An inductor (L =100mH ), a resistor (R = 100Ω) and a battery (E = 100 V) are initially connected in series as shown in the figure. After a long time the battery is disconnected after short circuiting the points A and B. The current in the circuit 1 ms after the short circuit is:

Question

An inductor (L =100mH ), a resistor (R = 100Ω) and a battery (E = 100 V) are initially connected in series as shown in the figure. After a long time the battery is disconnected after short circuiting the points A and B. The current in the circuit 1 ms after the short circuit is: (A) 1/e A (B) e A (C) 0.1 A (D) 1 A

Tardigrade Admin · Accepted Answer

This is a combined example of growth and decay of current in an LR circuit.
The current through circuit just before shorting the battery,

I_{0} === E / R = 1 A

[as inductor would be shorted in steady state] After this decay of current starts in the circuit according to the equation

I = I_{0} e^{- t / τ}

where

t = L / R .

I = 1 \times e^{- (1 \times 1 0^{- 3}) / (100 \times 1 0^{- 3} 100)}

= (1/ e) A

Q. An inductor (L =100mH ), a resistor (R = 100Ω) and a battery (E = 100 V) are initially connected in series as shown in the figure. After a long time the battery is disconnected after short circuiting the points A and B. The current in the circuit 1 ms after the short circuit is:

1/e A

e A

0.1 A

1 A

Q. An inductor (L =100mH ), a resistor (R = 100 $Ω$ ) and a battery (E = 100 V) are initially connected in series as shown in the figure. After a long time the battery is disconnected after short circuiting the points A and B. The current in the circuit 1 ms after the short circuit is: