Two identical batteries each have e.m.f. 6.0 V and internal resistance r. The batteries are connected to an external resistor of resistance 11 Ω, as shown.

Question 25

Two identical batteries each have e.m.f. 6.0 V and internal resistance r. The batteries are connected to an external resistor of resistance 11 Ω, as shown.

The current in the external resistor is 0.50 A.

What is the internal resistance r of each battery?

A 1.0 Ω           B 2.0 Ω           C 4.0 Ω           D 6.5 Ω

Reference: Past Exam Paper – June 2016 Paper 13 Q37

Solution:

Answer: B.

From Kirchhoff’s law, the sum of p.d. across any loop is equal to the sum of e.m.f. in that loop.

Consider any one of the two loops.

Ohm’s law: V = IR

p.d. across 11 Ω resistor = 11 × 0.50 = 5.5 V

5.5 V + p.d. across internal resistance = 6.0 V

So the p.d. across each internal resistance = 6.0 – 5.5 = 0.5 V.

Now, as the 0.50 A current enters the junction, it splits equally into the 2 branches since the internal resistances of both batteries are equal.

Current through each internal resistance = 0.25 A.

V = IR

Internal resistance r = 0.5 V / 0.25 A = 2.0 Ω.