Question

Five identical cells, each of emf 2 V and internal resistance 0.1 \( \Omega \), are connected in parallel. This combination in turn is connected to an external resistor of 9.98 \( \Omega \). The current flowing through the resistor is:

Answer 1

Since the five identical cells are connected in parallel, the total emf of the combination remains 2V. The equivalent internal resistance of the parallel combination is: \[ R_{eq} = \frac{r}{n} = \frac{0.1}{5} = 0.02\Omega \] Total resistance in the circuit: \[ R_{total} = R_{ext} + R_{eq} = 9.98 + 0.02 = 10\Omega \] Current in the circuit: \[ I = \frac{E}{R_{total}} = \frac{2}{10} = 0.2A \]