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Self Induction

Two different...

Question

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

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Solution

The correct options are
A

C

D

Power supplied (P) to an inductor of an inductance (L) is ${\frac{d}{d t}}^{(\frac{1}{2} L i^{2})} = \frac{L i d i}{d t}$
Say $i_{1}$ is the current in $L_{1}$ , $P_{1}$ is the power supplied to $L_{1}$ , $i_{2}$ is the current in $L_{2}$ and $P_{2}$ is power supplied to $L_{2}$ .
Given that $P_{1} = P_{2}$
$\Rightarrow L_{1} i_{1} \frac{d i_{1}}{d t} = L_{2} i_{2} \frac{d i_{2}}{d t}$
Also given that $\frac{d i_{1}}{d t} = \frac{d i_{2}}{d t}$
$\Rightarrow L_{1} i_{1} = L_{2} i_{2}$
$\Rightarrow \frac{i_{1}}{i_{2}} = \frac{L_{2}}{L_{1}} = \frac{1}{4}$
and $\frac{V_{1}}{V_{2}} = \frac{L_{1} \frac{d i_{1}}{d t}}{L_{2} \frac{d i_{2}}{d t}} = \frac{L_{1}}{L_{2}} = 4$
$\frac{W_{1}}{W_{2}} = \frac{\frac{1}{2} L_{1} i_{1}^{2}}{L_{2} i_{2}^{2}} = \frac{L_{1} i_{1}}{L_{2} i_{2}} \times \frac{i_{1}}{i_{2}}$
$= \frac{i_{1}}{i_{2}}$
$= \frac{1}{4}$

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