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(A) Let $i$ be the current through the resistance $R$,and $i_1$ and $i_2$ be the currents through inductors $L_1$ and $L_2$ respectively.At $t=0$,the

Vedclass · Accepted Answer

$A, B, C$

Vedclass · Answer

(A) Let $i$ be the current through the resistance $R$,and $i_1$ and $i_2$ be the currents through inductors $L_1$ and $L_2$ respectively.At $t=0$,the inductors act as open circuits because the current cannot change instantaneously. Thus,the current through the resistance $R$ is $0$.For $t > 0$,the voltage across the parallel combination of $L_1$ and $L_2$ is the same. Therefore,$V_{L1} = V_{L2} \implies L_1 \frac{di_1}{dt} = L_2 \frac{di_2}{dt}$.Integrating both sides with respect to time,we get $L_1 i_1 = L_2 i_2$ (assuming initial currents are zero). This implies $\frac{i_1}{i_2} = \frac{L_2}{L_1}$,so the ratio of currents is fixed at all times.After a long time,the inductors act as short circuits (ideal inductors). The total current $i = \frac{V}{R}$.Since $i_1 + i_2 = i = \frac{V}{R}$ and $L_1 i_1 = L_2 i_2$,we have $i_2 = \frac{L_1}{L_2} i_1$.Substituting this into the current equation: $i_1 + \frac{L_1}{L_2} i_1 = \frac{V}{R} \implies i_1 \left( \frac{L_1+L_2}{L_2} \right) = \frac{V}{R} \implies i_1 = \frac{V}{R} \frac{L_2}{L_1+L_2}$.Similarly,$i_2 = \frac{V}{R} \frac{L_1}{L_1+L_2}$.Thus,options $A, B,$ and $C$ are correct.

$List-I$	$List-II$
$(I)$ $\frac{1}{\sqrt{2}}(\sin \omega t \hat{j}+\cos \omega t \hat{k})$	$(P)$ $0$
$(II)$ $\frac{1}{\sqrt{2}}(\sin \omega t \hat{i}+\cos \omega t \hat{j})$	$(Q)$ $-\frac{\alpha}{4} \hat{i}$
$(III)$ $\frac{1}{\sqrt{2}}(\sin \omega t \hat{i}+\cos \omega t \hat{k})$	$(R)$ $\frac{3\alpha}{4} \hat{i}$
$(IV)$ $\frac{1}{\sqrt{2}}(\cos \omega t \hat{j}+\sin \omega t \hat{k})$	$(S)$ $\frac{\alpha}{4} \hat{j}$

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