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An ideal coil of 10 H is joined in serie...

An ideal coil of 10 H is joined in series with a resistance of `5Omega` and a battery of 5 V. 2 s after joining, the current flowing in ampere in the circuit will be

A

`e^(-1)`

B

`(1-e^(-1))`

C

`(1-e)`

D

`e`

Text Solution

Verified by Experts

The correct Answer is:
B
Given, `L=10H,R=5Omega,E=5V`
and t=2s
current R-L circuit
`I=I_(0)(1-e^(-Rt//L))`
`I=(E)/(R)(1-e^(-Rt//L))`
`=(5)/(5)(1-E^(-5XX2//10))` ltBRgt `I=(1-e^(-1))`
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