If a battery with internal resistance r is connected to an external circuit with resistance R_ext, the current is I = E/(R_ext + r). When R_ext is much larger than r, the current is approximately

If a battery with internal resistance r is connected to an external circuit with resistance R_ext, the current is I = E/(R_ext + r). When R_ext is much larger than r, the current is approximately

• A. I ≈ E/R_ext
• B. I ≈ E/r
• C. I ≈ R_ext / E
• D. I ≈ r / E

Answer: (A)

When a battery has its own internal resistance, the total resistance in the circuit is the sum of the external resistance and the internal resistance. The current is given by I = E / (R_ext + r). If the external resistance is much larger than the internal resistance, the + r becomes negligible, so the total resistance is effectively R_ext. That makes the current approximately I ≈ E / R_ext. The small internal resistance has little effect on the current in this case. The other forms would describe different limiting situations: for instance, I ≈ E / r would occur if the external resistance were tiny compared to the internal resistance, which isn’t the case here, and the inverted forms (R_ext / E or r / E) aren’t how current relates to resistance and emf in Ohm’s law.

When a battery has its own internal resistance, the total resistance in the circuit is the sum of the external resistance and the internal resistance. The current is given by I = E / (R_ext + r). If the external resistance is much larger than the internal resistance, the + r becomes negligible, so the total resistance is effectively R_ext. That makes the current approximately I ≈ E / R_ext. The small internal resistance has little effect on the current in this case.

The other forms would describe different limiting situations: for instance, I ≈ E / r would occur if the external resistance were tiny compared to the internal resistance, which isn’t the case here, and the inverted forms (R_ext / E or r / E) aren’t how current relates to resistance and emf in Ohm’s law.