An Inductor of 5 mh is Connected in Series

By Varadi Hema|Updated : July 29th, 2022

An inductor of inductance 5 mH is connected in series with 20 resistors and a battery of EMF 2 V, then the current through the circuit 10 msec after the switch is on is _________.
Fill in the blank with the correct answer.

 

Answer: 100 mAmp 

An inductor of inductance 5 mH is connected in series with 20 resistors and a battery of EMF 2 V, then the current through the circuit 10 msec after the switch is on is 32.96 mAmp 

100 mAmp 

Solution:

Given,

Inductance L=5 mH,

Resistance R=20 Ohm,

E=2 V 

At the instant t=0+(the instant just after the switch is closed), the inductor behaves as open circuit,

So, current through circuit I(0+)=I(0)=0 A

At the instant t=∞, inductor will act as short circuit, as the applied voltage is dc,

So current through the circuit I(∞)=E/R=2/20=0.1 A

The complete response of RL circuit I(t)=i(∞)+[i(0)-i(∞)]e-t/τ

Here, time constant τ=L/R=5mH /20 =0.25 msec

So, I(t)=0.1+[0-0.1]e-t/(0.25ms)

Then, current through inductor at t=10 msec is I(10 msec)=0.1+[0-0.1]e-(40)

I(10 msec)=100 mAmp

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