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Obtain the answers (a) to (b) in Exercise 7 .13 if the circuit is connected to a high frequency supply( 240 V, 10 kHz).

Q7.14 Obtain 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?

Answers (1)


The inductance of the coil L=50H

the resistance of the coil R=100\Omega

Supply voltage V=240V

Supply voltage frequencyf=10kHz


Now, as we know peak voltage = \sqrt2(RMS Voltage)

Peak voltage V_{peak}=\sqrt2*240=339.4V


The impedance of the circuit :

Z=\sqrt{R^2+(wL)^2}=\sqrt{100^2+(2\pi 10*10^3 *50)^2}

Now peak current in the circuit :

I_{peak}=\frac{V_{peak}}{Z}=\frac{339}{\sqrt{100^2+(2\pi 10*10^3 *50)^2}}=1.1*10^{-2}A

Hence peak current is 1.1*10^{-2}A 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 \phi  we have 

tan\phi =\frac{wL}{R}=\frac{2\pi *10*10^3*0.5}{100}=100\pi

\phi =89.82^0


t=\frac{\phi}{w}=\frac{89.82*\pi}{180*2\pi *10^3}=25\mu s

Hence time lag between the maximum voltage and the maximum current is 25\mu s.

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