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Current Electricity question

2021 · 17 Mar · Shift 2 · Q50
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Current Electricity question

2021 · 17 Mar · Shift 2 · Q50

JEE MainPhysicsCurrent ElectricityMCQ+4 / −1
Two cells of emf 2E and E with internal resistance r1 and r2 respectively are connected in series to an external resistor R (see figure). The value of R, at which the potential difference across the terminals of the first cell becomes zero is JEE Main 2021 (Online) 17th March Evening Shift Physics - Current Electricity Question 202 English
  1. A
    r1 −-− r2
  2. B
    r12−r2{{{r_1}} \over 2} - {r_2}2r1​​−r2​
  3. C
    r12+r2{{{r_1}} \over 2} + {r_2}2r1​​+r2​
  4. D
    r1 + r2
View written solutionFree

Correct answer: B

  1. Set up the circuit equation

Two cells of emfs 2E2E2E and EEE with internal resistances r1r_1r1​ and r2r_2r2​ are connected in series with an external resistance RRR.

Assuming both cells aid each other, total emf is 2E+E=3E2E+E=3E2E+E=3E

So current in the circuit is I=3ER+r1+r2I=\frac{3E}{R+r_1+r_2}I=R+r1​+r2​3E​

  1. Condition for terminal potential difference of the first cell to be zero

For the first cell (emf 2E2E2E, internal resistance r1r_1r1​), since it is delivering current, its terminal voltage is V1=2E−Ir1V_1=2E-Ir_1V1​=2E−Ir1​

Given that this terminal potential difference becomes zero, 2E−Ir1=02E-Ir_1=02E−Ir1​=0 I=2Er1I=\frac{2E}{r_1}I=r1​2E​

  1. Equate the two expressions for current

3ER+r1+r2=2Er1\frac{3E}{R+r_1+r_2}=\frac{2E}{r_1}R+r1​+r2​3E​=r1​2E​

Cancel EEE: 3R+r1+r2=2r1\frac{3}{R+r_1+r_2}=\frac{2}{r_1}R+r1​+r2​3​=r1​2​

Cross-multiplying, 3r1=2(R+r1+r2)3r_1=2(R+r_1+r_2)3r1​=2(R+r1​+r2​) 3r1=2R+2r1+2r23r_1=2R+2r_1+2r_23r1​=2R+2r1​+2r2​ r1=2R+2r2r_1=2R+2r_2r1​=2R+2r2​ 2R=r1−2r22R=r_1-2r_22R=r1​−2r2​ R=r12−r2R=\frac{r_1}{2}-r_2R=2r1​​−r2​

  1. Match with the options

R=r12−r2R=\frac{r_1}{2}-r_2R=2r1​​−r2​

This corresponds to Option B.

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