Obtain the answers (a) to (b) in Exercise 7.13 if the circuit is connected to a high frequency supply
Question:

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?

Solution:

Inductance of the inductor, L = 0.5 Hz

Resistance of the resistor, R = 100 Ω

Potential of the supply voltages, V = 240 V

Frequency of the supply,ν = 10 kHz = 104 Hz

Angular frequency, ω = 2πν= 2π × 104 rad/s

(a) Peak voltage, $V_{0}=\sqrt{2} \times V=240 \sqrt{2} \mathrm{~V}$

Maximum current, $I_{0}=\frac{V_{0}}{\sqrt{R^{2}+\omega^{2} L^{2}}}$

$=\frac{240 \sqrt{2}}{\sqrt{(100)^{2}+\left(2 \pi \times 10^{4}\right)^{2} \times(0.50)^{2}}}=1.1 \times 10^{-2} \mathrm{~A}$

(b) For phase differenceΦ, we have the relation:

$\tan \phi=\frac{\omega L}{R}$

$=\frac{2 \pi \times 10^{4} \times 0.5}{100}=100 \pi$

$\phi=89.82^{\circ}=\frac{89.82 \pi}{180} \mathrm{rad}$

$\omega t=\frac{89.82 \pi}{180}$

$t=\frac{89.82 \pi}{180 \times 2 \pi \times 10^{4}}=25 \mu \mathrm{s}$

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.