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Byju's Answer

Standard XII

Physics

RLC Circuit

In the given ...

Question

In the given circuit, initially switch $S_{1}$ is closed, and $S_{2}$ and $S_{3}$ are open. After charging of capacitor, at $t = 0, S_{1}$ is opened and $S_{2}$ and $S_{3}$ are closed. If the relation between inductance, capacitance and resistance is $L = 4 C R^{2}$ , then find the time (in s) after which current passing through capacitor and inductor will be same (given $R = l n 2 m Ω, L = 2 m H)$ :

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Solution

The correct option is A 1

Current through $S_{2}$ loop is $\frac{ε}{2 R} e^{\frac{- t}{2 C R}}$
current through $S_{3}$ loop is $\frac{ε}{2 R} (1 - e^{\frac{- t}{L / 2 R}})$

$\Rightarrow e^{\frac{- t}{2 C R}} = 1 - e^{\frac{- t}{L / 2 R}}$

$\Rightarrow e^{\frac{- t}{2 C R}} = 1 - e^{\frac{- t}{4 C R^{2} / 2 R}}$

$\Rightarrow 2 e^{\frac{- t}{2 C R}} = 1$

$\Rightarrow e^{\frac{- t}{2 C R}} = 1 / 2$

$\Rightarrow \frac{- t}{2 C R} = l n (1 / 2)$

$\Rightarrow t = 2 C R l n (2) = \frac{L}{2 R} l n (2) = 1 s$

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The correct option is A 1 Current through S2 loop is ε2Re−t2CRcurrent through S3 loop is ε2R(1−e−tL/2R) ⇒e−t2CR=1−e−tL/2R ⇒e−t2CR=1−e−t4CR2/2R ⇒2e−t2CR=1 ⇒e−t2CR=1/2 ⇒−t2CR=ln(1/2) ⇒t=2CRln(2)=L2Rln(2)=1s