An ideal coil of 10 H is connected in series with a resistance of 5Ω and a battery of 5V. 2 second after the connection is made, the current flowing in ampere in the circuit is
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a
e-1
b
(1-e-1)
c
(1-e)
d
e
answer is B.
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Detailed Solution
We know that i=imax1−e−t/τ Here τ=LR=105=2sec. So, at t=2sec. and imax=VR=55=1 we get i=11−e−2/2=1−e−1