NEET-XII-Physics
07: Alternating Current
- #16Obtain 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.
Ans : Capacitance of the capacitor, C = 100 μF = 100 × 10-6 F
Resistance of the resistor, R = 40 ``\Omega``
Supply voltage, V = 110 V
Frequency of the supply, ν = 12 kHz = 12 × 103 Hz
Angular Frequency, ω = 2 ``\pi``ν= 2 × ``\pi`` × 12 × 10303
= 24``\pi`` × 103 rad/s
Peak voltage,
Maximum current,
For an RC circuit, the voltage lags behind the current by a phase angle of Φ given as:
Hence, Φ 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 achieved, ω = 0. Hence, capacitor C amounts to an open circuit.