Question

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 ?

Answer 1

Here, `v = 10 kHz = 10^(4) Hz. Omega = 2 pi v = 2 pi xx 10^(4) rad//s`.
`I_(0) = (E_(0))/(sqrt(R^(2) + omega^(2) L^(2))) = (sqrt2 xx 240)/(sqrt(10^(4) + 4 pi^(2) xx 0.5^(2))) = 1.08 xx 10^(-2)` amp.
Here, contribution of term containing R is negligible
`tan phi = (omega L)/(R) = (2 pi xx 10^(4) xx 0.5)/(100) = 100 pi`, which is very large `:. phi rarr` tends to `90^(@)` or `(pi)/(2)` radian
On comparision, we find that at low frequencies, `I_(0) = 1.82` amp.
at high frequencies, `I_(0) = 0.0108` amp.
`:.` At high frequencies, L offers very high resistance which amounts to an open circuit (infinite resistance). In a d.c. circuit, after steady state `omega = 0`. Therefore, `X_(L) = omega L = 0`. Hence L acts like a pure conductor of negligible inductive reactance. It has no role to play.