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) what is the time lag between the current maximum and the voltage maximum ?

Explanation

(a) Given , capacitance , C 100 `mu` F = `100 xx 10^(-6) `F
Resistance , R = 40 `Omega`
rms value of voltage , `E_(rms) = 110` V
Peak value of voltage , `E_(0) = sqrt(2) xx E_(rms)`
Frequency , f = 60 Hz
Angular frequency,` omega = 2 pi ` f
Peak value of current is
` I_(0) = (E_(0))/(sqrt(R^(2) + (1)/(omega^(2)C^(2)) ) ) = (sqrt(2) xx 110)/( sqrt(1600 + (1)/(4 pi^(2) xx 3600 xx 10^(-8))))`
`rArr " " I_(0) ` = 3.24 A
(b) Phase angle is given by
` tan phi = (X_(C))/(R) = ((1)/(omega C))/(R) = (1)/(omega C R) = (1)/(2 pi f C R)`
= `(1)/(2 pi xx 60 xx 10^(-4) xx 40 ) = 0.664`
`rArr " " phi = tan^(-1) (0.664 ) = 33.5^(@)`
Instantaneous current in the circuit is
I = `I_(0) cos omega`t
Instantaneous voltage in the circuit is
` E = E_(0) cos (omega t - phi ) ` (Voltage lags behind current in RC ) circuit by phase angle `phi` )
current becomes maximum at t = 0 . similarly,
voltage become maximum at t = `(phi)/(omega)`
`therefore` Time lag = `(phi)/(omega) = (33.5 pi)/(180 xx 2 pi xx 60) ` = 1.55 ms