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 ) ( 240))/(sqrt((100)^(2) + ( 4) ( 3.14)^(2) ( 10000)^(2) ( 0.5)^(2)))`
`:. I_(m) = 0.011 A ~= 0`
`rArr` This is as good as an open circuit ( because there also current is zero )
(b) From equation (2) of above answer,
`tan phi = (( 2) ( 3.14) ( 10000) (0.5))/(100)`
` = 314`
`rArr tan phi ` is very large.
`rArr phi rarr ( pi )/( 2)` rad
In case of dc LR circuit, once the steady state is reached `I = I_(m) `= constant `rArr (dI)/(dt) = 0`
`rArr` self induced emf` epsilon = - L ( dI)/( dt) = 0`
`rArr` Inductor will act like a pure conducting wire ( without any inductance ) . Hence in this case there will be short circuit ( with minimum resistance ) .