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?
The inductance of the coil
the resistance of the coil
Supply voltage frequency
Now, as we know peak voltage = (RMS Voltage)
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.
For phase difference we have
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.