A lead-acid battery with 12 cells connected in series (no-load voltage = 2.1 volts per cell) furnishes 10 amperes to a load of 2-ohms resistance. The internal resistance of the battery in this instance is

• A. 0.52 ohm.
• B. 2.52 ohms.
• C. 5.0 ohms.
• D. 1.26 ohms.

Answer: (D)

The key idea is that a real battery has internal resistance that causes a voltage drop when delivering current. The no-load voltage is the emf of the cell stack, while under load the terminal voltage is reduced by I times the internal resistance.

First, find the total emf of the string: 12 cells × 2.1 V per cell = 25.2 V.

The load current is 10 A, so the voltage across the 2 Ω load is V_load = I × R_load = 10 A × 2 Ω = 20 V.

The voltage drop due to the battery’s internal resistance is E − V_load = 25.2 V − 20 V = 5.2 V.

Internal resistance r = (voltage drop) / (current) = 5.2 V / 10 A = 0.52 Ω.

So, the internal resistance is 0.52 ohm.