Two different coils have self inductance $8 m H$ and $2 m H$ . The current in both coils are increased at same constant rate. The ratio of the induced emf's in the coil is :

4 : 1

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1 : 4

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1 : 2

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2 : 1

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Solution

The correct option is A $4 : 1$
The induced emf, $e = \frac{L d i}{d t}$
Here, $L_{1} = 8 m H = 8 \times 10^{- 3} H$
and $L_{2} = 2 m H = 2 \times 10^{- 3} H$
So, ratio of induced emf's, $\frac{e_{1}}{e_{2}} = \frac{L_{1} \frac{d i}{d t}}{L_{2} \frac{d i}{d t}} = \frac{L_{1}}{L_{2}}$
(Since, current is increased at same rate)
$\Rightarrow \frac{e_{1}}{e_{2}} = \frac{8 \times 10^{- 3} H}{2 \times 10^{- 3} H} = \frac{4}{1}$

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