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

Here,`v = 12 kHz = 12 xx 10^(3) Hz, omega = 2 pi v = 24 pi xx 10^(3) rad//sec`.
`I_(0) = (E_(0))/(sqrt(R^(2) + (1)/(omega^(2) C^(2)))) = (sqrt(2) xx 110)/(sqrt(1600 + ((1)/((24 pi xx 10^(3) xx 10^(-4))^(2))))) =( sqrt2 xx 110)/(sqrt(1600 + 0.017)) = 3.9 A`
Here, contribuiton of term containing C is negligilbe.
In the RC circuit, voltage lags behind the current by a phase angle `phi`, where
`tan = phi = (1//omegaC)/(R ) = (1)/(omega CR) = (1)/(24 pi xx 10^(3) xx 10^(-4) xx 40) = (1)/(96 pi)`, Which is very small
`:. phi` tends to zero at high frequencies.
Hence at very high frequencies, capacitor acts like a pure conductor of negligilbe capacitive reactace. In d.c. circuits (after the steady state), `omega = 0. :. X_(C ) = (1)/(omega C) rarr oo`
Therefore, behaviour of capacitor towards d.c. amounts to an open circuit.