MENU
Question -

Obtain the answers (a) to (b) inExercise 7.13 if the circuit is connected to a high frequency supply (240 V, 10kHz). Hence, explain the statement that at very high frequency, an inductor ina circuit nearly amounts to an open circuit. How does an inductor behave in adc circuit after the steady state?



Answer -

Inductance of the inductor,┬аL┬а=0.5 Hz

Resistance of the resistor,┬аR┬а=100 ╬й

Potential of the supplyvoltages,┬аV┬а= 240 V

Frequency of the supply,╬╜┬а=┬а10 kHz = 104┬аHz

Angular frequency,┬а╧Й┬а=2╧А╬╜= 2╧А ├Ч 104┬аrad/s

(a)┬аPeak voltage,┬а
Maximum current,┬а

(b)┬аFor phasedifference╬ж, we have the relation:

It can be observed that┬аI0┬аisvery small in this case. Hence, at high frequencies, the inductor amounts to anopen circuit.

In a dc circuit, after a steadystate is achieved,┬а╧Й┬а= 0. Hence, inductor L behaves like apure conducting object.

Comment(S)

Show all Coment

Leave a Comment

Free - Previous Years Question Papers
Any questions? Ask us!
×