An A.C. Source of 250 V, 50 Hz is connected to a circuit consisting of an electric lamp rated 100 W, 50 V and a capacitor in series. What should be value of the capacitance of the capacitor to make the lamp work at the rated value?

Explanation

Given : `v_(rms)=250V,f=50Hz,P=100W,v=50` volt.
We know that, `P=(v^(2))/(R)`
`therefore R=(v^(2))/(P)=(50xx50)/(100)=25Omega`
but `P=iv`
i.e., `i=(100)/(50)=2A`
Also `i_(rms)=(v_(rms))/(Z)`,
Hence `2=(250)/(Z)` gives `Z=125Omega`, but `Z^(2)=R^(2)+X_(C)^(2),X_(C)^(2)=125^(2)-25^(2)=15000`
`therefore X_(C)=122.47Omega`
Also, `(1)/(2pifC)=122.47`
or `C=(1)/(2xx3.142xx50xx122.47)=25.987xx10^(-6)F`.