A tube light of `60 V, 60 W` rating is connected across an `AC` source of `100 V` and `50 Hz` frequency. Then,

Options

A. an inductance of `2/(5pi)` may be connected in series
B. a capacitor of `250/pimuF` may be connected in series to it
C. an inductor of `4/(5pi)H` may be connected in series
D. a resistance of 40 `Omega` may be connected in series

Answer

C, D

Explanation

`P_R=V_Ri`
`:. i=P_R/V_R=60/60=1A`
`Now, V_L=sqrt(V^2-V_R^2)`
`=sqrt((10)^2-(60)^2)`
`=80V=iX_L=i(2pifL)`
`L=80/(2pift)`
`=80/((2pi)(50)(1))=8/(5pi)H`
If we connect another resistance `R` is series, then it should consume `40 V`, so that remaining` 60 V` is used by the light.
`R=V/i=40/1=40Omega`