Electromagnetic Induction

JEE Main level, single correct, medium

110 of 153

Question 110 of 153

A coil of self inductance 10mH10mH and resistance of 0.1Ω0.1\Omega is connected through a switch to a battery of internal resistance 0.9Ω0.9\Omega . After the switch is closed, the time taken for the current to attain 8080% of the saturation value is: [ln⁡5=1.6][\ln 5=1.6]

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Answer: (D) 0.016s0.016s

The instantaneous current in an LR circuit is given by the equation :
(instantaneous current) I=Is(1−e−RtL)I={I}_{s}\left(1-{e}^{-\dfrac{Rt}{L}}\right)
Where I,Is,R,t,LI,{I}_{s},R,t,L are the instantaneous current, saturation current, resistance, time and inductance respectively.
Given that the current becomes so percent of saturation current.
⇒(0.8)Is=Is(1−e−RtL)\Rightarrow \left(0.8\right){I}_{s}={I}_{s}\left(1-{e}^{-\dfrac{Rt}{L}}\right)
R=0.1Ω+0.9Ω=1ΩR=0.1\text{Ω}+0.9\text{Ω}=1\text{Ω}
⇒0.8=(1−e−1×t10×10−3)\Rightarrow 0.8=\left(1-{e}^{-\dfrac{1\times t}{10\times {10}^{-3}}}\right)
⇒0.8=1−e−100t\Rightarrow 0.8=1-{e}^{-100t}
Therefore, the time is:
t=ln⁡(11−0.8)100=ln⁡⁡(5)100=0.016st=\dfrac{ln⁡\left(\dfrac{1}{1-0.8}\right)}{100}=\dfrac{\ln ⁡\left(5\right)}{100}=0.016s

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