question_answer 14)

Obtain the answer (a) to (b) in Exercise 7.13 if the 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 supply state? 

Answer:

For the given high frequency,                 Now                                                                                 i.e., is closed to, then time lag                                 in this case is too small, so it can be concluded that at high frequencies an inductance behaves as on open circuit.                 In a steady d.c. circuit v = 0, so inductance acts as a simple conductor.