
                                                                    1. the equivalent resistance of the circuit.<\/li>\n
                                                                    2. total current in the circuit.<\/li>\n
                                                                    3. potential difference across the 7.2 c resistor.<\/li>\n<\/ol>\n

                                                                      Answer:<\/strong><\/span>
                                                                      \n(a)
                                                                      \n<\/strong>\"A<\/strong><\/p>\n

                                                                        \n
                                                                      1. r1<\/sub>, r2<\/sub> are in series
                                                                        \nRs<\/sub> = 10 + 10 = 20 \u2126
                                                                        \nRs<\/sub> and r3<\/sub> are in parallel
                                                                        \n1\/Rp<\/sub> between A and B = 1\/20+ 1\/5 = 5\/20= 1\/4
                                                                        \n\u2234 Rp<\/sub> = 4
                                                                        \nNow r4, Rp and r5 are in series
                                                                        \n\u2234\u00a0R = 3 + 4 + 3 = 10\u2126
                                                                        \n(b)<\/strong> When cell is sending current in an external circuit i.e. current is drawn from the cell, its TERMINAL VOLTAGE \u2018V\u2019 is less than its e.m.f. (E) by an amount equal to the voltage drop inside the cell.
                                                                        \n\"A<\/li>\n<\/ol>\n

                                                                        2013<\/strong><\/span><\/p>\n

                                                                        Question 19.<\/strong><\/span>
                                                                        \n(a)<\/strong> Calculate the equivalent resistance between the points A and B for the following combination of resistors :
                                                                        \n\"A
                                                                        \n(b)<\/strong><\/p>\n

                                                                          \n
                                                                        1. State Ohm\u2019s law.<\/li>\n
                                                                        2. A metal wire of resistance 6\u2126 is stretched so that its length is increased lo twice the original length. Calculate Ike new resistance.<\/li>\n<\/ol>\n

                                                                          (c)<\/strong> The figure shows a circuit when the circuit is switched on, the ammeter reads 0.5 A. 6.0 V<\/p>\n

                                                                            \n
                                                                          1. Calculate the value of the unknown resistor R.<\/li>\n
                                                                          2. Calculate the charge passing through the 3\u2126 resistor in 120 s.<\/li>\n
                                                                          3. Calculate the power dissipated in the 3\u2126 resistor.<\/li>\n<\/ol>\n

                                                                            Answer:<\/strong><\/span>
                                                                            \n( a) Resistance of three 4\u2126 resistors in series = 4\u00a0\u00d7 3 = 12\u2126 Resistance of three 2\u2126 resistors in series = 2 \u00d7 3 = 6\u2126
                                                                            \n\u2234 Equivalent resistance of 12\u2126, 6\u2126 and 4\u2126 in parallel.
                                                                            \n1\/Rp<\/sub> = 1\/12 + 1\/6 + 1\/4 = 1 + 2 + 3\/12 = 1\/2 \u21d2 Rp<\/sub> = 2 \u2126
                                                                            \nEquivalent resistance of 5\u2126, 2\u2126 and 6\u2126 in series.
                                                                            \nR = (5 + 2 + 6) 0 = 13\u2126
                                                                            \n\"A
                                                                            \n(b)<\/p>\n

                                                                              \n
                                                                            1. Ohm\u2019s Law : It states, all physical conditions of a conductor remaining same, the current flowing through it is directly proportional to the potential difference at its ends.<\/li>\n
                                                                            2. Let the original length be (l) and area of cross-section (a), such that its resistance is 6\u2126
                                                                              \nApplying, R = K 1\/a \u21d26 = kl\/a …(i)
                                                                              \nWhen the length 2 l, its area of cross-section becomes a\/2. If
                                                                              \nR \u00a1s the new resistance of conductor then :
                                                                              \nR = k 2 l\/a\/2 = 4 k l\/a …..(ii)
                                                                              \nDividing (ii) by (i) R\/6 = 4 R= 24 \u2126<\/li>\n<\/ol>\n

                                                                              (c)<\/strong><\/p>\n

                                                                                \n
                                                                              1. I = v\/R 0.5 = 6\/R +3
                                                                                \n0.5 R + 1.5 = 6 \u21d2 0.5 R = 4.5 \u21d2 R = 9 \u2126<\/li>\n
                                                                              2. Charge Q = I x t = 0.5 x 120 = 60 Coulombs<\/li>\n
                                                                              3. Power dissipated, P = I x V = 0.5 x 6 = 3 Watt.<\/li>\n<\/ol>\n

                                                                                2014<\/strong><\/span><\/p>\n

                                                                                Question 20.<\/strong><\/span>
                                                                                \nFind the equivalent resistance between points A and B in the following figure.<\/p>\n

                                                                                Answer:<\/strong><\/span><\/p>\n

                                                                                \"A<\/p>\n

                                                                                \"A<\/p>\n

                                                                                Question 21.<\/strong><\/span>
                                                                                \n(a)<\/strong> Two resistors of 4 \u2126 and 6 \u2126 are connected in parallel to a cell to draw a current of 0.5 A from the cell.<\/p>\n

                                                                                  \n
                                                                                1. Draw a labelled circuit diagram showing the above arrangement.<\/li>\n
                                                                                2. Calculate the current in each resistor.<\/li>\n<\/ol>\n

                                                                                  (b)<\/strong><\/p>\n

                                                                                    \n
                                                                                  1. What is an Ohmic resistance ?<\/li>\n
                                                                                  2. Two copper wires are of the same length, but one is thicker than the other.
                                                                                    \n(i) Which wire will have more resistance?
                                                                                    \n(ii) Which wire will have more specific resistance?<\/li>\n<\/ol>\n

                                                                                     <\/p>\n

                                                                                    Answer:<\/strong><\/span>
                                                                                    \n(a)<\/p>\n

                                                                                      \n
                                                                                    1. I = V\/R V = I R
                                                                                      \n\"A<\/li>\n
                                                                                    2. Let current through 4 \u2126 resistance is I then current through 6\u2126 resistance is (0.5 \u2014 1)
                                                                                      \n… 1 \u00d7 4 = (0.5-1) \u00d7 6
                                                                                      \n4 I = 3 – 61
                                                                                      \n4 I + 6 I= 3
                                                                                      \n10I = 3
                                                                                      \n1 = 0.3 A
                                                                                      \n\u2234 Current through 4 \u2126 resistance = 0.3 A
                                                                                      \nand current through 6 \u2126 resistance = 0.5 – 0.3 = 0.2 A<\/li>\n<\/ol>\n

                                                                                      (b)<\/p>\n

                                                                                        \n
                                                                                      1. Ohmic Resistors :<\/strong> The resistors which obey Ohm\u2019s law are called the Ohmic resistors or linear resistances. For such resistors, a graph plotted for the potential difference V against current I is a straight line.<\/li>\n
                                                                                      2. (i) Thin wire will have more resistance.
                                                                                        \n(ii) Specific resistance of both wire is same.<\/li>\n<\/ol>\n

                                                                                        2015
                                                                                        \nQuestion 22.<\/strong><\/span>
                                                                                        \n(a)<\/strong> What happens to the resistivity of semi-conductor with the increase in temperature?
                                                                                        \n(b)<\/strong> Fig. the equialent resistance between point A and B.
                                                                                        \n\"A
                                                                                        \nAnswer:<\/strong><\/span>
                                                                                        \n(a)<\/strong> The resistivity of a semiconductor decreases with increase in temperature.
                                                                                        \n(b)<\/strong> Let RP be the equivalent resistance of the resistors 12\u2126, 6 \u2126 and 4 \u2126 connected in parallel. Hence, we have
                                                                                        \nI\/Rp<\/sub> = 1\/12 + 1\/6 + 1\/4 = 1 + 2 + 3\/12 = 1\/2
                                                                                        \nRp<\/sub> = 2 \u2126
                                                                                        \nTherefore, the equivalent resistance of the circuit is 2
                                                                                        \n2 \u2126 + Rp<\/sub> + 5 \u2126 = 2 \u2126 + 5 \u2126 = 9 \u2126
                                                                                        \nThus, the equivalent resistance between points A and B is 9 \u2126<\/p>\n

                                                                                        Question 23.<\/strong><\/span>
                                                                                        \n(a)<\/strong> The relationship between the potential difference and the current iii a conductor is stated in the form of a law.<\/p>\n

                                                                                          \n
                                                                                        1. Name the law.<\/li>\n
                                                                                        2. What does the slope of V-I graph for a conductor represent?<\/li>\n
                                                                                        3. Name the material used for making the connecting wire.<\/li>\n<\/ol>\n

                                                                                          (b)<\/strong> A cell of Emf 2 V and internal resistance 1.2 \u2126 is connected with an ammeter of resistance 0.8 \u2126 and two resistors of 4.5 \u2126 and 9 \u2126 as shown in the diagram below:
                                                                                          \n\"A<\/p>\n

                                                                                            \n
                                                                                          1. What would be the reading on the Ammeter?<\/li>\n
                                                                                          2. What is the potential difference across the terminals of the cell?<\/li>\n<\/ol>\n

                                                                                            Answer:<\/strong><\/span>
                                                                                            \n(a)<\/strong><\/p>\n

                                                                                              \n
                                                                                            1. The relationship between the potential difference and the current in a conductor is given by Ohm\u2019s law.<\/li>\n
                                                                                            2. The slope of the V\u2014I graph gives the resistance of the conductor.
                                                                                              \nSlope R = v\/I = R
                                                                                              \nThe material used for making connecting wires is copper.<\/li>\n<\/ol>\n

                                                                                              (b)<\/strong> Given that =2 V, r = I.2 \u2126 , RA = O.8 \u2126 ,R1<\/sub> = 4.5 \u2126 , R2<\/sub>=9 \u2126<\/p>\n

                                                                                                \n
                                                                                              1. We know that for the circuit
                                                                                                \n= IR total
                                                                                                \nNow, the total resistance of the circuit is
                                                                                                \nR total = r + RA + Rp<\/sub>
                                                                                                \n1\/Rp<\/sub> = 1\/4.5 + 1\/9 = 3\/9
                                                                                                \nRp<\/sub> = 3 \u2126
                                                                                                \nR total = 1.2 +0.8 +3 = 5 \u2126
                                                                                                \nHence, the current through the ammeter is
                                                                                                \n\"A
                                                                                                \nI = v\/R total = 2\/5 = 0.4 AB<\/li>\n<\/ol>\n

                                                                                                \u21d2 0.48 = 2 – V-I
                                                                                                \nV= 2 – 4.8 – 1.52 V
                                                                                                \n:. Potential difference Vcell 1.52 V<\/p>\n

                                                                                                2016
                                                                                                \nQuestion 24.<\/strong><\/span>
                                                                                                \n(a)<\/strong> The V-I graph for a series combination and for a parallel combination of two resistors is shown in the figure below. Which of the two A or B. represents the parallel combination? Give reasons for your answer.
                                                                                                \n\"A
                                                                                                \n(b)<\/strong> A music system draws a current of 400 m A when connected to a 12 V battery.<\/p>\n

                                                                                                  \n
                                                                                                1. What is the resistance of the music system?<\/li>\n
                                                                                                2. The music system f left playing for several hours and finally the battery voltage drops and the music system stops playing when current drops to 320 m A. At what voltage the music
                                                                                                  \nsystem stops playing?<\/li>\n<\/ol>\n

                                                                                                  (c)<\/strong> A battery of emf 12 V and internal resistance 2\u2126 is connected with two resistors A and B of resistance 4 \u2126 and 6 \u2126 respectively joined in series.
                                                                                                  \n\"A
                                                                                                  \nFind:<\/strong><\/p>\n

                                                                                                    \n
                                                                                                  1. Current in circuit<\/li>\n
                                                                                                  2. The terminal voltage of the cell<\/li>\n
                                                                                                  3. P.D. across 6 \u03a9 resistor.<\/li>\n
                                                                                                  4. Electrical energy spent per minute in 4 \u03a9 resistor.<\/li>\n<\/ol>\n

                                                                                                    Answer:<\/strong><\/span>
                                                                                                    \n(a)<\/strong> A represents Parallel Combination
                                                                                                    \nReason :<\/strong> More current flows in parallel combination as compared to series combination.
                                                                                                    \n(b)<\/strong><\/p>\n

                                                                                                      \n
                                                                                                    1. (i) Given : I = 400 m A = 400 \u00d7 10-3<\/sup> A
                                                                                                      \nV=12 V
                                                                                                      \nV=IR
                                                                                                      \nR = v\/1 = 12 v\/400 \u00d7 10-3<\/sup> A
                                                                                                      \nR- 400\u00d7 10-3<\/sup> A
                                                                                                      \nR = 30\u2126<\/li>\n
                                                                                                    2. Current drops to I = 320 m A = 320 \u00d7 10-3<\/sup> A
                                                                                                      \nThe music stops playing at
                                                                                                      \nV = IR
                                                                                                      \nV = 320 \u00d7 10-3<\/sup> \u00d7 30
                                                                                                      \nV = 9.6 V<\/li>\n<\/ol>\n

                                                                                                      (C) Given, Emf (E) = 12 V; r1<\/sub> = 2\u2126 ; RA = 4\u2126 ; RB = 6\u2126<\/p>\n

                                                                                                        \n
                                                                                                      1. The current in the circuit is
                                                                                                        \n1= E\/Rtotal<\/sub>\u00a0= E\/R1<\/sub> + RA<\/sub> + RB<\/sub>
                                                                                                        \nI = 12\/2 + 4 + 6 = IA<\/li>\n
                                                                                                      2. The terminal voltage of the cell is
                                                                                                        \nTerminal Voltage = Emf- Ir,
                                                                                                        \nTerminal Voltage= 12 – (1 \u00d7 2) = 12 – 2 = 10 V<\/li>\n
                                                                                                      3. The potential difference across the 6 \u03a9 resistor is
                                                                                                        \nVB<\/sub> = IRB<\/sub>
                                                                                                        \n\u2234 VB = 1 x 6 = 6 V<\/li>\n
                                                                                                      4. The electrical energy spent per minute (= 60 s) is
                                                                                                        \nE = I2<\/sup>Rt
                                                                                                        \nE= 12<\/sup> \u00d7 4 \u00d7 60 = 240 J<\/li>\n<\/ol>\n","protected":false},"excerpt":{"rendered":"

                                                                                                        A New Approach to ICSE Physics Part 2 Class 10 Solutions Electric Circuits, Resistance & Ohm\u2019s Law Exercise – 1 Question 1. In which direction conventional current and electronic current flow from a source of electricity ? Answer: Electronic current is always in opposite direction to conventional current. When both the bodies are positively charged … Read more<\/a><\/p>\n","protected":false},"author":8,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[3034],"tags":[35255,35256],"yoast_head":"\nA New Approach to ICSE Physics Part 2 Class 10 Solutions Electric Circuits, Resistance & Ohm\u2019s Law - CBSE Library<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/cbselibrary.com\/a-new-approach-to-icse-physics-part-2-class-10-solutions-electric-circuits-resistance-ohms-law\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"A New Approach to ICSE Physics Part 2 Class 10 Solutions Electric Circuits, Resistance & Ohm\u2019s Law\" \/>\n<meta property=\"og:description\" content=\"A New Approach to ICSE Physics Part 2 Class 10 Solutions Electric Circuits, Resistance & Ohm\u2019s Law Exercise – 1 Question 1. In which direction conventional current and electronic current flow from a source of electricity ? Answer: Electronic current is always in opposite direction to conventional current. When both the bodies are positively charged ... 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