{"id":118379,"date":"2022-04-26T18:27:26","date_gmt":"2022-04-26T12:57:26","guid":{"rendered":"https:\/\/www.mapsofindia.com\/my-india\/?p=118379"},"modified":"2022-04-26T18:27:26","modified_gmt":"2022-04-26T12:57:26","slug":"chapter-3-current-electricity-questions-and-answers-ncert-solutions-for-class-12-physics","status":"publish","type":"post","link":"https:\/\/www.mapsofindia.com\/my-india\/education\/chapter-3-current-electricity-questions-and-answers-ncert-solutions-for-class-12-physics","title":{"rendered":"Chapter 3 – Current Electricity Questions and Answers: NCERT Solutions for Class 12 Physics"},"content":{"rendered":"

1. The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is
\n0.4\u03a90.4\u03a9
\n, what is the maximum current that can be drawn from the battery?<\/h2>\n

Ans: In the above question it is given that:
\nEmf of the battery, E=12VE=12V
\nInternal resistance of the battery, r=0.4\u03a9r=0.4\u03a9
\nConsider the maximum current drawn from the battery to be II.
\nTherefore, using Ohm\u2019s law,
\nE=IrE=Ir
\n\u21d2I=Er\u21d2I=Er
\n\u21d2I=120.4\u21d2I=120.4
\n\u21d2I=30A\u21d2I=30A
\nClearly, the maximum current drawn from the given battery is 30A30A .<\/h3>\n

2. A battery of emf 10 V and internal resistance 3\u03a93\u03a9 is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?<\/h2>\n

Ans: In the above question it is given that:
\nEmf of the battery, E = 10 V
\nInternal resistance of the battery, r=3\u03a9r=3\u03a9
\nCurrent in the circuit, I=0.5AI=0.5A
\nConsider the resistance of the resistor to be RR.
\nTherefore, using Ohm\u2019s law,
\nI=ER+rI=ER+r
\nR+r=EIR+r=EI
\n\u21d2R+r=100.5\u21d2R+r=100.5
\n\u21d2R+r=20\u21d2R+r=20
\n\u21d2R=20\u22123=17\u03a9\u21d2R=20\u22123=17\u03a9
\nLet the terminal voltage of the resistor be VV.
\nUsing the Ohm\u2019s law,
\nV=IRV=IR
\n\u21d2V=0.5\u00d717=8.5V\u21d2V=0.5\u00d717=8.5V
\nThus, the resistance of the resistor is 17\u03a917\u03a9 and the terminal voltage is 8.5V8.5V .<\/h3>\n

3.
\na) Three resistors 1\u03a91\u03a9 , 2\u03a92\u03a9 and 3\u03a93\u03a9 are combined in series. What is the total resistance of the combination?<\/h2>\n

Ans: In the above question it is given that three resistors of resistances\u00a0 1\u03a91\u03a9 , 2\u03a92\u03a9 and 3\u03a93\u03a9\u00a0 are combined in series.
\nThe total resistance of a series combination of resistors is the algebraic sum of individual resistances.
\nHence the total resistance is given by:
\nTotal Resistance =1+2+3=6\u03a9=1+2+3=6\u03a9<\/h3>\n

b) If the combination is connected to a battery of emf 12 V and negligible internal resistance, obtain the potential drop across each resistor.<\/h2>\n

Ans: Consider the current flowing through the circuit to be II.
\nEmf of the battery, E=12VE=12V
\nTotal resistance of the circuit, R=6\u03a9R=6\u03a9
\nThe relation for current using Ohm\u2019s law is given by:
\nI=ERI=ER
\n\u21d2I=126=2A\u21d2I=126=2A
\nConsider potential drop across 1\u03a91\u03a9 resistor to be V1V1 .
\nUsing Ohm\u2019s law, the value of V1V1 can be obtained as:
\nV1=2\u00d71=2VV1=2\u00d71=2V \u2026\u2026 (1)
\nConsider potential drop across 2\u03a92\u03a9 resistor to be V2V2 .
\nAgain, using Ohm\u2019s law, the value of V2V2 can be obtained as:
\nV2=2\u00d72=4VV2=2\u00d72=4V \u2026\u2026 (2)
\nConsider potential drop across 3\u03a93\u03a9 resistor to be V3V3 .
\nV3=2\u00d73=6VV3=2\u00d73=6V …… (3)
\nClearly, the potential drops across 1\u03a91\u03a9 , 2\u03a92\u03a9 and 3\u03a93\u03a9\u00a0 resistors are 2V2V , 4V4V and 6V6V respectively.<\/h3>\n

4.
\na) Three resistors 2\u03a92\u03a9 , 4\u03a94\u03a9 and 5\u03a95\u03a9\u00a0 are combined in parallel. What is the total resistance of the combination?<\/h2>\n

Ans: In the above question it is given that there are three resistors of resistances 2\u03a92\u03a9 , 4\u03a94\u03a9 and 5\u03a95\u03a9 connected in parallel as shown below:
\n(Image will be uploaded soon)
\nLet
\nR1=2\u03a9;R2=4\u03a9;R3=5\u03a9R1=2\u03a9;R2=4\u03a9;R3=5\u03a9
\nNow, using the rule for parallel combination of resistors, total resistance RR of the combination will be:
\n1R=1R1+1R2+1R31R=1R1+1R2+1R3
\n\u21d21R=12+14+15=1920\u21d21R=12+14+15=1920
\n\u21d2R=2019\u03a9\u21d2R=2019\u03a9
\nClearly, total resistance of the given parallel combination is 2019\u03a92019\u03a9 .<\/h3>\n

b) If the combination is connected to a battery of emf 20 V and negligible internal resistance, determine the current through each resistor, and the total current drawn from the battery.<\/h2>\n

Ans: In the above question, it is given that the parallel combination mentioned in (a) is connected to a battery and negligible internal resistance as shown:
\n(Image will be uploaded soon)
\nIt is known that R1=2\u03a9;R2=4\u03a9;R3=5\u03a9R1=2\u03a9;R2=4\u03a9;R3=5\u03a9.
\nAlso given that,
\nEmf of the battery, V=20VV=20V
\nOhm\u2019s law can be used to determine the individual current flowing through each resistor and hence, the total current in the circuit.
\nIf current I1I1 is flowing through resistor R1R1, it is given by:
\nI1=VR1I1=VR1
\n\u21d2I1=202=10A\u21d2I1=202=10A
\nIf current I2I2 is flowing through resistor R2R2, it is given by:
\nI2=VR2I2=VR2
\n\u21d2I2=204=5A\u21d2I2=204=5A
\nIf current I3I3 is flowing through resistor R3R3, it is given by:
\nI3=VR3I3=VR3
\n\u21d2I3=205=4A\u21d2I3=205=4A
\nNow, the total current will be:
\nI=I1+I2+I3I=I1+I2+I3
\n\u21d2I=10+5+4=19A\u21d2I=10+5+4=19A
\nClearly, the current through resistors 2\u03a92\u03a9 , 4\u03a94\u03a9 and 5\u03a95\u03a9 are 10A10A, 5A5A and 4A4A respectively while the total current in the circuit is 19A19A.<\/h3>\n

5. At room temperature 27.0\u2218C27.0\u2218C, the resistance of a heating element is 100\u03a9100\u03a9. What is the temperature of the element if the resistance is found to be 117\u03a9117\u03a9, given that the temperature coefficient of the material of the resistor is 1.70\u00d710\u22124\u2218C\u221211.70\u00d710\u22124\u2218C\u22121 ?<\/h2>\n

Ans: In the above question it is given that at room temperature (T=27.0\u2218C)(T=27.0\u2218C), the resistance of the heating element is 100\u03a9100\u03a9 (say R).
\nAlso, the heating element\u2019s temperature coefficient is given to be \u03b1=1.70\u00d710\u22124\u2218C\u22121\u03b1=1.70\u00d710\u22124\u2218C\u22121.
\nNow, it is said that the resistance of the heating element at an increased temperature (say T1T1) is 117\u03a9117\u03a9 (say R1R1). To compute this unknown increased temperature T1T1, the formula for temperature coefficient of a material can be used. It is known that temperature co-efficient of a material provides information on the nature of that material with respect to its change in resistance with temperature. Mathematically,
\n\u03b1=R1\u2212RR(T1\u2212T)\u03b1=R1\u2212RR(T1\u2212T)
\n\u21d2T1\u2212T=R1\u2212RR\u03b1\u21d2T1\u2212T=R1\u2212RR\u03b1
\nSubstituting the given values,
\n\u21d2T1\u221227=117\u2212100100\u00d71.70\u00d710\u22124\u21d2T1\u221227=117\u2212100100\u00d71.70\u00d710\u22124
\n\u21d2T1\u221227=1000\u21d2T1\u221227=1000
\n\u21d2T1=1027\u2218C\u21d2T1=1027\u2218C
\nClearly, it is at 1027\u2218C1027\u2218C when the resistance of the element is 117\u03a9117\u03a9.<\/h3>\n

6. A negligibly small current is passed through a wire of length 15 m and uniform cross-section 6.0\u00d710\u22127m26.0\u00d710\u22127m2 , and its resistance is measured to be 5.0\u03a95.0\u03a9 . What is the resistivity of the material at the temperature of the experiment?<\/h2>\n

Ans: In the above question it is given that:
\nLength of the wire, l=15ml=15m
\nArea of cross-section of the wire, a=6.0\u00d710\u22127m2a=6.0\u00d710\u22127m2
\nResistance of the material of the wire, R=5.0\u03a9R=5.0\u03a9
\nLet resistivity of the material of the wire be \u03c1\u03c1
\nIt is known that, resistance is related with the resistivity as:
\nR=\u03c1lAR=\u03c1lA
\n\u21d2\u03c1=RAl\u21d2\u03c1=RAl
\n\u21d2\u03c1=5\u00d76.0\u00d710\u2212715\u21d2\u03c1=5\u00d76.0\u00d710\u2212715
\n\u21d2\u03c1=2\u00d710\u22127m2\u21d2\u03c1=2\u00d710\u22127m2
\nTherefore, the resistivity of the material is 2\u00d710\u22127m22\u00d710\u22127m2 .<\/h3>\n

7. A silver wire has a resistance of
\n2.1\u03a92.1\u03a9
\nat 27.5\u2218C27.5\u2218C , and a resistance of 2.7\u03a92.7\u03a9 at 100\u2218C100\u2218C. Determine the temperature coefficient of resistivity of silver.<\/h2>\n

Ans: In the above question it is given that:
\nTemperature, T1=27.5\u2218CT1=27.5\u2218C.
\nResistance of the silver wire at T1T1 is R1=2.1\u03a9R1=2.1\u03a9 .
\nTemperature, T2=100\u2218CT2=100\u2218C .
\nResistance of the silver wire at T2T2 is R2=2.7\u03a9R2=2.7\u03a9 .
\nLet the temperature coefficient of silver be \u03b1\u03b1 . It is known that temperature co-efficient of a material provides information on the nature of that material with respect to its change in resistance with temperature. Mathematically, it is related with temperature and resistance by the formula:
\n\u03b1=R2\u2212R1R1(T2\u2212T1)\u03b1=R2\u2212R1R1(T2\u2212T1)
\n\u21d2\u03b1=2.7\u22122.12.1(100\u221227.5)=0.0039\u2218C\u22121\u21d2\u03b1=2.7\u22122.12.1(100\u221227.5)=0.0039\u2218C\u22121
\nClearly, the temperature coefficient of silver is 0.0039\u2218C\u221210.0039\u2218C\u22121.<\/h3>\n

8. A heating element using nichrome connected to a 230 V supply draws an initial current of 3.2A which settles after a few seconds to a steady value of 2.8 A. What is the steady temperature of the heating element if the room temperature is 27\u2218C27\u2218C ? Temperature coefficient of resistance of nichrome averaged over the temperature range involved is 1.70\u00d710\u22124\u2218C\u221211.70\u00d710\u22124\u2218C\u22121 .<\/h2>\n

Ans: In the above question it is given that:
\nSupply voltage is V=230VV=230V
\nInitial current drawn is I1=3.2AI1=3.2A.
\nLet the initial resistance be R1R1 .
\nTherefore, using Ohm\u2019s law,
\nR1=VI1R1=VI1
\n\u21d2R1=2303.2=71.87\u03a9\u21d2R1=2303.2=71.87\u03a9
\nSteady state value of the current is I2=2.8AI2=2.8A.
\nLet the resistance of the steady state be R2R2 .
\nTherefore, using Ohm\u2019s law.
\nR2=VI2R2=VI2
\n\u21d2R2=2302.8=82.14\u03a9\u21d2R2=2302.8=82.14\u03a9
\nTemperature co-efficient of nichrome is \u03b1=1.70\u00d710\u22124\u2218C\u22121\u03b1=1.70\u00d710\u22124\u2218C\u22121 .
\nInitial temperature of nichrome is T1=27\u2218CT1=27\u2218C.
\nLet steady state temperature reached by nichrome be T2T2 .
\nNow, it is known that temperature co-efficient of a material provides information on the nature of that material with respect to its change in resistance with temperature. Mathematically, it is given by
\n\u03b1=R2\u2212R1R1(T2\u2212T1)\u03b1=R2\u2212R1R1(T2\u2212T1)
\n\u21d2(T2\u2212T1)=R2\u2212R1R1\u03b1\u21d2(T2\u2212T1)=R2\u2212R1R1\u03b1
\nSubstituting the given values,
\n\u21d2(T2\u221227)=82.14\u221271.8771.87\u00d71.70\u00d710\u22124\u21d2(T2\u221227)=82.14\u221271.8771.87\u00d71.70\u00d710\u22124
\n\u21d2T2\u221227=840.5\u21d2T2\u221227=840.5
\n\u21d2T2=867.5\u2218C\u21d2T2=867.5\u2218C
\nClearly, the steady temperature of the heating element is 867.5\u2218C867.5\u2218C.<\/h3>\n

9. Determine the current in each branch of the network shown in figure:
\n(Image will be uploaded soon)<\/h2>\n

Ans: Current flowing through various branches of the circuit is represented in the given figure.
\n(Image will be uploaded soon)
\nConsider
\nI1=I1=Current flowing through the outer circuit
\nI2=I2=Current flowing through branch AB
\nI3=I3=Current flowing through branch AD
\nI2\u2212I4=I2\u2212I4=Current flowing through branch BC
\nI3+I4=I3+I4=Current flowing through branch CD
\nI4=I4=Current flowing through branch BD
\nFor the closed circuit ABDA, potential is zero i.e.,
\n10I2+5I4\u22125I3=010I2+5I4\u22125I3=0
\n2I2+I4\u2212I3=02I2+I4\u2212I3=0
\nI3=2I2+I4I3=2I2+I4 \u2026\u2026 (1)
\nFor the closed circuit BCDB, potential is zero i.e.,
\n5(I2\u2212I4)\u221210(I3+I4)\u22125I4=05(I2\u2212I4)\u221210(I3+I4)\u22125I4=0
\n5I2+5I4\u221210I3\u221210I4\u22125I4=05I2+5I4\u221210I3\u221210I4\u22125I4=0
\n5I2\u221210I3\u221220I4=05I2\u221210I3\u221220I4=0
\nI2=2I3+4I4I2=2I3+4I4 \u2026\u2026 (2)
\nFor the closed circuit ABCFEA, potential is zero i.e.,
\n\u221210+10(I1)+10(I2)+5(I2\u2212I4)=0\u221210+10(I1)+10(I2)+5(I2\u2212I4)=0
\n10=15I2+10I1\u22125I410=15I2+10I1\u22125I4
\n3I3+2I1\u2212I4=23I3+2I1\u2212I4=2 \u2026\u2026 (3)
\nFrom equations (1) and (2), we obtain
\nI3=2(2I3+4I4)+I4I3=2(2I3+4I4)+I4
\nI3=4I3+8I4+I4I3=4I3+8I4+I4
\n\u22123I3=9I4\u22123I3=9I4
\n\u22123I4=+I3\u22123I4=+I3 \u2026\u2026 (4)
\nPutting equation (4) in equation (1), we obtain
\nI3=2I2+I4I3=2I2+I4
\n\u22124I4=2I2\u22124I4=2I2 \u2026\u2026 (5)
\nIt is evident from the given figure that,
\nI1=I3+I2I1=I3+I2 \u2026\u2026. (6)
\nPutting equation (6) in equation (1), we obtain
\n3I2+2(I3+I2)\u2212I4=23I2+2(I3+I2)\u2212I4=2
\n5I2+2I3\u2212I4=25I2+2I3\u2212I4=2 \u2026\u2026 (7)
\nPutting equations (4) and (5) in equation (7), we obtain
\n5(\u22122I4)+2(\u22123I4)\u2212I4=25(\u22122I4)+2(\u22123I4)\u2212I4=2
\n\u221210I4\u22126I4\u2212I4=2\u221210I4\u22126I4\u2212I4=2
\n17I4=\u2212217I4=\u22122
\nI4=\u2212217AI4=\u2212217A
\nEquation (4) reduces to
\nI3=\u22123(I4)I3=\u22123(I4)
\nI3=\u22123(\u2212217)=617AI3=\u22123(\u2212217)=617A
\nI2=\u22122(I4)I2=\u22122(I4)
\nI2=\u22122(\u2212217)=417AI2=\u22122(\u2212217)=417A
\nI2\u2212I4=417\u2212(\u2212217)=617I2\u2212I4=417\u2212(\u2212217)=617
\nI3+I4=617+(\u2212217)=417AI3+I4=617+(\u2212217)=417A
\nI1=I3+I2I1=I3+I2
\n\u2234I1=617+417=1017A\u2234I1=617+417=1017A
\nTherefore, current in branch AB =417A=417A
\nCurrent in branch BC =617A=617A
\nCurrent in branch CD =\u2212417A=\u2212417A
\nCurrent in branch AD =617A=617A
\nCurrent in branch BD =(\u2212217)A=(\u2212217)A
\nTotal current =417+617+\u2212417+617+\u2212217=1017A=417+617+\u2212417+617+\u2212217=1017A .<\/h3>\n

10.
\na) In a metre bridge the balance point is found to be at 39.5 cm from the end A, when the resistor Y is of 12.5\u03a912.5\u03a9 . Determine the resistance of X. Why are the connections between resistors in a Wheatstone or meter bridge made of thick copper strips?<\/h2>\n

Ans: A metre bridge with resistors XX and YY is represented in the given figure.
\n(Image will be uploaded soon)
\nA meter bridge works with the same principle as that of a Wheatstone bridge, which is an electrical circuit used to measure unknown resistances. It forms a bridge circuit, wherein the two legs of the circuit are balanced, out of which, one leg has the unknown resistance. Here, the concept of balance point is used, when there is no deflection in the galvanometer (no current flow condition).
\nIn the above question it is given that:
\nBalance point from end A is at distance, l1=39.5cml1=39.5cm
\nResistance of the resistor
\nY=12.5\u03a9Y=12.5\u03a9
\nCondition for the balance with respect to the given meter bridge is given by
\nXY=100\u2212l1l1XY=100\u2212l1l1
\n\u21d2X=100\u221239.539.5\u00d712.5=8.2\u03a9\u21d2X=100\u221239.539.5\u00d712.5=8.2\u03a9
\nThus, the resistance of resistor XX is 8.2\u03a98.2\u03a9.
\nThe connection between resistors in a Wheatstone or metre bridge is made of thick copper strips which helps to minimize the resistance. Hence it is not taken into consideration in the bridge formula.<\/h3>\n

b) Determine the balance point of the bridge above if X and Y are interchanged.<\/h2>\n

Ans: When XX and YYare interchanged, then l1l1 and 100\u2212l1100\u2212l1 also get interchanged.
\nThus, the balance point of the bridge would be 100\u2212l1100\u2212l1 from AA.
\n\u2234100\u2212l1=100\u221239.5=60.5cm\u2234100\u2212l1=100\u221239.5=60.5cm
\nClearly, the balance point is 60.5cm60.5cm from AA.<\/h3>\n

c) What happens if the galvanometer and cell are interchanged at the balance point of the bridge? Would the galvanometer show any current?<\/h2>\n

Ans: When the galvanometer and cell are swapped at the balance point of the bridge, the galvanometer would show zero deflection. As there is null deflection, zero current would flow through the galvanometer.<\/h3>\n

11. A storage battery of emf 8.0 V and internal resistance 0.5\u03a90.5\u03a9 is being charged by a 120 V DC supply using a series resistor of 15.5\u03a915.5\u03a9. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit?<\/h2>\n

Ans: In the above question it is given that:
\nEmf of the storage battery is E=0.8VE=0.8V.
\nInternal resistance of the battery is r=0.5\u03a9r=0.5\u03a9 .
\nDC supply voltage is V=120VV=120V
\nResistance of the resistor is R=15.5\u03a9R=15.5\u03a9.
\nConsider effective voltage in the circuit to be V\u2032V\u2032, which would be the difference in the supply voltage and the emf of the battery.
\nV\u2032=V\u2212EV\u2032=V\u2212E
\n\u21d2V\u2032=120\u22128=112V\u21d2V\u2032=120\u22128=112V
\nNow, current flowing in the circuit is II and the resistance RR is connected in series to the storage battery.
\nTherefore, using Ohm\u2019s law,
\nI=V\u2032R+rI=V\u2032R+r
\n\u21d2I=11215.5+0.5=7A\u21d2I=11215.5+0.5=7A
\nThus, voltage across resistor RRwould be:
\nIR=7\u00d715.5=108.5VIR=7\u00d715.5=108.5V
\nDC supply voltage = Terminal voltage of battery + Voltage drop across RR
\nTerminal voltage of battery =120\u2212108.5=11.5V=120\u2212108.5=11.5V
\nA series resistor in a charging circuit takes the responsibility for controlling the current drawn from the external source. Excluding this series resistor is dangerous as the current flow would be extremely high if so.<\/h3>\n

12. In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35.0 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63.0 cm, what is the emf of the second cell?<\/h2>\n

Ans: A potentiometer arrangement facilitates adjustable voltage dividing. It can be used to compare the emf\u2019s of two different cells with the help of balance points in each case.
\nBalance point or null point is the point when the galvanometer in the circuit shows no deflection. i.e., when there is no current flowing in the circuit.
\nIn the above question it is given that when the cell has an emf E1=1.25VE1=1.25V, the balance point of the potentiometer is given to be at a distance, l1=35cml1=35cm.
\nNow, when the cell is replaced by another cell of emf E2E2, the balance point of the potentiometer is given to be at a distance l2=63cml2=63cm.
\nThe balance condition to compare the emf\u2019s of two cells using a potentiometer setup is given by the relation,
\nE1E2=l1l2E1E2=l1l2
\n\u21d2E2=E1\u00d7l1l2\u21d2E2=E1\u00d7l1l2
\n\u21d2E2=1.25\u00d76335=2.25V\u21d2E2=1.25\u00d76335=2.25V
\nClearly, the emf of the second cell is 2.25V2.25V .<\/h3>\n

13. The number density of free electrons in a copper conductor estimated in Example 3.1 is 8.5\u00d71028m\u221238.5\u00d71028m\u22123 . How long does an electron take to drift from one end of a wire 3.0 m long to its other end? The area of cross-section of the wire is 2.0\u00d710\u22126m22.0\u00d710\u22126m2 and it is carrying a current of 3.0 A.<\/h2>\n

Ans: In the above question it is given that:
\nNumber density of free electrons in a copper conductor is n=8.5\u00d71028m\u22123n=8.5\u00d71028m\u22123.
\nLength of the copper wire is l=3.0ml=3.0m.
\nArea of cross-section of the wire is A=2.0\u00d710\u22126m2A=2.0\u00d710\u22126m2.
\nCurrent carried by the wire is I=3.0AI=3.0A.
\nNow we know that:
\nI=nAeVdI=nAeVd
\nWhere,
\nee is the electric charge of magnitude 1.6\u00d710\u221219C1.6\u00d710\u221219C.
\nVdVd is the drift velocity and
\nDrift velocity=Length of the wire (l)Time taken to cover (t)Drift velocity=Length of the wire (l)Time taken to cover (t)
\nI=nAeltI=nAelt
\n\u21d2t=nAelI\u21d2t=nAelI
\n\u21d2t=3\u00d78.5\u00d71028\u00d72\u00d710\u22126\u00d71.6\u00d710\u2212193.0\u21d2t=3\u00d78.5\u00d71028\u00d72\u00d710\u22126\u00d71.6\u00d710\u2212193.0
\n\u2234t=2.7\u00d7104s\u2234t=2.7\u00d7104s .
\nHence the time taken by an electron to drift from one end of the wire to the other is 2.7\u00d7104s2.7\u00d7104s.<\/h3>\n

14. The earth\u2019s surface has a negative surface charge density of 10\u22129Cm\u2212210\u22129Cm\u22122 . The potential difference of 400 kV between the top of the atmosphere and the surface results (due to the low conductivity of the lower atmosphere) in a current of only 1800 A over the entire globe. If there were no mechanism of sustaining atmospheric electric field, how much time (roughly) would be required to neutralize the earth\u2019s surface? (This never happens in practice because there is a mechanism to replenish electric charges, namely the continual thunderstorms and lightning in different parts of the globe). (Radius of earth = 6.37\u00d7106m6.37\u00d7106m.)<\/h2>\n

Ans: In the above question it is given that:
\nSurface charge density of the earth is \u03c3=10\u22129Cm\u22122\u03c3=10\u22129Cm\u22122.
\nCurrent over the entire globe is I=1800AI=1800A.
\nRadius of the earth is r=6.37\u00d7106mr=6.37\u00d7106m.
\nSurface area of the earth is given by:
\nA=4\u03c0r2A=4\u03c0r2
\n\u21d2A=4\u03c0\u00d7(6.37\u00d7106)2\u21d2A=4\u03c0\u00d7(6.37\u00d7106)2
\n\u21d2A=5.09\u00d71014m2\u21d2A=5.09\u00d71014m2
\nCharge on the earth surface is given by:
\nq=\u03c3Aq=\u03c3A
\n\u21d2q=10\u22129\u00d75.09\u00d71014\u21d2q=10\u22129\u00d75.09\u00d71014
\n\u21d2q=5.09\u00d7105C\u21d2q=5.09\u00d7105C
\nNow, when the time taken to neutralize the surface of the earth is taken to be tt, then,
\nCurrent, I=qtI=qt
\n\u21d2t=qI\u21d2t=qI
\n\u21d2t=5.09\u00d71051800=282.77s\u21d2t=5.09\u00d71051800=282.77s
\nClearly, the time taken to neutralize the surface of the earth is 282.77s282.77s.<\/h3>\n

15.
\na) Six lead-acid type of secondary cells each of emf 2.0 V and internal resistance 0.015\u03a90.015\u03a9 are joined in series to provide a supply to a resistance of 8.5\u03a98.5\u03a9 . What is the current drawn from the supply and its terminal voltage?<\/h2>\n

Ans: In the above question, it is given that six lead-acid type of secondary cells are joined in series as shown below:
\n(Image will be uploaded soon)
\nHere, number of secondary cells is n=6n=6.
\nEmf of each secondary cell is E=2.0VE=2.0V.
\nInternal resistance of each cell is r=0.015\u03a9r=0.015\u03a9 .
\nSeries resistor is connected to the combination of cells.
\nResistance of the resistor RR is 8.5\u03a98.5\u03a9.
\nIf the current drawn from the supply is II, then
\nI=nER+nrI=nER+nr
\n\u21d2I=6\u00d728.5+6\u00d70.015\u21d2I=6\u00d728.5+6\u00d70.015
\n\u21d2I=1.39A\u21d2I=1.39A
\nNow the terminal voltage would be
\nV=IR=1.39\u00d78.5=11.87VV=IR=1.39\u00d78.5=11.87V
\nClearly, the current drawn from the supply is 1.39A1.39A and the terminal voltage is 11.87V11.87V.<\/h3>\n

b) A secondary cell after long use has an emf of 1.9 V and a large internal resistance of 380\u03a9380\u03a9. What maximum current can be drawn from the cell? Could the cell drive the starting motor of a car?<\/h2>\n

Ans: It is given that after using a secondary cell for long, emf of the secondary cell is E=1.9VE=1.9V.
\nAlso, the internal resistance of the cell is r=380\u03a9r=380\u03a9.
\nHere, using Ohm\u2019s law, the maximum current =Er=Er
\n\u21d2Er=1.9380=0.005A\u21d2Er=1.9380=0.005A.
\nClearly, the maximum current used from the cell is 0.005A0.005A. To start the motor of a car, a large amount of current is required. Thus, this cell which produces just 0.005A cannot be used for this purpose.<\/h3>\n

16. Two wires of equal length, one of aluminium and the other of copper have the same resistance. Which of the two wires is lighter? Hence explain why aluminium wires are preferred for overhead power cables.
\n\u03c1Al=2.63\u00d7108\u03a9m;\u03c1Cu=1.72\u00d710\u22128\u03a9m;\u03c1Al=2.63\u00d7108\u03a9m;\u03c1Cu=1.72\u00d710\u22128\u03a9m;Relative density of Al=2.7;Relative density of Cu=8.9Relative density of Al=2.7;Relative density of Cu=8.9<\/h2>\n

Ans: In the above question it is given that:
\nResistivity of aluminium is \u03c1Al=2.63\u00d7108\u03a9m\u03c1Al=2.63\u00d7108\u03a9m.
\nRelative density of aluminium is d1=2.7d1=2.7.
\nConsider l1l1 to be the length of aluminium wire, m1m1 as its mass, the resistance of the copper wire as R2R2, and area of cross-section of the copper wire as A2A2.
\nTherefore, using the relation between resistance and resistivity,
\nR1=\u03c11l1A1R1=\u03c11l1A1 \u2026\u2026 (1)
\nAnd
\nR2=\u03c12l2A2R2=\u03c12l2A2 \u2026\u2026 (2)
\nHowever, we have R1=R2R1=R2
\n\u21d2\u03c11l1A1=\u03c12l2A2\u21d2\u03c11l1A1=\u03c12l2A2
\nAlso, we have l1=l2l1=l2.
\n\u21d2\u03c11A1=\u03c12A2\u21d2\u03c11A1=\u03c12A2
\n\u21d2A1A2=\u03c11\u03c12\u21d2A1A2=\u03c11\u03c12
\n\u21d2A1A2=2.63\u00d71081.72\u00d710\u22128=2.631.72\u21d2A1A2=2.63\u00d71081.72\u00d710\u22128=2.631.72
\nNow, mass of the aluminium wire is given by:
\nm1=Volume\u00d7Densitym1=Volume\u00d7Density
\n\u21d2m1=A1l1\u00d7d1=A1l1d1\u21d2m1=A1l1\u00d7d1=A1l1d1 \u2026\u2026 (3)
\nSimilarly mass of the copper wire is given by:
\nm2=Volume\u00d7Densitym2=Volume\u00d7Density
\n\u21d2m2=A2l2\u00d7d2=A2l2d2\u21d2m2=A2l2\u00d7d2=A2l2d2 \u2026\u2026 (4)
\nDividing equation (3) by equation (4), we get:
\nm1m2=A1l1d1A2l2d2m1m2=A1l1d1A2l2d2
\nAs l1=l2l1=l2,
\nm1m2=A1d1A2d2m1m2=A1d1A2d2
\nAs
\nA1A2=2.631.72A1A2=2.631.72
\nm1m2=2.631.72\u00d72.78.9=0.46m1m2=2.631.72\u00d72.78.9=0.46
\nIt indicated that m1<m2m1<m2.
\nClearly, aluminium is lighter than copper.
\nSince aluminium is lighter, it is prioritized for overhead power cables instead of copper.<\/h3>\n

17. What conclusion can you draw from the following observations on a resistor made of alloy manganin?<\/h2>\n\n\n\n\n\n<\/colgroup>\n\n\n\n\n\n\n\n\n
Current (A)<\/b><\/td>\nVoltage (V)<\/b><\/td>\nCurrent (A)<\/b><\/td>\nVoltage (V)<\/b><\/td>\n<\/tr>\n
0.2<\/b><\/td>\n3.94<\/b><\/td>\n3.0<\/b><\/td>\n59.2<\/b><\/td>\n<\/tr>\n
0.4<\/b><\/td>\n7.87<\/b><\/td>\n4.0<\/b><\/td>\n78.8<\/b><\/td>\n<\/tr>\n
0.6<\/b><\/td>\n11.8<\/b><\/td>\n5.0<\/b><\/td>\n98.6<\/b><\/td>\n<\/tr>\n
0.8<\/b><\/td>\n15.7<\/b><\/td>\n6.0<\/b><\/td>\n118.5<\/b><\/td>\n<\/tr>\n
1.0<\/b><\/td>\n19.7<\/b><\/td>\n7.0<\/b><\/td>\n138.2<\/b><\/td>\n<\/tr>\n
2.0<\/b><\/td>\n39.4<\/b><\/td>\n8.0<\/b><\/td>\n158.0<\/b><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Ans: From the given table,
\nThe ratio of voltage with current is a constant equal to 19.7.
\nAlso, manganin is an ohmic conductor, which follows Ohm\u2019s law.
\nOhm\u2019s law states that the ratio of voltage with current can be equated to the resistance of the conductor.
\nClearly, the resistance of manganin is
\n19.7\u03a919.7\u03a9
\n.<\/h3>\n

18. Answer the following questions:
\na) A steady current flow in a metallic conductor of non-uniform cross-section. Which of these quantities is constant along the conductor: current, current density, electric field, drift speed?<\/h2>\n

Ans: In the above question it is given that a steady current flows in a metallic conductor of non-uniform cross-section. Clearly, the current which flows through the conductor must be constant.
\nHowever, current density, electric field, and drift speed are known to be inversely proportional to the cross-sectional area of the conductor. Thus, these quantities cannot be constant.<\/h3>\n

b) Is ohm\u2019s law universally applicable for all conducting elements? If not, give examples of elements which do not obey Ohm\u2019s law.<\/h2>\n

Ans: All conducting elements need not satisfy Ohm\u2019s law. For instance, a vacuum diode semiconductor is a conductor which does not follow Ohm\u2019s law. Here, voltage is not proportional to either the current or the resistance.<\/h3>\n

c) A low voltage supply from which one needs high currents must have very low internal resistance. Why?<\/h2>\n

Ans: From Ohm\u2019s law, it is known that V=IRV=IR.
\nIt states that Voltage (V)(V) is directly proportional to current (I)(I) .
\nClearly,
\nI=VRI=VR
\nwhere, RR is the internal resistance of the source.
\nNow, when V is less, then R must be very less for a high current used from the source.<\/h3>\n

d) A high tension (HT) supply of, say, 6 kV must have a very large internal resistance. Why?<\/h2>\n

Ans: A high tension supply of a very large internal resistance is needed in order to control the current from exceeding the safety limit.
\nWhen the internal resistance is very less, then the current used exceeds the safety limits in case of short circuits.<\/h3>\n

19. Choose the correct alternative:
\na) Alloys of metals usually have (greater\/less) resistivity than that of their constituent metals.<\/h2>\n

Ans: Metal alloys generally have greater resistivity than that of their constituent metals.<\/h3>\n

b) Alloys usually have much (lower\/higher) temperature coefficients of resistance than pure metals.<\/h2>\n

Ans: Metal alloys generally have lower temperature coefficients of resistance than pure metals.<\/h3>\n

c) The resistivity of the alloy manganin is (nearly independent of\/increases rapidly) with increase of temperature.<\/h2>\n

Ans: When the temperature is increased, the resistivity of manganin gets no effect. Thus, its resistivity can be considered nearly independent with an increase in temperature.<\/h3>\n

d) The resistivity of a typical insulator like amber) is greater than that of a metal by a factor of the order of (1022\/103)(1022\/103).<\/h2>\n

Ans: A typical insulator\u2019s resistivity is greater than that of a metal by a factor of the order of 10221022.<\/h3>\n

20.
\na) Given n resistors each of resistance R, how will you combine them to get the (i) maximum (ii) minimum effective resistance? What is the ratio of the maximum to minimum resistance?<\/h2>\n

Ans: In the above question, it is provided that the total number of resistors are nn where the resistance of each resistor is RR.
\n(i) When nn resistors are connected in series, the effective resistance would be maximum and is considered as the product nRnR. Mathematically, the maximum resistance of the series combination is represented as R1=nRR1=nR.
\n(ii) The effective resistance would be minimum when resistors are connected parallelly. Mathematically, the maximum resistance of the parallel combination is represented R2=RnR2=Rn.
\n(iii) The ratio of the maximum to the minimum resistance would be
\nR1R2=nRR\/n=n2R1R2=nRR\/n=n2<\/h3>\n

b) Given the resistances of 1\u03a9,2\u03a9,3\u03a91\u03a9,2\u03a9,3\u03a9 , how will you combine them to get an equivalent resistance of (i) (11\/3)\u03a9(11\/3)\u03a9 (ii) (11\/5)\u03a9(11\/5)\u03a9 (iii) 6\u03a96\u03a9 (iv) (6\/11)\u03a9(6\/11)\u03a9 ?<\/h2>\n

Ans: In the above question it is provided that:
\nR1=1\u03a9,R2=2\u03a9,R3=3\u03a9R1=1\u03a9,R2=2\u03a9,R3=3\u03a9
\n(i) For an equivalent resistance of (11\/5)\u03a9(11\/5)\u03a9:
\nConsider the circuit diagram given below.
\n(Image will be uploaded soon)
\nEquivalent resistance for this circuit would be:
\nR\u2032=2\u00d712+1+3=23+3=113\u03a9R\u2032=2\u00d712+1+3=23+3=113\u03a9<\/h3>\n

(ii) For an equivalent resistance of (11\/3)\u03a9(11\/3)\u03a9:
\nConsider the circuit diagram given below.
\n(Image will be uploaded soon)
\nEquivalent resistance for this circuit would be:
\nR\u2032=2\u00d732+3+1=65+1=115\u03a9R\u2032=2\u00d732+3+1=65+1=115\u03a9<\/h3>\n

(iii) For an equivalent resistance of 6\u03a96\u03a9:
\nConsider the circuit diagram given below.
\n(Image will be uploaded soon)
\nEquivalent resistance for this circuit would be:
\nR\u2032=1+2+3=6\u03a9R\u2032=1+2+3=6\u03a9<\/h3>\n

(iv) For an equivalent resistance of (6\/11)\u03a9(6\/11)\u03a9 :
\nConsider the circuit diagram given below.
\n(Image will be uploaded soon)
\nEquivalent resistance for the circuit would be:
\nR\u2032=1\u00d72\u00d731\u00d72+2\u00d73+3\u00d71=611\u03a9R\u2032=1\u00d72\u00d731\u00d72+2\u00d73+3\u00d71=611\u03a9<\/h3>\n

c) Calculate the equivalent resistance of the following circuit:
\n(Image will be uploaded soon)<\/h2>\n

Ans: Consider the primary small loop on the left. Two resistors of resistance 1\u03a91\u03a9 each are connected in series in the upper arm of the loop.
\nClearly, their series equivalent resistance =1+1=2\u03a9=1+1=2\u03a9.
\nAlso, two resistors of resistance 2\u03a92\u03a9 each are connected in series in the lower arm of the loop.
\nClearly, their series equivalent resistance =2+2=4\u03a9=2+2=4\u03a9 .
\nNow, the circuit can be rearranged to:
\n(Image will be uploaded soon)
\nHere 2\u03a92\u03a9 and 4\u03a94\u03a9 resistors are connected in parallel in all the newly formed four loops.
\nClearly, their parallel equivalent resistance R\u2032R\u2032 is given by,
\nR\u2032=2\u00d742+4=86=43\u03a9R\u2032=2\u00d742+4=86=43\u03a9
\nThe circuit further gets rearranged to:
\n(Image will be uploaded soon)
\nNow, four resistors are connected in series.
\nClearly, their series equivalent resistance (or the equivalent resistance of the whole circuit) would be 43\u00d74=163\u03a943\u00d74=163\u03a9.<\/h3>\n

d) Calculate the equivalent resistance of the following circuit:
\n(Image will be uploaded soon)<\/h2>\n

Ans: From the given circuit, it is clear that five resistors of resistance RR are connected in series.
\nThus, their series equivalent resistance (or the equivalent resistance of the whole circuit)=R+R+R+R+R=5R=R+R+R+R+R=5R.<\/h3>\n

21. Determine the current drawn from a 12 V supply with internal resistance 0.5\u03a90.5\u03a9 by the infinite network shown in Figure. Each resistor has 1\u03a91\u03a9 resistance.
\n(Image will be uploaded soon)<\/h2>\n

Ans: In the above question, it is provided that the resistance of each resistor connected in the given circuit is R=1\u03a9R=1\u03a9.
\nConsider the equivalent resistance of the given circuit to be R\u2032R\u2032.
\nBecause the network is infinite, the equivalent resistance is given by the relation,
\nR\u2032=2+R\u2032(R\u2032+1)R\u2032=2+R\u2032(R\u2032+1)
\n\u21d2(R\u2032)2\u22122R\u2032\u22122=0\u21d2(R\u2032)2\u22122R\u2032\u22122=0
\n\u21d2R\u2032=2\u00b112\u221a2=1\u00b13\u2013\u221a\u21d2R\u2032=2\u00b1122=1\u00b13
\nAs only positive value is acceptable,
\n\u21d2R\u2032=1+3\u2013\u221a\u21d2R\u2032=1+3
\nInternal resistance of the circuit is r=0.5\u03a9r=0.5\u03a9.
\nAlso, total resistance =2.73+0.5=3.23\u03a9=2.73+0.5=3.23\u03a9.
\nNow, with respect to Ohm\u2019s law,
\nI=VR=123.23=3.72AI=VR=123.23=3.72A.
\nClearly, the current drawn is 3.72A3.72A.<\/h3>\n

22. Figure shows a potentiometer with a cell of 2.0 V and internal resistance 0.40\u03a90.40\u03a9 maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02 V (for very moderate currents up to a few mA) gives a balance point at 67.3 cm length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of 600k\u03a9600k\u03a9 is put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown emf \u03b5\u03b5 and the balance point found similarly, turns out to be at 82.3 cm length of the wire.
\n(Image will be uploaded soon)
\na) What is the value of \u03b5\u03b5?<\/h2>\n

Ans: A potentiometer can be used to compare the emf\u2019s of two cells with the help of balance points, at which there is no current flow in the circuit.
\nIn the above question, it is provided that constant emf of the given standard cell is E1=1.02VE1=1.02V.
\nThe balance point on the wire (when the galvanometer shows null deflection) is at a distance l1=67.3cml1=67.3cm.
\nNow, when the cell of unknown emf \u03b5\u03b5 gets replaced with the standard cell, the new balance point shifts to l=82.3cml=82.3cm.
\nThe relationship between emf and balance point in this potentiometer arrangement is given by
\nE1l1=\u03b5lE1l1=\u03b5l
\n\u21d2\u03b5=ll1\u00d7E1=82.367.3\u00d71.02=1.247V\u21d2\u03b5=ll1\u00d7E1=82.367.3\u00d71.02=1.247V
\nClearly, the value of unknown emf is 1.247V1.247V.<\/h3>\n

b) What purpose does the high resistance of 600k\u03a9600k\u03a9 have?<\/h2>\n

Ans: A high resistance of 600k\u03a9600k\u03a9 is used in order to lessen the current through the galvanometer when the movable contact is distant away from the balance point.<\/h3>\n

c) Is the balance point affected by this high resistance?<\/h2>\n

Ans: No, the balance point would not get affected by the presence of high resistance.<\/h3>\n

d) Is the balance point affected by the internal resistance of the driver cell?<\/h2>\n

Ans: No, the balance point would not get affected by the internal resistance of the driver cell.<\/h3>\n

e) Would the method work in the above situation if the driver cell of the potentiometer had an emf of 1.0 V instead of 2.0 V?<\/h2>\n

Ans: When the driver cell of the potentiometer had an emf of 1.0V1.0V in place of 2.0V2.0V, the given method ceases to work because when the emf of the driver cell of the potentiometer is less than the emf of the other cell, then the balance point refuses to exist on the wire.<\/h3>\n

f) Would the circuit work well for determining an extremely small emf, say of the order of a few mV (such as the typical emf of a thermos-couple)? If not, how will you modify the circuit?<\/h2>\n

Ans: To determine an extremely small emf, this arrangement cannot be used. Because of the instability in the circuit, the balance point would be near to end A. Thus, there would be large percentage errors in measurements.
\nHowever, the given arrangement can be modified by connecting a series resistance with the wire AB.
\nWhen done so, the potential drop across AB would be slightly greater than the measured emf. This facilitates the percentage error to become less.<\/h3>\n

23. Figure shows a Potentiometer circuit for the comparison of two resistances. The balance point with a standard resistor R=10.0\u03a9R=10.0\u03a9 is found to be 58.3 cm, while that with the unknown resistance X is 68.5 cm. Determine the value of X. What might you do if you failed to find a balance point with the given cell of emf \u03b5\u03b5 ?
\n(Image will be uploaded soon)<\/h2>\n

Ans: In the potentiometer setup, it is provided that:
\nThe resistance of the standard resistor is R=10.0\u03a9R=10.0\u03a9.
\nThe balance point for this resistance is at a distance l1=58.3cml1=58.3cm.
\nCurrent in the potentiometer wire is ii.
\nHence, potential drop across RR is E1=iRE1=iR.
\nThe resistance of the unknown resistor is XX.
\nThe balance point for this resistance is at a distance l2=68.5cml2=68.5cm.
\nHence, potential drop across XX is E2=iXE2=iX.
\nThe relation connecting emf and balance point in a potentiometer bridge circuit is given by
\nE1E2=l1l2E1E2=l1l2
\nRelating this with our problem,
\n\u21d2iRiX=l1l2\u21d2iRiX=l1l2
\n\u21d2X=l1l2\u00d7R=68.558.3\u00d710=11.749\u03a9\u21d2X=l1l2\u00d7R=68.558.3\u00d710=11.749\u03a9 .
\nClearly, the value of the unknown resistance XX is 11.749\u03a911.749\u03a9.
\nNow, if we couldn\u2019t find a balance point with the given cell of emf \u03b5\u03b5, then the potential drop across RR and XX is lessened by putting a resistance in series with it.
\nA balance point can be noted only when the potential drop across RR or XX is smaller than the potential drop across the potentiometer wire AB.<\/h3>\n

24. Figure shows a 2.0 V potentiometer used for the determination of internal resistance of 1.5 V cell. The balance point of the cell in open circuit is 76.3 cm. When a resistor of 9.5\u03a99.5\u03a9 is used in the external circuit of the cell, the balance point shifts to 64.8 cm length of the potentiometer wire. Determine the internal resistance of the cell.
\n(Image will be uploaded soon)<\/h2>\n

Ans: In the potentiometer setup, it is provided that:
\nBalance point of the cell in the open circuit is l1=76.3cml1=76.3cm.
\nAn external resistance RR of resistance 9.5\u03a99.5\u03a9 is connected to the circuit.
\nNew balance point of the circuit is at a distance l2=64.8cml2=64.8cm .
\nCurrent flowing through the circuit =I=I .
\nIn this potentiometer arrangement, the relationship connecting internal resistance of the cell and the two balance points of the bridge setup is:
\nr=(l1\u2212l2l2)Rr=(l1\u2212l2l2)R
\n\u21d2r=(76.3\u221264.864.8)9.5=1.68\u03a9\u21d2r=(76.3\u221264.864.8)9.5=1.68\u03a9
\nClearly, the internal resistance of the cell is 1.68\u03a91.68\u03a9 .<\/h3>\n","protected":false},"excerpt":{"rendered":"

Class 12 Physics NCERT book solutions for Chapter 3 – Current Electricity Questions and Answers.<\/p>\n","protected":false},"author":21830,"featured_media":118347,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[7],"tags":[],"class_list":{"0":"post-118379","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-education"},"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/118379","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/users\/21830"}],"replies":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/comments?post=118379"}],"version-history":[{"count":1,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/118379\/revisions"}],"predecessor-version":[{"id":118381,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/118379\/revisions\/118381"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/media\/118347"}],"wp:attachment":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/media?parent=118379"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/categories?post=118379"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/tags?post=118379"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}