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

Capacitance of the capacitor, `C=100muF=100xx10^(-6)F`
Resistance of the resistor, R=40`Omega`
Supply voltage, V=110V
Frequency of the supply. V=12KHz=`12xx10^(3)Hz`
`= omega=2nv=2xxnxx12xx10^(3)03`
=24nxx `10^(3)` rad/s
Peak voltage, `V_(@)=V sqrt(2)=110sqrt(2)V`
`I_(@)=(V_(@))/ (sqrt(R^(2)+(1)/(omega^(2)C^(2)))`
Maximum current.
`=(110sqrt(2))/(sqrt((40)^(2)+(24pixx10^(3)xx1 00 xx10^(-6))^(2))`
For an d RC circuit the voltage lags behind the current by a phase angle of `phi` given as:
`tan phi=((1)/( omegaC))/(R)=(1)/(omegaCR)`
`=(0.2pi)/(180)`rad
`therefore` Time lag `=(phi)/(omega)`
`=(0.2pi)/(180xx24pixx10^(3))=1.55x x10^(-3)s=0.04mus`
Hence `phi` tends to become zero at high frequencies. At a high frequency capacitor C acts as a conductor.
In a dc circuit, after the steady state is acheived `omega=0` . Hence, capacitor C amounts to an open circuit.