Two ideal inductors are connected in parallel as shown in figure. A time - varying current flows as shown. The ratio $\frac{I_{1}}{I_{2}}$ at any time t is

\frac{L_{1}}{L_{2}}

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\frac{L_{2}}{L_{1}}

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\sqrt{\frac{L_{1}}{L_{2}}}

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\sqrt{\frac{L_{2}}{L_{1}}}

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Solution

The correct option is B $\frac{L_{2}}{L_{1}}$
Since the inductors are connected in parallel, the potential difference across $L_{1}$ = potential difference across $L_{2}$ at any time t. Hence,
$L_{1} \frac{d I_{1}}{d t} = L_{2} \frac{d I_{2}}{d t}$
$\Rightarrow L_{1} d I_{1} = L_{2} d I_{2}$
Integrating, we get $L_{1} I_{1} = L_{2} I_{2}$
Which give $\frac{I_{1}}{I_{2}} = \frac{L_{2}}{L_{1}}$ .

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