Three bulbs of `40 W , 60 W and 100 W` are connected in series with a `240 V` source.

Options

A. The potential difference will be maximum across the `40 W` bulb .
B. The current difference will be maximum in `100 W` bulb.
C. The resistance of the `40 W` bulb is minimum.
D. The current through the `60 W` bulb will be `0.1 A`.

Answer

A

Explanation

Since all the three resistors are in series , the same current will flow through them . Their resistances are given by `R = V^(2)//P`.
`R_(40) = ( 240 xx 240 )/( 40) = 1440 Omega`
`R_( 60) = 960 Omega and R_(100) = 576 Omega`
Total resistance is `R = 2976 Omega`
`I = ( 240)/( 2976) = 0.0806 A`
Potential difference across the `40 W` bulb is
`1440 xx 0.0806 = 116 V`