Obtain the answers to (a) and (b) in Q .15, if the circuit is connected to 110 V, 12 kHz supply. Hence explain the statement that a capacitor is a conductor at very high frequencies. Compare this behaviour with that of a capacitor in d.c. after the steady state.

Explanation

(a) In this case,
`E_(rms) = 110` V
`E_(0) = sqrt(2) xx 100` V
f = 12 kHz = 12`xx 10^(3)` Hz
`omega = 2 pi xx 12 xx 10^(3) `rad/s
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 144 xx 10^(6) xx 10^(-8))))`
(b) Phse angle, tan` phi = (1)/(2 pi "fCR") = (1)/(2 pi xx 12 xx 10^(3) xx 10^(-4) xx 40)`
`rArr " " phi = tan^(-1) (0.0033) = 0.189 = 0.2 `
This is very small, thus, we can say that `phi` tends to zero for higher frequencies.
we see that at high frequency, capacitor of capacitance C acts like a conductor as reactance
offered by capacitor, `X_(c) = (1)/(omega C)` tends to zero. for a DC circuit after steady state. i.e., `omega` = 0 ,
capacitor behaves as an open circuit as it offers infinite resistance.