The open-circuit emf of a storage cell is 2.2 volts. The terminal voltage measured when the current is 12 A is found to be 1.98 volts. The internal resistance of the cell is
- A. 0.00183 ohm
- B. 0.0183 ohm
- C. 0.183 ohm
- D. 1.83 ohm
Answer: B. 0.0183 ohm
Explanation: Internal voltage drop $= E - V = 2.2 - 1.98 = 0.22$ V. Internal resistance $r = \frac{\text{Voltage Drop}}{I} = \frac{0.22}{12} = 0.0183~\Omega$.