In the figure below, the switches S1 and S2 are closed simultaneously at...

In the figure below, the switches $S_{1} a n d S_{2}$ 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 $I_{m a x}$ time $t = τ$ . Which of the following statements is (are) true?

I_{m a x} = \frac{V}{2 R}

I_{m a x} = \frac{V}{4 R}

τ = \frac{L}{R} \ln 2

τ = \frac{2 L}{R} l \ln 2

Solution:

The current in the middle, $I$ , as shown in the figure above has a maximum value,
$i_{m a x} = {(i_{2} - i_{1})}_{m a x}$

Using the equation of current decay in an $L - R$ circuit, for the given values in the figure above,
$i = (i_{2} - i_{1}) = \frac{V}{R} [1 - e^{- (\frac{R}{2 L}) t}] - \frac{V}{R} [1 - e^{(- \frac{R}{L}) t}]$
$\frac{V}{R} [e^{(- \frac{R}{L}) t} - e^{(- \frac{R}{2 L}) t}]$
For ${(i)}_{m a x}, \frac{d (i)}{d t} = 0$
$\Rightarrow \frac{V}{R} [- \frac{R}{L} e^{- (\frac{R}{L}) t} - (- \frac{R}{2 L}) e^{- (\frac{R}{2 L}) t}] = 0$
$\Rightarrow e^{- (\frac{R}{L}) t} = \frac{1}{2} e^{- (\frac{R}{2 L}) t}$
$\Rightarrow e^{- (\frac{R}{2 L}) t} = \frac{1}{2}$
$\Rightarrow (\frac{R}{2 L}) t = l n 2$
$t = \frac{2 L}{R} l n 2 \to$ time when I is maximum
$i_{m a x} = \frac{V}{R} [e^{- \frac{R}{L} (\frac{2 L}{R} l n 2)} - e^{- (\frac{R}{2 L}) (\frac{2 L}{R} l n 2)}]$
$\Rightarrow |i_{m a x}| = \frac{V}{R} |[\frac{1}{4} - \frac{1}{2}]| = \frac{1}{4} \frac{V}{R}$

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In the figure below, the switches S1 and S2 are closed simultaneously at...

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