Question

Obtain the answers to (a) and (b) in Exercise 7.15 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 dc circuit after the steady state.

Answer 1

Given: The capacitance is 100 μF, the resistance is 40 Ω, the applied potential difference is 110 V and the frequency of signal is 12 kHz.

The peak voltage is given as,

V 0 =V 2

Where, the applied potential difference is V.

By substituting the given values in the above equation, we get

V 0 =110 2  V

The angular frequency is given as,

ω=2πν

Where, the frequency of the signal is ν.

By substituting the given values in the given equation, we get

ω=2π( 12× 10 3 ) =24π× 10 3   rad/s

The maximum current is given as,

I 0 = V 0 ( R 2 + 1 ω 2 C 2 )

By substituting the given values in the above equation, we get

I 0 = 110 2 40 2 + 1 ( 24π× 10 3 ×100× 10 −6 ) 2 = 110 2 1600+ ( 10 24π ) 2 = 110 2 1600.01 =3.9 A

The phase angle is given as,

tanϕ= 1 ωCR

By substituting the given values in the above equation, we get

tanϕ= 1 24π× 10 3 ×100× 10 −6 ×40 tanϕ= 1 96π ϕ= tan −1 ( 1 96π ) =0.2°

The phase angle is,

ϕ= 0.2π 180  rad

The time lag is given as,

Time lag= ϕ ω

By substituting the values in the above equation, we get

Time lag= 0.2π 180×24π× 10 3 =0.04 μs

Thus, the phase angle is very low and tends to become zero at high frequencies. So at a high frequency capacitor acts as a conductor and in a dc circuit after the steady state is achieved ω=0 so the capacitor will behave as an open circuit.