A torch bulb rated as $4.5\, W$, $1.5\, V$ is connected as shown in the figure. The $e.m.f.$ of the cell needed to make the bulb glow at full intensity is ................ $V$
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(d) Current in the bulb $ = \frac{P}{V} = \frac{{4.5}}{{1.5}} = 3\,A$
Current in $1$ $\Omega$ resistance $ = \frac{{1.5}}{1} = 1.5\,A$
Hence total current from the cell $i = 3 + 1.5 = 4.5\,A$
By using $E = V + ir$==> $E = 1.5 + 4.5 \times (2.67) = 13.5\,V$
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