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
Answer & explanation
Correct answer: option 4
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: $16P$