Two different coils have self-inductance $L_{1} = 8 mH$ and $L_{2} = 2 mH$ . At a certain instant the current in the two coil is increasing at the same constant rate and the power supplied to the two coils is the same. Find the ratio of energies stored in the two coils at that instant.

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$\frac{1}{8}$

$\frac{1}{16}$

$\frac{1}{2}$

$\frac{1}{4}$

answer is B.

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Detailed Solution

As emf induced in a coil, $ε = L \frac{di}{dt}$

So, $\frac{ε_{1}}{ε_{2}} = \frac{L_{1}}{L_{2}} = \frac{8 mH}{2 mH} = 4 (as \frac{di}{dt} = constant)$

A, power supplied $P = εi = constant$

So, $\frac{i_{1}}{i_{2}} = \frac{ε_{1}}{ε_{2}} = \frac{1}{4}$ , As energy, stored in coil, $U = \frac{1}{2} {LI}^{2}$

i.e. $\frac{U_{1}}{U_{2}} = \frac{L_{1}}{L_{2}} {[\frac{I_{1}}{I_{2}}]}^{2} = \frac{8}{2} {[\frac{1}{4}]}^{2} = \frac{1}{4}$

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