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

Inductance of the inductor,L=0.4Hz
Resistance of the resistor, R=100`Omega`
Potential of the supply voltages, V=240V
Frequency of the supply v=10KHz=`10^(4)Hz`
Angular frequ ency, `omega=2nv=2nxx10^(4)` rad/s
(a) Peak voltage, `V_(@)=sqrt(2)xxV=240sqrt(2)V `
Maximum currnet, `I_(@)=(V_(@))/sqrt(R^(2)+omega^(2)L^(2))`
`=(240sqrt(2))/(sqrt((100)^(2)+(2pixx10^(4))^(2)xx(0.50)^(2))=1.1xx10^(-2)A`
(b) For phase difference `phi`, we have the relation:
`tan phi=(omegaL)/(R)`
`-(2pixx10^(4)xx0.5)/(100)=100pi`
`phi=89.82^(@)=(89.82phi)/(180)` rad
`omegat=(89.82pi)/(180)`
`t=(89.82pi)/(180xx2pixx10^(4))=25mus`
It can be observed that `I_(@)` is very small in this case, Hence, at the fr equencies, the inductor amounts to an open circuit.
In a dc circuit, after a steadyy state is achieved, `omega=0` hence, inductor bhehave like a pure conducting object.