Two inductors $L_{1}$ and $L_{2}$ are connected in parallel and a time varying current flows as shown in figure. Then the ratio of current s $i_{1} / i_{2}$ at any time t is

L_{1} / L_{2}

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L_{2} / L_{1}

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{\frac{L_{1}^{2}}{(L_{1} + L_{2})}}^{2}

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{\frac{L_{2}^{2}}{(L_{1} + L_{2})}}^{2}

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Solution

The correct option is B $L_{2} / L_{1}$
Since $L_{1}$ and $L_{2}$ are connected in parallel. Potential difference across $L_{1}$ is equal to that across $L_{2}$ i.e. $E_{1} = E_{2}$
Or $- L_{1} \frac{d i_{1}}{d t} = - L_{2} \frac{d I_{2}}{d t}$
Or $L_{1} \int d i_{1} = L_{2} \int d i_{2}$
Or $L_{2} i_{2} = L_{1} i_{1}$
Or $\frac{i_{1}}{i_{2}} = \frac{L_{2}}{L_{1}}$

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