Electrical Engineering - Time Response of Reactive Circuits - Discussion

It has a square wave as o/p at the top it will get full vtg. At the low, it will get 0.

I cannot understand this. Please explain.

It is clear the time constant=L/R=2 microseconds and Time period=1/f=10micro seconds as the time period is 5 times of time constant say 5 torr.

So in this time period, the inductor is fully charged when pulse ends the inductor is fully charged and the voltage across inductor will be zero because inductor is a device which stores energy in terms of current. At the start the inductor behaves like open and when fully charged it behaves like short so that the voltage will be zero.

I hope you will be understand.

@Anil is right. And Jash if it would have been an open circuit then the output voltage would have been Vm (i.e. 6V ).

Inductor initially acts as a open circuit. After some time (steady state) it will act as a short circuit, that's why voltage across it zero.

Please explain with equation. How it becomes zero?

Time constant = L/R = 400/200 = 2 microseconds.

Time period = 1/F = 1/100 = 10 microseconds.

It is a open circuit thereby the output becomes zero.

How did you say that time constant is 2 microseconds and 10 micro seconds?

The time constant of the circuit is 2 microseconds and the time period of the pulse is 10 microseconds. Therefore the output will become zero at the end of each pulse.