A torch bulb rated as 6 W, 1.5 V is connected in a circuit as shown in the figure. The <i>e.m.f.</i> of the cell needed to make the bulb glow normally is

Solution

The bulb and the $$1.5\ \Omega$$ resistor are in parallel across the cell, so both must have 1.5 V across them. The bulb then draws $$\frac{6}{1.5} = 4$$ A and the resistor draws $$\frac{1.5}{1.5} = 1$$ A, so the cell supplies 5 A. Including the internal resistance, $$E = 1.5 + 5 \times 0.5 = 4.0\ \text{V}$$.

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