Answer
(b) 4
Work Step by Step
let $R $be the resistance of bulb.
Equivalent resistance in circuit A
$ \frac{1}{R_{eqA} }= \frac {1}{R}+\frac {1}{R}$
$ \implies R_{eqA} = \frac{R}{2}$
Equivalent resistance in circuit A
$R_{eqB} = R +R = 2R$
We have
$ P = \frac{V^2}{R}$
$\therefore \frac{P_A }{P_B } = \frac{ V^2}{R_{eqA }} * \frac{R_{eqB} }{V^2 } =\frac{R_{eqB} }{R_{eqA }} = \frac{ 2R}{ \frac{R}{2}} = 4$
