The coefficients of self induction of two coils are $L_{1} =$ 8 mH and $L_{2} =$ 2 mH respectively. The current rises in the two coils at the same rate. The power given to the two coils at any instant is same. The ratio of currents flowing in the coils will be :

\frac{i_{1}}{i_{2}} = \frac{1}{4}

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\frac{i_{1}}{i_{2}} = \frac{4}{1}

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\frac{i_{1}}{i_{2}} = \frac{3}{4}

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\frac{i_{1}}{i_{2}} = \frac{4}{3}

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Solution

The correct option is B $\frac{i_{1}}{i_{2}} = \frac{1}{4}$
Power to 1st coil = $V_{1} i_{1} = L_{1} \frac{d i_{1}}{d t} i_{i}$
Power to 2nd coil = $V_{2} i_{2} = L_{2} \frac{d i_{2}}{d t} i_{2}$
As the powers are same,
$L_{1} \frac{d i_{1}}{d t} i_{i} = L_{2} \frac{d i_{2}}{d t} i_{2}$
rate of change of current is given to be same,
$L_{1} i_{i} = L_{2} i_{2}$
or, $\frac{i_{1}}{i_{4}} = \frac{L_{2}}{L_{1}} = \frac{1}{4}$

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Two different coils have self-inductance $L_{1}$ = 8mH and $L_{2}$ = 2mH respectively. The current in both the coils is increased at a same rate. At a certain instant of time, the power being supplied to both the coils is equal. At that instant, the current, the induced voltage, and the energy stored in $L_{1}$ and $L_{2}$ are $I_{1}$ and $I_{2}$ , $V_{1}$ and $V_{2}$ , and $W_{1}$ and $W_{2}$ respectively. Then