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(B) Since the inductors $L_1$ and $L_2$ are connected in parallel,the potential difference across them must be equal.Let the potential difference acro

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$L_2/L_1$

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(B) Since the inductors $L_1$ and $L_2$ are connected in parallel,the potential difference across them must be equal.Let the potential difference across $L_1$ be $V_1$ and across $L_2$ be $V_2$. Then $V_1 = V_2$.The induced electromotive force $(EMF)$ across an inductor is given by $\mathcal{E} = -L \frac{di}{dt}$.Equating the magnitudes of the induced EMFs across both inductors:$L_1 \frac{di_1}{dt} = L_2 \frac{di_2}{dt}$Integrating both sides with respect to time $t$ (assuming initial currents are zero at $t=0$):$L_1 \int di_1 = L_2 \int di_2$This gives:$L_1 i_1 = L_2 i_2$Therefore,the ratio of the currents is:$\frac{i_1}{i_2} = \frac{L_2}{L_1}$

Two inductors $L_1$ and $L_2$ are connected in parallel,and a time-varying current flows as shown. The ratio of currents $i_1/i_2$ at any time $t$ is:

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