A 50 W, 100V lamp is to be connected to an ac mains of 200V, 50 Hz. What capacitor is essential to be put in series with the lamp?

Explanation

As resistance of the lamp `R=V_(s)^(2)/W=100^(2)/50=200Omega`
and the maximum current `i=V/R=100/200=1/2A,`
so when the lamp is put in series with a capacitance
and run at 200 V ac, from V =iZ we have,
`Z=V/i=200/((1//2))=400Omega`
Now as in case of C-R circuit, `Z=sqrt(R^(2)+(1/(omegaC))^(2)),`
i.e., `R^(2)+(1/(omegaC))^(2)=160000`
or, `(1/(omegaC))^(2)=16 times 10^(4)-(200)^(2)=12 times 10^(4)`
So, `1/(omegaC)=sqrt(12) times 10^(2)`
or `C=1/(100pi times sqrt(12) times 10^(2))F`
i.e., `C=100/(pisqrt(12))muF=9.2muF`