A battery has an internal resistance of 0.2 Ω. If the external circuit is 10 Ω and the battery supplies 9 V open-circuit, what is the current when connected?

A battery has an internal resistance of 0.2 Ω. If the external circuit is 10 Ω and the battery supplies 9 V open-circuit, what is the current when connected?

• A. 0.5 A
• B. 0.882 A
• C. 9 A
• D. 0.2 A

Answer: (B)

When a real battery is used, its own internal resistance adds to the external resistance in the circuit, so the current is determined by the total resistance. The emf is 9 V, external resistance is 10 Ω, and internal resistance is 0.2 Ω, giving a total resistance of 10.2 Ω. The current is I = E / (R + r) = 9 / 10.2 ≈ 0.882 A. The voltage across the external resistor is V = I × 10 ≈ 8.82 V, while the drop across the internal resistance is I × r ≈ 0.176 V, summing to 9 V. So the current is about 0.882 A.

When a real battery is used, its own internal resistance adds to the external resistance in the circuit, so the current is determined by the total resistance. The emf is 9 V, external resistance is 10 Ω, and internal resistance is 0.2 Ω, giving a total resistance of 10.2 Ω. The current is I = E / (R + r) = 9 / 10.2 ≈ 0.882 A. The voltage across the external resistor is V = I × 10 ≈ 8.82 V, while the drop across the internal resistance is I × r ≈ 0.176 V, summing to 9 V. So the current is about 0.882 A.