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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e−1

(1−e−1)

(1−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

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