In the figure below, the switches $S_{1} {and S}_{2}$ are closed simultaneously at t = 0 and a current starts to flow in the circuit. Both the batteries have the same magnitude of the electromotive force $(emf)$ and the polarities are as indicated in the figure. Ignore mutual inductance between the inductors . The current $I$ in the middle wire reaches its maximum magnitude $I_{\max}$ at time $t = τ$ .which of the following statements is ( are) true ?

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$I_{max} = \frac{V}{4R}$

$I_{max} = \frac{V}{2R}$

$τ = \frac{2L}{R} In2$

$τ = \frac{L}{R} In2$

answer is B, D.

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Detailed Solution

Here ${I + I}_{2} {= I}_{1} ∴ {I = I}_{1} {- I}_{2}$

$∴ I = \frac{V}{R} [1 - e^{\frac{- R t}{2 L}}] - \frac{V}{R} [1 - e^{\frac{- R t}{L}}]$

$\Rightarrow I = \frac{V}{R} [e^{\frac{- R t}{L}} - e \frac{- R t}{2 L}]$

For I to be maximum , $\frac{d l}{d t} = 0$

$\begin{aligned} ∴ \frac{V}{R} [\frac{- R}{L} e^{\frac{- R t}{L}} - (\frac{- R}{2 L}) e^{\frac{- R t}{2 L}}] = 0 \\ ∴ e \frac{- R I}{2 L} = \frac{1}{2} \Rightarrow (\frac{R}{2 L}) t = In 2 \\ ∴ t = \frac{2 L}{R} In 2 \end{aligned}$

This is the time when I is maximum

Further $I_{M a x} = \frac{V}{R} [e^{\frac{- R}{L} (\frac{2 L}{R} l n 2)} - e^{\frac{- R}{2 L} (\frac{2 L}{R} l n 2)}]$

$\begin{aligned} \Rightarrow I_{M a x} = \frac{V}{R} [\frac{1}{4} - \frac{1}{2}] \\ ∴ I_{M a x} = \frac{V}{R} [\frac{1}{4} - \frac{1}{2}] \\ ∴ I_{M a x} = \frac{V}{4 R} \end{aligned}$

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