Two cells, each with electromotive force (e.m.f.) E, but different internal resistances r1 and r2, are connected in series to a resistor R.

Question 17

Two cells, each with electromotive force (e.m.f.) E, but different internal resistances r1 and r2, are connected in series to a resistor R. The reading on the voltmeter is 0 V.

What is the resistance of R?

A 0                  B r1 – r2                               C r1 + r2                               D r1r2 / r1 + r2

Reference: Past Exam Paper – November 2017 Paper 13 Q36

Solution:

Answer: B.

From Kirchhoff’s law, the sum of p.d. in a circuit is equal to the sum of e.m.f. in the circuit.

Sum of e.m.f. = E + E = 2E

Let the current in the circuit be I.

Sum of p.d. = Ir₁ + Ir₂ + IR = I(r₁ + r₂ + R)

So, I(r₁ + r₂ + R) = 2E             ------------------ (1)

Let the potential on the right be 0 V.

Across each cell, there would be some lost volts.

Lost volt = current I × internal resistance

The terminal p.d. across the cell on the right is (E – Ir₂).

p.d. across A and B = E – Ir₂

Potential at B – potential at A = E – Ir₂

Potential at B – 0 = E – Ir₂

Potential at B = E – Ir₂

Since the voltmeter reads 0 V,

Potential at C – Potential at B = 0

Potential at C = Potential at B = E – Ir₂

But the difference in potential between B and C is also equal to the terminal p.d. across the left cell.

Terminal p.d. across left cell = E – Ir₁ = 0

So, E = Ir₁                   ------------------ (2)

Replacing E by Ir₁ in eqn (1).

I(r₁ + r₂ + R) = 2Ir₁

r₁ + r₂ + R = 2r₁

R = 2r₁ – r₁ – r₂ = r₁ – r₂

R = r₁ – r₂