A lead-acid battery with 12 cells in series (no-load voltage 2.1 V per cell) delivers 10 A to a 2 Ω load. The internal resistance of the battery is:

A lead-acid battery with 12 cells in series (no-load voltage 2.1 V per cell) delivers 10 A to a 2 Ω load. The internal resistance of the battery is:

• A. 0.25 Ω
• B. 1.00 Ω
• C. 0.52 Ω
• D. 2.00 Ω

Answer: (C)

When a battery delivers current, the voltage you measure at its terminals is the emf minus the drop across the internal resistance (V_terminal = E − I·r). Here, the emf comes from 12 cells in series, each 2.1 V, so E = 12 × 2.1 = 25.2 V. The load is 2 Ω and the current is 10 A, so the terminal voltage across the load is V_terminal = I × R_load = 10 × 2 = 20 V. The internal drop is the difference between emf and terminal voltage: 25.2 − 20 = 5.2 V. With a current of 10 A, the internal resistance is r = V_drop / I = 5.2 / 10 = 0.52 Ω. Therefore the internal resistance is 0.52 Ω. (If you test other values for r, they wouldn’t yield 10 A with the same emf and load, as the resulting currents would differ.)

When a battery delivers current, the voltage you measure at its terminals is the emf minus the drop across the internal resistance (V_terminal = E − I·r). Here, the emf comes from 12 cells in series, each 2.1 V, so E = 12 × 2.1 = 25.2 V. The load is 2 Ω and the current is 10 A, so the terminal voltage across the load is V_terminal = I × R_load = 10 × 2 = 20 V. The internal drop is the difference between emf and terminal voltage: 25.2 − 20 = 5.2 V. With a current of 10 A, the internal resistance is r = V_drop / I = 5.2 / 10 = 0.52 Ω. Therefore the internal resistance is 0.52 Ω. (If you test other values for r, they wouldn’t yield 10 A with the same emf and load, as the resulting currents would differ.)