A 100 `muF` capacitor in series with a `40Omega` resistance is connected to a 110V, 60Hz supply.
(a) What is the maximum current in the circuit? (b) What is the time lag between the current maximum and the voltage maximum?

Answer

`3.24A; 1.55xx10^(-3)s`

Explanation

Impedance of circuit is
`Z=sqrt((40)^(2)+(1)/((120 pi xx 100 xx10^(-6))^(2)) =48 Omega`
Maximum current: `I_(0) =(E_0)/(Z) =(110 sqrt(2))/(48) = 3.24 A`
[ we have `E=E_(0) sin omega t and I=(I_0) sin (omega t + phi]`
`tan(phi)=(X_C)/(R )=(1)/(120 pi (100xx10^(-6)40)`
`implies phi =33.5^(@)=0.186 pi`
requried time lag: `(t_0)=(phi)/(omega)=(0.186 pi)/(2 pi 60) =1.55xx10^(-3)s`.