Obtain the answers to (a) and (b) Q.13, if the circuit is connected to a high frequency supply (240 V , 10 kHz). Hence explain statement that at very high frequency. Inductor in circuit nearly amount to open circuit. How does an indcutor behave in a d.c. circuit after the steady state ?

Explanation

Given,
Inductance, L = 0.50 H
Resistance, R = 100 `Omega`
rms value of voltage, `E_(rms) = 240` V
Frequency, `omega = 2 pi ` F ` = 2 pi xx 10^(4) "rad" s^(-1)`
Peak value of current is
`I_(0) = (sqrt(2) xx 240)/(sqrt(10^(4) + (0.50)^(2) xx 4 pi^(2) xx 10^(8) )) = 0.01 ` A
In the above equation, the magnitude of R is negligible
`tan phi = (X_(L))/(R) = (2 pi fL)/(R) = (2 pi xx 10^(4) xx 0.5)/(100) = 100 pi`
`phi = 89^(@) 48`. which is close to `pi`/2 `(i.e., 90^(@))`
When f = 50 Hz, `I_(0) = 1.8` A
When f = 10,000 Hz, `I_(0) ` = 0.01 A
showing there by that at high frequencies, reactance offered by become very large and nearly amounts to an open circuit. In a DC circuit (after steady state) ,
`omega = 0 `
`therefore " " X_(L) = 2 omega `L = 0
Thus, the inductor with inductance L acts like a pure conductor.