Question

A torch bulb rated as $4.5W,1.5V$ is connected as shown in the figure. The e.m.f. of the cell(in Volts) needed to make the bulb glow at full intensity is

Answer 1

Current in the bulb $=\frac{P}{V}=\frac{4.5}{1.5}=3A$
As bulb is connected is parallel to resistance($1\Omega$) Potential difference across resistance($1\Omega$) is $1.5V$

Current in $1\Omega$ resistance =$\frac{1.5}{1}=1.5A$

Hence total current from the cell $i=3+1.5=4.5A$

By using relation between emf and potential difference
$E=V+ir\Rightarrow E=1.5+4.5\times \left(2.67\right)=13.5V$