Alternating Current

JEE Main level, single correct, medium

132 of 190

Question 132 of 190

Consider the LR circuit shown in the figure. If the switch S is closed at t=0t=0 then the amount of charge that passes through the battery between t=0t=0 and t=LRt=\dfrac{L}{R} is:

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Answer: (D) EL2.7R2\dfrac{ EL}{2.7{R}^{2}}

The current in the LR circuit at any time tt is
I=ER(1−e−RtL)I=\dfrac{E}{R}\left(1-{e}^{-\dfrac{Rt}{L}}\right)
Where E,R,L,IE,R,L,I are the EMF, resistance, inductance and the current respectively.
∴∴ Charge flown through the circuit from time 0 to LR\dfrac{L}{R} is
(charge) ∆q=∫0LRI.dt=∫0LRER(1−e−RtL)dt∆q=\int _{0}^{\dfrac{L}{R}}I.dt=\int _{0}^{\dfrac{L}{R}}\dfrac{E}{R}\left(1-{e}^{-\dfrac{Rt}{L}}\right)dt
⇒∆q=ER[t+LRe−RtL]0LR\Rightarrow ∆q=\dfrac{E}{R}{\left[t+\dfrac{L}{R}{e}^{-\dfrac{Rt}{L}}\right]}_{0}^{\dfrac{L}{R}}
⇒∆q=ER[LR+LRe−1−LR]\Rightarrow ∆q=\dfrac{E}{R}\left[\dfrac{L}{R}+\dfrac{L}{R}{e}^{-1}-\dfrac{L}{R}\right]
⇒∆q=ER×LeR\Rightarrow ∆q=\dfrac{E}{R}\times \dfrac{L}{eR}
⇒∆q=ELeR2=EL2.7R2\Rightarrow ∆q=\dfrac{EL}{e{R}^{2}}=\dfrac{EL}{2.7{R}^{2}}

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