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: $$I = \frac{E}{R_{total}} = \frac{2}{10} = 0.2A$$