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

a) Peak valueof voltage `E_(0) = sqrt(2)E_(rms) = sqrt(2) xx 110`V
Let `I_(0)`=maximum current in the circuit
`therefore I_(0) = E_(0)/Z = (E_(0))/sqrt(R^(2)+(1/(omegaC)^(2))^(2))`
or `155.1/(sqrt(1600 + 703.84))=3.23`A
b) In an R-C circuit, the voltage lags behind hte current by the phase anlge `phi` is given be
`tanphi = (X_(C)/R) = 1/(omegaCR) = 1/(120pi xx 10^(-4) xx 40)`
`=0.6628 = tan 33.5^(@)`
`therefore phi = 33.5^(@) = 33.5 xx pi/180 rad`
`therefore` Time between the current maximum and the voltage maximum is given by
`t= phi/omega = ((33.5pi)/(180))/((120 pi))`
`=1.55 xx 10^(-3)`s
=1.55 ms