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 d.c. circuit after the steady state ?

Explanation

Here `L = 0.5 H, R = 100 Omega, V_(rms) = 240 V` and v= 10 kHz `=10^(4)` Hz
`therefore omega = 2pi v = 2 xx 3.14 xx 10^(4) = 6.28 xx 10^(4)s^(-1)`
(a) `therefore` Impedance `Z = sqrt(R^(2) + (Lomega)^(2)) = sqrt((100)^(2) + (0.5 xx 6.28 xx 10^(4))^(2)) = 3.14 xx 10^(4) Omega`
`therefore I_(m) = V_(m)/Z =(sqrt(2)V_(rms))/Z = sqrt(2) xx 240 = 1.1 xx 10^(-2) A`
(b) `phi = tan^(-1) (OLomega)/R = tan^(-1) (0.5 xx 6.28 xx 10^(4))/100 = tan^(-1)( 31.4) = pi/2` rad
`therefore t = phi/ (2pi v) =pi/(2 xx 2pi xx 6.28 xx 10^(4)) = 4.0 xx 10^(-6)` s.
From these calculations, it is clear that for a high frequency, current is very very small i.e., at high frequencies the inductor behaves as if the circuit is an open circuit or its impedance is extremely high.
An inductor offers zero resistance in a d.c. circuit after the steady state is reached.