# Q7.14 Obtain the answers (a) to (b) in Exercise 7.13 if the circuit is connected to a high-frequency supply . 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?

Given,

The inductance of the coil

the resistance of the coil

Supply voltage

Supply voltage frequency

a)

Now, as we know peak voltage = (RMS Voltage)

Peak voltage

Now,

The impedance of the circuit :

Now peak current in the circuit :

Hence peak current is  in the circuit.

The current in the circuit is very small, which is one of the indications of inductor working as a nearly open circuit in the case of high frequency.

b)

For phase difference   we have

Now

Hence time lag between the maximum voltage and the maximum current is .

In the DC circuit, after attaining the steady state, inductor behaves line short circuit as  is Zero.

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