A `100(mu)F` capacitor in series with a `40 Omega` resistor is connected to a 110V, 60 HZ supply. What is the time lag between current maximum and voltage maximum?

A `100(mu)F` capacitor in series with a `40 Omega` resistor is connect

1.	A `100(mu)F` capacitor in series with a `40 Omega` resistor is connected to a 110V, 60 HZ supply. What is the time lag between current maximum and voltage maximum?
Answer» Correct Answer - `3.24A; 1.55xx10^(-3)s` 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`.

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A `100(mu)F` capacitor in series with a `40 Omega` resistor is connected to a 110V, 60 HZ supply. What is the time lag between current maximum and voltage maximum?

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