A L-R circuit is shown in figure and switch is closed at $t = 0$ . Select INCORRECT option(s).

Question

A L-R circuit is shown in figure and switch is closed at t = 0 . Select INCORRECT option(s).

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Ans (A,B,C,D)

In the square network, the plain-wire left and right edges and the plain-wire diagonal short all its nodes together, so $R_{e q} = 0$ .

The circuit reduces to V across $2 L ∥ L$ : $L_{e q} = \frac{2 L}{3}$

For each inductor $V = L \frac{d i}{d t}$ with $i (0) = 0$ : $i_{2 L} = \frac{V t}{2 L}$ , $i_{L} = \frac{V t}{L}$

Battery current $i = \frac{3 V t}{2 L}$ (linear in t, not exponential) $\Rightarrow$ statement (A) is incorrect.

$\frac{d i}{d t} = \frac{3 V}{2 L}$ is constant for both inductors, neither decreasing nor increasing $\Rightarrow$ statements (B) and (C) are incorrect.

$i_{2 L} : i_{L} = \frac{1}{2 L} : \frac{1}{L} = 1 : 2$ , not $2 : 1$ $\Rightarrow$ statement (D) is incorrect.

All four statements are incorrect.

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Options:(select one or more)

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