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 :

Capacitor C = 100μF = 100 x 10⁻⁶ F

Resistor R = 40 ohms

Voltage V = 110 V

Frequency v = 12 KHz = 12 x 10³ Hz

Angular frequency ω = 2πv = 2 x π x 12 x 10³= 24 π x 10³ rad/s

Peak voltage V0 = V√2 = 110√2

Maximum current  I0 = V0/√ R2 + 1/ω²L²  

                                      = 110√2/ √(40)² + 1/ (24 π x 10³ x 100 x 10⁻⁶ )²

=3.9 A

For an RC circuit, the voltage lags behind the current by a phase angle

tan ɸ = 1/ωCR = 1/  24

π x 10³ x 100 x 10⁻⁶ x 40

= 1/96 π

ɸ = 0.2 ° = 0.2 π /180 rad

Time lag=t = ɸ/ω =  0.2 π /180 x 24 π x 10³ = 1.55 x 10⁻³ s = 0.04 μs

Its clear that  ɸ tends to zero at high frequencies. So capacitor C acts as conductor.

In Dc circuit, after the steady state is achieved ω=0 So, Capacitor C amounts to an open circuit.

Answer 2

Answer:

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