Two identical batteries of emf E and internal resistance r are connected to a 6 Ω resistor. Current is the same when batteries are connected:

• In series
• In parallel

Find the value of internal resistance r.

Solution:

Series combination:

Equivalent emf = 2E

Equivalent resistance = 6 + 2r

Current:

I_s = 2E/(6 + 2r)

= E/(3 + r)

Parallel combination:

Equivalent emf = E

Equivalent internal resistance = r/2

Current:

I_p = E/(6 + r/2)

Given:

I_s = I_p

E/(3+r) = E/(6+r/2)

3 + r = 6 + r/2

r/2 = 3

r = 6 Ω

Answer: 6 Ω