NEET-XII-Physics
07: Alternating Current
- #14Obtain the answers (a) to (b) in Exercise 7.13 if the circuit is connected to a high frequency supply (240 V, 10 kHz). Hence, explain the statement that at very high frequency, an inductor in a circuit nearly amounts to an open circuit. How does an inductor behave in a dc circuit after the steady state?
Ans : Inductance of the inductor, L = 0.5 Hz
Resistance of the resistor, R = 100 ``\Omega``
Potential of the supply voltages, V = 240 V
Frequency of the supply,ν = 10 kHz = 104 Hz
Angular frequency, ω = 2``\pi``ν= 2``\pi`` × 104 rad/s
(a) Peak voltage,
Maximum current,
(b) For phase differenceΦ, we have the relation:
It can be observed that I0 is very small in this case. Hence, at high frequencies, the inductor amounts to an open circuit.
In a dc circuit, after a steady state is achieved, ω = 0. Hence, inductor L behaves like a pure conducting object.