`n` identical light bulbs, each designed to draw `P` power from a certain voltage supply , are joined in series across that supply. The total power which they will draw is

Options

A. `nP`
B. `P`
C. `(P)/(n)`
D. `(P)/(n^(2))`

Answer

C

Explanation

Let V = voltage of the source,
R = resistance of each bulb.
`therefore R=(V^(2))/(P)`
When n bulbs are joined in series across V, current in each bulb `=I=(V)/(nR)`
Power drawn by each bulb
`=I^(2)R=(V^(2))/(n^(2)R^(2))R=(V^(2))/(n^(2)R)=(P)/(n^(2))`
Total power drawn = `nxx(P)/(n^(2))=(P)/(n)`