A) \[nP\]
B) \[P\]
C) \[\frac{P}{n}\]
D) \[\frac{P}{{{n}^{2}}}\]
Correct Answer: C
Solution :
Let \[V=\]voltage of the source, R = resistance of each bulb. \[\therefore \]\[R={{V}^{2}}/P.\] When n bulbs are joined in series across F, current in each bulb\[=i=\frac{V}{nR}.\] Power drawn by each bulb\[={{i}^{2}}R=\frac{{{V}^{2}}}{{{n}^{2}}{{R}^{2}}}.R\frac{{{V}^{2}}}{{{n}^{2}}R}=\frac{P}{{{n}^{2}}}.\] Total power drawn \[=n\times P/{{n}^{2}}=P/n.\] Hence, the correction option is (c).
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