A bulb is rated at `100V`, `100W`. It can be treated as a resistor. Find out the inductance of an inductor (called choke coil) that should be connected in series with the bulb at its rated power with the help of an ac source of `200V` and `50Hz`.

Options

A. `(pi)/(sqrt(3)) H`
B. `100 H`
C. `(sqrt(2))/(pi) H`
D. `(sqrt(3))/(pi) H`

Answer

4

Explanation

Resistance of bulb is `R = (100 xx 100)/(100) = Omega`
Rated current is `(100)/(100) = 1A`
In ac, `I_("rms") = (V_("rms"))/(Z) , Z = 200 Omega`
`sqrt(100^(2) + (omega L)^(2)) = 200 implies omega^(2) L^(2) = 30000` and
`L = sqrt((30000)/((100 pi)^(2))) = (sqrt(3))/(pi)` henry.