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Current Electricity question

2022 · 25 Jun · Shift 2 · Q54
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  5. /2022 · 25 Jun · Shift 2 · Q54

Current Electricity question

2022 · 25 Jun · Shift 2 · Q54

JEE MainPhysicsCurrent ElectricityMCQ+4 / −1
Two cells of same emf but different internal resistances r1 and r2 are connected in series with a resistance R. The value of resistance R, for which the potential difference across second cell is zero, is :
  1. A
    r2 −-− r1
  2. B
    r1 −-− r2
  3. C
    r1
  4. D
    r2
View written solutionFree

Correct answer: A

  1. Set up the circuit

Let each cell have emf EEE.

Their internal resistances are r1r_1r1​ and r2r_2r2​, and they are connected in series with an external resistance RRR.

So total resistance in the circuit is R+r1+r2R + r_1 + r_2R+r1​+r2​ and total emf is 2E.2E.2E.

Hence the current in the circuit is I=2ER+r1+r2.I = \frac{2E}{R + r_1 + r_2}.I=R+r1​+r2​2E​.

  1. Condition for zero potential difference across the second cell

For a cell delivering current, terminal potential difference is V=E−Ir.V = E - Ir.V=E−Ir.

For the second cell, internal resistance is r2r_2r2​. Given that its potential difference is zero, E−Ir2=0.E - I r_2 = 0.E−Ir2​=0.

So, I=Er2.I = \frac{E}{r_2}.I=r2​E​.

  1. Equate the two expressions for current

From the circuit, 2ER+r1+r2=Er2.\frac{2E}{R + r_1 + r_2} = \frac{E}{r_2}.R+r1​+r2​2E​=r2​E​.

Cancel EEE: 2R+r1+r2=1r2.\frac{2}{R + r_1 + r_2} = \frac{1}{r_2}.R+r1​+r2​2​=r2​1​.

Cross-multiplying, R+r1+r2=2r2.R + r_1 + r_2 = 2r_2.R+r1​+r2​=2r2​.

Thus, R=r2−r1.R = r_2 - r_1.R=r2​−r1​.

  1. Check options
  • A: r2−r1r_2 - r_1r2​−r1​ ✅
  • B: r1−r2r_1 - r_2r1​−r2​ ❌
  • C: r1r_1r1​ ❌
  • D: r2r_2r2​ ❌

Therefore, the correct answer is r2−r1.\boxed{r_2 - r_1}.r2​−r1​​.

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