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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 ? |
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Answer» Solution :(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 ) . |
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