if the circuit is connected to a 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 a de circuit after the steady state.

Explanation

`I_(max) = (E_(max) )/(sqrt(R^2 + X_C^2) )`
`E_(max) = 110 sqrt2 , R = 40 Omega , v = 12 kHz, C = 100 mu F`
`I_(max) = 3.88A`
b. `tan phi = (-1)/(omega CR) = - 0.2` or `phi` is nearly zero at high frequency .C. term being negligible at high frequency ,it acts like resistor .
For stready dc, `v = 0` and `X_C = oo` . so it acts like an open circuit for dc.