Four identical bulbs are connected in series with a source. These together dissipate a power P. The power dissipated, if the four bulbs are connected in parallel with the same source, will be

16 P

The correct answer is Option (4) → 16 P

Let resistance of each bulb = $R$

Series combination resistance = $4R$

Power in series: $P = \frac{V^2}{4R}$

Parallel combination resistance = $\frac{R}{4}$

Power in parallel: $P' = \frac{V^2}{R/4} = \frac{4V^2}{R}$

Now, $\frac{P'}{P} = \frac{\frac{4V^2}{R}}{\frac{V^2}{4R}} = \frac{4V^2}{R} \cdot \frac{4R}{V^2} = 16$

Therefore, $P' = 16P$

Answer: $16P$