Question

Two batteries (6V, 0.8Ω and 3V, 0.2Ω) are connected in series with external resistance R. Find R so that the potential difference across the 6V battery is zero.

Hint:

Potential difference and internal resistance

Answer 1

Total emf = 6V + 3V = 9V. Total internal resistance = 0.8Ω + 0.2Ω = 1Ω. Current (i) = 9/(R+1). Potential difference across 6V battery: 6 - i(0.8) = 0. Substituting i: 6 - (9/(R+1)) * 0.8 = 0. Solving for R gives R = 0.2Ω.