An inductor (L=100mH), a resistor (R=100Ω) and a battery (E=100V) 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 circuit is
(A)1 A
(B)1/eA✓ Correct
(C)e A
(D)0.1 A
Answer: (B)
Solution
After the source has been connected a long time the inductor carries the steady current I0=E/R=100/100=1A. When the battery is removed and A\text{--}B are shorted, the L\text{--}R loop decays as I=I0e−Rt/L. The time constant is L/R=100Ω100mH=1ms, so after exactly 1ms the current has fallen to I0e−1=e1A.
JEE Main · 2006 · 30 Apr 2006 · Offline · Single correct option