Obtain the answers (a) to (b) in Exercise 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 ?

Explanation

(a) From equation (1) of above answer,
`I_(m) = ( ( 1.414 ) ( 110))/( sqrt(( 40)^(2) + ( 1)/( ( 4) (3.14 )^(2) ( 12000)^(2) ( 100 xx 10^(-6) )^(2))))`
`:. I_(m) = 3.89 A ` ...(1)
Now, in the absence of capacitor,
`I_(m)^(.) = ( V_(m))/( R ) = ( sqrt( 2) V_(rms))/( R )`
`:. I._(m) = (( 1.414) ( 110))/( ( 40))`
`:. I._(m) = 3.8885A` ...(2)
`rArr` Equation (1) and ( 2) indicate that
`I._(m) ~~ I_(m)`
`rArr` At very high frequency of ac, capacitor acts like a simple conductor and hence the circuit becomes short circuit with minimum resistance.
For dc supply, `f = 0 rarr X_(C ) = (1)/( 2 pi f C ) =oo`
`rArr` Capacitor in a dc circuit will make it open circuit with infinite resistance ( after the steady state is reached ) .