In the figure below, the switches and are closed simultaneously at 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. [2018]

Question

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 [2018]

Smart Achievers · Accepted Answer

(2, 4)

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

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

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

For $I_{\max}$ , $\frac{d I}{d t} = 0$

$∴$ $\frac{V}{R} [- \frac{R}{L} e^{- \frac{R t}{L}} - (- \frac{R}{2 L}) e^{- \frac{R t}{2 L}}] = 0$

$∴$ $e^{- \frac{R t}{2 L}} = \frac{1}{2}$ $\Rightarrow (\frac{R}{2 L}) t = \ln 2$ $∴$ $t = \frac{2 L}{R} \ln 2$

This is the time when $I$ is maximum.

Putting this value of time in Eq.(i),

Further $I_{\max} = \frac{V}{R} [e^{- \frac{R}{L} (\frac{2 L}{R} \ln 2)} - e^{- \frac{R}{2 L} (\frac{2 L}{R} \ln 2)}]$

$\Rightarrow I_{\max} = \frac{V}{R} [\frac{1}{4} - \frac{1}{2}]$

Taking the magnitude,

$I_{\max} = \frac{V}{R} [\frac{1}{2} - \frac{1}{4}] = \frac{V}{4 R}$

Solution

1 $I_{\max} = \frac{V}{2 R}$

2 $I_{\max} = \frac{V}{4 R}$

3 $τ = \frac{L}{R} \ln 2$

4 $τ = \frac{2 L}{R} \ln 2$

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Solution

1 Imax=V2R

2 Imax=V4R

3 τ=LRln2

4 τ=2LRln2

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1 $I_{\max} = \frac{V}{2 R}$

2 $I_{\max} = \frac{V}{4 R}$

3 $τ = \frac{L}{R} \ln 2$

4 $τ = \frac{2 L}{R} \ln 2$