Question 1

Question 15If the length l of wire is increased to 3l by stretching, its resistance R increases to ...............3R4R6R9R

Accepted Answer

9RReason — When the wire is stretched to thrice it's length, it's area of cross section becomes 1/3 and it's length becomes 3l.Let, a be the area of initial cross section and ρ be the specific resistance of the material of wire.Then,length = l,Resistance = R Ω,new length = 3l,new area = a3\dfrac{a}{3}3a​From relationR = ρ la\dfrac{l}{a}al​ = ρ lπr2\dfrac{l}{πr^2}πr2l​Initial resistance R1 = R = ρ la\dfrac{l}{a}al​    [Equation 1]New resistance Rn = ρ 3la3\dfrac{3l}{\dfrac{a}{3}}3a​3l​ = ρ 9la\dfrac{9l}{a}a9l​    [Equation 2]On dividing eqn (ii) by (i), we get,RnR=ρ9laρla⇒RnR=9⇒Rn=9R\dfrac{R_n}{R} = \dfrac{ ρ \dfrac{9l}{a}}{ρ \dfrac{l}{a}} \[1em] \Rightarrow \dfrac{R_n}{R} = 9 \[1em] \Rightarrow R_n = 9RRRn​​=ρal​ρa9l​​⇒RRn​​=9⇒Rn​=9RHence, the new resistance = 9R.

Question 2

Question 1In a conductor, 6.25 × 1016 electrons flow from it's end A to B in 2 s. Find the current flowing through the conductor. (e = 1.6 × 10-19 C)

Accepted Answer

Given,Number of electrons flowing (n) = 6.25 × 1016Time taken (t) = 2 se = 1.6 × 10-19 CCurrent in conductor is given by,I = net\dfrac{\text{ne}}{\text{t}}tne​Substituting the values in the formula above we get,I=6.25×1016×1.6×10−192I=6.25×1.6×10−32I=6.25×0.8×10−3I=5×10−3AI = \dfrac{6.25 \times 10^{16} \times 1.6 \times 10^{-19}}{2} \[0.5em] I = \dfrac{6.25 \times 1.6 \times 10^{-3}}{2} \[0.5em] I = 6.25 \times 0.8 \times 10^{-3} \[0.5em] I = 5 \times 10^{-3} AI=26.25×1016×1.6×10−19​I=26.25×1.6×10−3​I=6.25×0.8×10−3I=5×10−3AHence, the current flowing through the conductor = 5 mA from B to A

Question 3

Question 9A given wire of resistance 1 Ω is stretched to double it's length. What will be it's new resistance?

Accepted Answer

When the wire is stretched to double it's length, it's area of cross section becomes half and it's length becomes double.Let, a be the area of initial cross section and ρ be the specific resistance of the material of wire.Then,length = l,R = 1 Ω,new length = 2l,new area = a2\dfrac{a}{2}2a​From relationR = ρ la\dfrac{l}{a}al​ = ρ lπr2\dfrac{l}{πr^2}πr2l​Initial resistance R1 = 1 = ρ la\dfrac{l}{a}al​    [Equation 1]New resistance Rn = ρ 2la2\dfrac{2l}{\dfrac{a}{2}}2a​2l​ = ρ 4la\dfrac{4l}{a}a4l​    [Equation 2]On dividing eqn (ii) by (i), we get,Rn1=ρ4laρla⇒Rn=4Ω\dfrac{R_n}{1} = \dfrac{ ρ \dfrac{4l}{a}}{ρ \dfrac{l}{a}} \[0.5em] \Rightarrow R_n = 4 \varOmega1Rn​​=ρal​ρa4l​​⇒Rn​=4ΩHence, the new resistance = 4 Ω.

Question 4

Question 10A wire of resistance 3 ohm and length 10 cm is stretched to length 30 cm. Assuming that it has a uniform cross-section, what will be it's new resistance?

Accepted Answer

Given,Resistance (R) = 3 ΩLength (l) = 10 cmLet, a be the area of initial cross section and ρ be the specific resistance of the material of wire.Then,R = 3 Ωlength = 10 cm,new length = 30 cm,new area a = a3\dfrac{a}{3}3a​From relationR = ρ la\dfrac{l}{a}al​ = ρ lπr2\dfrac{l}{πr^2}πr2l​Initial resistance 3 = ρ 10a\dfrac{10}{a}a10​    [Equation 1]New resistance R2 = ρ 30a3\dfrac{30}{\dfrac{a}{3}}3a​30​ = ρ 90a\dfrac{90}{a}a90​    [Equation 2]On dividing eqn (i) by (ii), we get,3R2=ρ10aρ90a3R2=10903R2=19⇒R2=3×9⇒R2=27Ω\dfrac{3}{R_2} = \dfrac{ρ \dfrac{10}{a}}{ ρ \dfrac{90}{a}} \[0.5em] \dfrac{3}{R_2} = \dfrac{10}{90} \[0.5em] \dfrac{3}{R_2} = \dfrac{1}{9} \[0.5em] \Rightarrow R_2 = 3 \times 9 \[0.5em] \Rightarrow R_2 = 27 \varOmega \[0.5em]R2​3​=ρa90​ρa10​​R2​3​=9010​R2​3​=91​⇒R2​=3×9⇒R2​=27ΩHence, the new resistance = 27 Ω.

Question 5

Question 4Show by a diagram how two resistors R1 and R2 are joined in parallel. Obtain an expression for the total resistance of combination.

Accepted Answer

Below diagram shows two resistors connected in parallel:Let I1 and I2 be the currents through the resistances R1 and R2 respectively, then total current drawn from the battery isI = I1 + I2    [Equation 1]If potential difference between the two ends A and B is V, then by Ohm's lawcurrent in R1 is I1 = VR1\dfrac{V}{R_1}R1​V​current in R2 is I2 = VR2\dfrac{V}{R_2}R2​V​On adding these,I1 + I2 = VR1\dfrac{V}{R_1}R1​V​ + VR2\dfrac{V}{R_2}R2​V​    [Equation 2]If the equivalent resistance of the combination between the points A and C is Rp, then total current drawn from the source isI = VRp\dfrac{V}{R_p}Rp​V​    [Equation 3]Substituting the values of I and I1 + I2 from equation 3 and 2 in 1, we get,VRp=V(1R1+1R2)⇒1Rp=1R1+1R2\dfrac{V}{R_p} = V\Big(\dfrac{1}{R_1} + \dfrac{1}{R_2}\Big) \[0.5em] \Rightarrow \dfrac{1}{R_p} = \dfrac{1}{R_1} + \dfrac{1}{R_2}Rp​V​=V(R1​1​+R2​1​)⇒Rp​1​=R1​1​+R2​1​Thus, in the parallel combination, the reciprocal of the equivalent resistance is equal to the sum of the reciprocals of the individual resistances.

Question 6

Question 7Four resistors each of resistance 5 Ω are connected in parallel. What is the effective resistance?

Accepted Answer

Give,R1 = R2 = R3 = R4 = 5 ΩIn parallel, if the equivalent resistance is RP, then1RP=1R1+1R2+1R3+1R41RP=15+15+15+151RP=45⇒RP=54⇒RP=1.25Ω\dfrac{1}{R_P} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3} + \dfrac{1}{R_4} \[1em] \dfrac{1}{R_P} = \dfrac{1}{5} + \dfrac{1}{5} + \dfrac{1}{5} + \dfrac{1}{5} \[1em] \dfrac{1}{R_P} = \dfrac{4}{5} \[1em] \Rightarrow R_P = \dfrac{5}{4} \[1em] \Rightarrow R_P = 1.25 ΩRP​1​=R1​1​+R2​1​+R3​1​+R4​1​RP​1​=51​+51​+51​+51​RP​1​=54​⇒RP​=45​⇒RP​=1.25ΩHence, equivalent resistance = 1.25 Ω

Question 7

Question 8You have three resistors of values 2 Ω, 3 Ω and 5 Ω . How will you join them so that the total resistance is less than 1 Ω ? Draw diagram and find the total resistance.

Accepted Answer

In order to get a total resistance less than 1 Ω, the three resistors of values 2 Ω, 3 Ω and 5 Ω, should be connected in parallel as shown in the diagram below:In parallel, if the equivalent resistance is RP, then1RP=1R1+1R2+1R31RP=12+13+151RP=15+10+6301RP=3130⇒RP=0.97Ω\dfrac{1}{R_P} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3} \[0.5em] \dfrac{1}{R_P} = \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{5} \[0.5em] \dfrac{1}{R_P} = \dfrac{15 + 10+ 6}{30} \[0.5em] \dfrac{1}{R_P} = \dfrac{31}{30} \[0.5em] \Rightarrow R_P = 0.97 ΩRP​1​=R1​1​+R2​1​+R3​1​RP​1​=21​+31​+51​RP​1​=3015+10+6​RP​1​=3031​⇒RP​=0.97ΩHence, the three resistors should be connected in parallel and total resistance = 0.97 Ω

Question 8

Question 11A combination consists of three resistors in series. Four similar sets are connected in parallel. If the resistance of each resistor is 2 ohm, find the resistance of the combination.

Accepted Answer

Let the four set of resistors be R1, R2, R3, R4 .As each set consists of three resistors of 2 Ω each in series:Resistance of each set = 2 + 2 + 2 = 6 ΩAs four similar sets of equivalent resistance 6 Ω are arranged in parallel, so the equivalent resistance (RP) will be:1RP=1R1+1R2+1R3+1R41RP=16+16+16+161RP=46RP=64⇒RP=1.5Ω\dfrac{1}{R_P} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3} + \dfrac{1}{R_4} \[0.5em] \dfrac{1}{R_P} = \dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{6} \[0.5em] \dfrac{1}{R_P} = \dfrac{4}{6} \[0.5em] R_P = \dfrac{6}{4} \[0.5em] \Rightarrow R_P = 1.5 ΩRP​1​=R1​1​+R2​1​+R3​1​+R4​1​RP​1​=61​+61​+61​+61​RP​1​=64​RP​=46​⇒RP​=1.5ΩHence, equivalent resistance = 1.5 Ω

Question 9

Question 13Calculate the effective resistance between the points A and B in the circuit shown in figure.

Accepted Answer

In the circuit, there are four parts. In the first part, three resistors of 1 Ω each are connected in series. If the equivalent resistance of this part is R's thenR's = (1 + 1 + 1) Ω = 3 ΩIn the second part, three resistors of 2 Ω each are connected in series. If the equivalent resistance of this part is R''s thenR''s = (2 + 2 + 2) Ω = 6 ΩIn the third part, the two parts of resistance R's = 3 Ω and R''s = 6 Ω and 2 Ω are connected in parallel. If the equivalent resistance is Rp then1Rp=1Rs′+12+1Rs′′1Rp=13+12+161Rp=2+3+161Rp=66⇒Rp=1Ω\dfrac{1}{R_p} = \dfrac{1}{R'_s} + \dfrac{1}{2} + \dfrac{1}{R''_s} \[0.5em] \dfrac{1}{R_p} = \dfrac{1}{3} + \dfrac{1}{2} + \dfrac{1}{6} \[0.5em] \dfrac{1}{R_p} = \dfrac{2 + 3 + 1}{6} \[0.5em] \dfrac{1}{R_p} = \dfrac{6}{6} \[0.5em] \Rightarrow R_p = 1 Ω \[0.5em]Rp​1​=Rs′​1​+21​+Rs′′​1​Rp​1​=31​+21​+61​Rp​1​=62+3+1​Rp​1​=66​⇒Rp​=1Ω∴ Rp = 1 ΩIn the fourth part 1 Ω, (Rp = 1 Ω) and 1 Ω are connected in series in between points A and B.Hence, the equivalent resistance is 1 + 1 + 1 = 3 Ω

Question 10

Question 14A uniform wire with a resistance of 27 Ω is divided into three equal pieces and then they are joined in parallel. Find the equivalent resistance of the parallel combination.

Accepted Answer

Given, uniform wire with a resistance of 27 Ω is divided into three equal pieces. Hence, resistance of each piece = 9 Ω.Three such pieces are joined in parallel.In parallel, if the equivalent resistance is RP, then1RP=1R1+1R2+1R31RP=19+19+19=39=13⇒RP=3Ω\dfrac{1}{R_P} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3} \[0.5em] \dfrac{1}{R_P} = \dfrac{1}{9} + \dfrac{1}{9} + \dfrac{1}{9} = \dfrac{3}{9} = \dfrac{1}{3} \[0.5em] \Rightarrow R_P = 3 ΩRP​1​=R1​1​+R2​1​+R3​1​RP​1​=91​+91​+91​=93​=31​⇒RP​=3ΩHence, equivalent resistance = 3 Ω

Question 11

Question 17Calculate the equivalent resistance between the points A and B in figure.

Accepted Answer

In the circuit, there are three parts. In the first part, two resistors of 3 Ω, 2 Ω are connected in series. If the equivalent resistance is R's thenR's = 3 + 2 = 5 ΩIn the second part, two resistors of 6 Ω, 4 Ω are connected in series. If the equivalent resistance is R''s thenR''s = 6 + 4 = 10 ΩIn the third part R's, 30 Ω and R''s are connected in parallel. If the equivalent resistance is Rp then1RP=1Rs′+130+1Rs′′1RP=15+130+1101RP=6+1+3301RP=10301RP=13⇒RP=3Ω\dfrac{1}{R_P} = \dfrac{1}{R'_s} + \dfrac{1}{30} + \dfrac{1}{R''_s} \[0.5em] \dfrac{1}{R_P} = \dfrac{1}{5} + \dfrac{1}{30} + \dfrac{1}{10} \[0.5em] \dfrac{1}{R_P} = \dfrac{6 + 1 + 3}{30} \[0.5em] \dfrac{1}{R_P} = \dfrac{10}{30} \[0.5em] \dfrac{1}{R_P} = \dfrac{1}{3} \[0.5em] \Rightarrow R_P = 3 ΩRP​1​=Rs′​1​+301​+Rs′′​1​RP​1​=51​+301​+101​RP​1​=306+1+3​RP​1​=3010​RP​1​=31​⇒RP​=3ΩHence, RP = 3 Ω

Question 12

Question 18In the network shown in figure, calculate the equivalent resistance between the points (a) A and B (b) A and C

Accepted Answer

In the circuit, there are two parts. In the first part, three resistors of 2 Ω each are connected in series. If the equivalent resistance of this part is R's thenR's = (2 + 2 + 2) Ω = 6 ΩIn the second part, the resistance of first part (R's = 6 Ω ) and 2 Ω are connected in parallel. If the equivalent resistance of this part is Rp then1Rp=16+121Rp=1+361Rp=46⇒Rp=1.5Ω\dfrac{1}{R_p} = \dfrac{1}{6} + \dfrac{1}{2} \[0.5em] \dfrac{1}{R_p} = \dfrac{1 + 3}{6} \[0.5em] \dfrac{1}{R_p} = \dfrac{4}{6} \[0.5em] \Rightarrow R_p = 1.5 Ω \[0.5em]Rp​1​=61​+21​Rp​1​=61+3​Rp​1​=64​⇒Rp​=1.5Ω∴ Rp = 1.5 Ω(b) In the circuit, there are three parts. In the first part, two resistors of 2 Ω each are connected in series. If the equivalent resistance of this part is R's thenR's = (2 + 2) Ω = 4 ΩIn the second part, two resistors of 2 Ω each are connected in series. If the equivalent resistance of this part is R''s thenR''s = (2 + 2) Ω = 4 ΩIn the third part, the two parts of resistance R's = 4 Ω and R''s = 4 Ω are connected in parallel. If the equivalent resistance between points A and C is Rp then1Rp=14+141Rp=241Rp=12⇒Rp=2Ω\dfrac{1}{R_p} = \dfrac{1}{4} + \dfrac{1}{4} \[0.5em] \dfrac{1}{R_p} = \dfrac{2}{4} \[0.5em] \dfrac{1}{R_p} = \dfrac{1}{2} \[0.5em] \Rightarrow R_p = 2 Ω \[0.5em]Rp​1​=41​+41​Rp​1​=42​Rp​1​=21​⇒Rp​=2Ω∴ Equivalent Resistance between A and C = 2 Ω

Question 13

Question 19Five resistors, each of 3 ohm, are connected as shown in figure. Calculate the resistance (a) between the points P and Q, and (b) between the points X and Y.

Accepted Answer

(a) In the circuit, there are two parts. In the first part, two resistors of 3 Ω each are connected in series. If the equivalent resistance of this part is R's thenR's = (3 + 3) Ω = 6 ΩIn the second part, resistance R's = 6 Ω and 3 Ω are connected in parallel. If the equivalent resistance between points P and Q is Rp then1Rp=16+131Rp=1+261Rp=36⇒Rp=2Ω\dfrac{1}{R_p} = \dfrac{1}{6} + \dfrac{1}{3} \[0.5em] \dfrac{1}{R_p} = \dfrac{1 + 2}{6} \[0.5em] \dfrac{1}{R_p} = \dfrac{3}{6} \[0.5em] \Rightarrow R_p = 2 Ω \[0.5em]Rp​1​=61​+31​Rp​1​=61+2​Rp​1​=63​⇒Rp​=2Ω∴ Equivalent resistance between the points P and Q = 2 Ω(b) In the circuit, 3 Ω, Rp = 2 Ω and 3 Ω are connected in series. If the equivalent resistance of this part is Rs thenRs = (3 + 2 + 3) Ω = 8 Ω∴ Equivalent resistance between the points X and Y = 8 Ω

Question 14

Question 20Two resistors of 4.0 Ω and 6.0 Ω are connected (a) in series, (b) in parallel, with a battery of 6.0 V and negligible internal resistance. For each case draw a circuit diagram and calculate the current through the battery.

Accepted Answer

(a) Circuit diagram showing two resistors of 4.0 Ω and 6.0 Ω connected in series with a battery of 6.0 V and negligible internal resistance is shown below:Given,Two resistors of 4 Ω and 6 Ω are connected in series. If the equivalent resistance of this part is R's thenR's = (4 + 6) Ω = 10 ΩPotential Difference V = 6 VCurrent I = ?From Ohm's lawV = IRSubstituting the values in the formula above, we get,6 = I x 10⇒ I = 6 / 10 = 0.6 AHence, in series, current through the battery = 0.6 A(b) Circuit diagram showing two resistors of 4.0 Ω and 6.0 Ω connected in parallel with a battery of 6.0 V and negligible internal resistance is shown below:Given,Two resistors of 4 Ω and 6 Ω are connected in parallel. If the equivalent resistance of this part is Rp then1Rp=14+161Rp=3+212=512Rp=125=2.4Ω\dfrac{1}{R_p} = \dfrac{1}{4} + \dfrac{1}{6} \[0.5em] \dfrac{1}{R_p} = \dfrac{3 + 2}{12} = \dfrac{5}{12} \[0.5em] R_p = \dfrac{12}{5} = 2.4 ΩRp​1​=41​+61​Rp​1​=123+2​=125​Rp​=512​=2.4ΩPotential Difference V = 6 VCurrent I = ?From Ohm's lawV = IRSubstituting the values in the formula above, we get,6 = I x 2.4 ⇒ I = 6 / 2.4 = 2.5 AHence, in parallel, current through the battery = 2.5 A

Question 15

Question 22Two resistors of resistance 2 Ω and 3 Ω are connected in parallel to a cell to draw 0.5 A current from the cell.(a) Draw a labelled diagram of the arrangement.(b) Calculate the current in each resistor.

Accepted Answer

(a) Labelled diagram of the arrangement is shown below:(b) Two resistors of 2 Ω and 3 Ω are connected in parallel. If the equivalent resistance of this part is Rp then1Rp=12+131Rp=3+261Rp=56Rp=65⇒Rp=1.2Ω\dfrac{1}{R_p} = \dfrac{1}{2} + \dfrac{1}{3} \[0.5em] \dfrac{1}{R_p} = \dfrac{3 + 2}{6} \[0.5em] \dfrac{1}{R_p} = \dfrac{5}{6} \[0.5em] R_p = \dfrac{6}{5} \[0.5em] \Rightarrow R_p = 1.2 ΩRp​1​=21​+31​Rp​1​=63+2​Rp​1​=65​Rp​=56​⇒Rp​=1.2ΩHence, equivalent resistance = 1.2 ΩCurrent I = 0.5 APotential Difference V = ?From Ohm's lawV = IRSubstituting the values in the formula above, we get,V = 0.5 x 1.2 = 0.6 VCurrent through 2 Ω = ?From Ohm's lawV = IRSubstituting the values in the formula above, we get,0.6 = I x 2 ⇒ I = 0.6 / 2 = 0.3 AHence, current through 2 Ω resistor = 0.3 ACurrent through 3 Ω = ?From Ohm's lawV = IRSubstituting the values in the formula above, we get,0.6 = I x 3 ⇒ I = 0.6 / 3 = 0.2 AHence, current through 3 Ω resistor = 0.2 A

Question 16

Question 23Calculate the current flowing through each of the resistors A and B in the circuit shown in figure.

Accepted Answer

(a) Given,R = 1 ΩV = 2 VCurrent through 1 Ω = ?From Ohm's lawV = IRSubstituting the values in the formula above, we get,2 = I x 1⇒ I = 2 / 1 = 2 AHence, current through Resistor A = 2 A(b) Given,R = 2 ΩV = 2 VCurrent through 2 Ω = ?From Ohm's lawV = IRSubstituting the values in the formula above, we get,2 = I x 2⇒ I = 2 / 2 = 1 AHence, current through Resistor B = 1 A

Question 17

Question 24In figure, calculate —(a) the total resistance of the circuit,(b) the value of R, and(c) the current flowing in R

Accepted Answer

(a) Given,V = 4 VI = 0.4 AFrom Ohm's lawV = IR4 = 0.4 x R⇒ R = 4 / 0.4 = 10 ΩHence, total resistance of the circuit = 10 Ω(b) As total resistance of the circuit is equal to 10 Ω and the two resistors R and 20 Ω are connected in parallel, hence we get,110=1R+1201R=110−1201R=2−1201R=120R=20Ω\dfrac{1}{10} = \dfrac{1}{R} + \dfrac{1}{20} \[0.5em] \dfrac{1}{R} = \dfrac{1}{10} - \dfrac{1}{20} \[0.5em] \dfrac{1}{R} = \dfrac{2 - 1}{20} \[0.5em] \dfrac{1}{R} = \dfrac{1}{20} \[0.5em] R = 20 Ω \[0.5em]101​=R1​+201​R1​=101​−201​R1​=202−1​R1​=201​R=20ΩHence, the value of R = 20 Ω(c) current flowing in R = ?V = 4 VR = 20 ΩI = ?From Ohm's lawV = IR⇒ 4 = I x 20⇒ I = 4 / 20 = 1 / 5 = 0.2 AHence, the current flowing in R = 0.2 A

Question 18

Question 25A particular resistance wire has a resistance of 3.0 ohm per meter. Find —(a) The total resistance of three lengths of this wire each 1.5 m long, joined in parallel.(b) The potential difference of the battery which gives a current of 2.0 A in each of the 1.5 m length when connected in parallel to the battery (assume that resistance of the battery is negligible).(c) The resistance of 5 m length of a wire of the same material, but with twice the area of cross section.

Accepted Answer

(a) Resistance of 1 m of wire = 3 Ω.Resistance of 1.5 m of wire = 3 x 1.5 = 4.5 Ω. As three such wires are joined in parallel and if the equivalent resistance of this part is Rp then1Rp=14.5+14.5+14.51Rp=34.5⇒Rp=4.53⇒Rp=1.5Ω\dfrac{1}{R_p} = \dfrac{1}{4.5} + \dfrac{1}{4.5} + \dfrac{1}{4.5} \[0.5em] \dfrac{1}{R_p} = \dfrac{3}{4.5} \[0.5em] \Rightarrow R_p = \dfrac{4.5}{3} \[0.5em] \Rightarrow R_p = 1.5 ΩRp​1​=4.51​+4.51​+4.51​Rp​1​=4.53​⇒Rp​=34.5​⇒Rp​=1.5ΩHence, total resistance of circuit = 1.5 Ω(b) I = 2 AFrom Ohm's LawV = IRSubstituting the values in the formula above we get,V = 2 x 4.5 = 9 VHence, potential difference = 9 V(c) R = 3 Ω for 1 meter wireTherefore, for 5 mR = 3 x 5 = 15 ΩHere the area is twice and resistance is inversely proportional to area.Thus, resistance becomes halfR = 15 / 2 = 7.5 ΩHence, resistance = 7.5 Ω

Question 19

Question 26A cell supplies a current of 1.2 A through two resistors each of 2 Ω connected in parallel. When the resistors are connected in series, it supplies a current of 0.4 A. Calculate (i) the internal resistance and (ii) e.m.f. of the cell.

Accepted Answer

(i) Given,Two resistors each of 2 Ω connected in parallel. If the equivalent resistance of this part is Rp then1Rp=12+121Rp=22⇒Rp=1Ω\dfrac{1}{R_p} = \dfrac{1}{2} + \dfrac{1}{2} \[0.5em] \dfrac{1}{R_p} = \dfrac{2}{2} \[0.5em] \Rightarrow R_p = 1 ΩRp​1​=21​+21​Rp​1​=22​⇒Rp​=1ΩHence, equivalent resistance Rp = 1 ΩGiven,I = 1.2 AFrom relation,ε = I (R + r)ε = 1.2 (1 + r)ε = 1.2 + 1.2r    [Equation 1]When the resistors are connected in series, it supplies a current of 0.4 A,If the equivalent resistance of this part is Rs thenRs = 2 + 2 = 4 ΩHence, equivalent resistance Rs = 4 ΩGiven, I = 0.4 AFrom relation,ε = I (R + r)ε = 0.4 (4 + r)ε = 1.6 + 0.4r    [Equation 2]Equating 1 and 2, we get,1.2 + 1.2r = 1.6 + 0.4r⇒ 1.2r – 0.4r = 1.6 - 1.2⇒ 0.8r = 0.4⇒ r = 0.40.8\dfrac{0.4}{0.8}0.80.4​⇒ r = 0.5 ΩHence, Internal resistance r = 0.5 Ω(ii) Substituting the value in equation 1 we get,ε = 1.2(1 + r)= 1.2 (1 + 0.5)= 1.2 x 1.5= 1.8 VHence, e.m.f. of the cell = 1.8 V

Question 20

Question 27A battery of e.m.f. 16 V and internal resistance 2 Ω is connected to two resistors 3 Ω and 6 Ω connected in parallel. Find (a) the current through the battery (b) p.d. between the terminals of the battery (c) the current in 3 Ω resistor (d) the current in 6 Ω resistor.

Accepted Answer

(a) Given,e.m.f. = 16 Vinternal resistance r = 2 Ωcurrent through battery = ?If Rp is the equivalent resistance of resistors 3 Ω and 6 Ω connected in parallel, then1Rp=13+161Rp=2+161Rp=361Rp=12⇒Rp=2Ω\dfrac{1}{R_p} = \dfrac{1}{3} + \dfrac{1}{6} \[0.5em] \dfrac{1}{R_p} = \dfrac{2 + 1}{6} \[0.5em] \dfrac{1}{R_p} = \dfrac{3}{6} \[0.5em] \dfrac{1}{R_p} = \dfrac{1}{2} \[0.5em] \Rightarrow R_p = 2 ΩRp​1​=31​+61​Rp​1​=62+1​Rp​1​=63​Rp​1​=21​⇒Rp​=2ΩFrom relation,ε = I (R + r)Substituting the value in the formula above we get,16 = I(2 + 2)⇒ 16 = I x 4 ⇒ I = 16 / 4 = 4 AHence, current through the battery = 4 A(b) Potential difference between the terminals of the battery = ?Using Ohm's lawV = IRR = 2 ΩI = 4 ASubstituting the values in the formula above we get,V = 4 x 2 = 8 VHence, potential difference between the terminals of the battery = 8 V(c) Current in 3 Ω resistor = ?Using Ohm's lawV = IRR = 3 ΩV = 8 VI = ?Substituting the values in the formula above we get,8 = I × 3I = 83\dfrac{8}{3}38​⇒ I = 2.66 AHence, current in 3 Ω resistor is 2.66 A(d) Current in 6 Ω resistor = ?Using Ohm's lawV = IRR = 6 ΩV = 8 VI = ?Substituting the values in the formula above we get,8 = I × 6⇒ I = 8 / 6 = 1.333 AHence, current in 6 Ω resistor is 1.34 A

Current Electricity

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