Question

A `10 mu F` capacitor in series with a `40 Omega` resistance is connected to a 110 V, 60Hz supply. (a) What is the maximum current in the circuit ? (b) What is the time lag between current maximum and voltage maximum ?

Answer 1

Here, `C = 10 mu F = 100 cc 10^(-6) F = 10^(-4) F, R = 40 Omega, E_(v) = 110` volt, `E_(0) = sqrt2 E_(v) = sqrt2 xx 110 V`
`v = 60 Hz. Omega = 2 pi v = 120 pi rad//s, I_(0) = ?`
In RC circuit, as `Z = sqrt(R^(2) + X_(C )^(2)) = sqrt(R^(2) + (1)/(omega^(2) C^(2))`
`:. I_(0) = (E_(0))/(sqrt(R^(2) + (1)/(omega^(2) C^(2)))) = (sqrt2 xx 110)/(sqrt(1600 + (1)/((120 pi xx 10^(-4))^(2)))) = 3.24` amp
In RC circuit, voltage lags behind the current by phase angle `phi`.
When `tan phi = (1//omega C)/(R ) = (1)/(omega CR) = (1)/(120 pi xx 10^(-4) xx 40) = 0.6628`
`phi = tan^(-1) (0.6628) = 33.5^(@) = (33.5 pi)/(180)` rad.
Time lag `= (phi)/(omega) = (33.5 pi)/(180 xx 120 pi) = 1.55 xx 10^(-3)` sec