An ideal coil of 10 henry is joined in series with a resistance of 5 ohm and a battery of 5 volt. 2 second after joining, the current flowing in ampere in the circuit will be
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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
From i=i01−e−Rt/L, where i0=55=1amp∴i=11−e−5×210=1−e−1amp