Two inductor coils of self inductance 3H and 6H respectively are connected with a resistance $10 Ω$ and a battery 10V as shown in figure. The ratio of total energy stored at steady state in the inductors to that of heat developed in resistance in 10 seconds at the steady state is (neglect mutual inductance between $L_{1}$ and $L_{2}$ )

$\frac{1}{10}$

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$\frac{1}{100}$

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$\frac{1}{1000}$

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$1$

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Solution

The correct option is B
$\frac{1}{100}$

$L = \frac{L_{1} L_{2}}{L_{1} + L_{2}} = 2 H i_{0} = \frac{E}{R} = 1 A N o w, \frac{\frac{1}{2} L i_{0}^{2}}{i_{0}^{2} R t} = \frac{L}{2 R t} = \frac{2}{2 \times 10 \times 10} = \frac{1}{100}$

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The correct option is B 1100 L=L1L2L1+L2=2Hi0=ER=1ANow,12Li20i20Rt=L2Rt=22×10×10=1100

The correct option is B
$\frac{1}{100}$

$L = \frac{L_{1} L_{2}}{L_{1} + L_{2}} = 2 H i_{0} = \frac{E}{R} = 1 A N o w, \frac{\frac{1}{2} L i_{0}^{2}}{i_{0}^{2} R t} = \frac{L}{2 R t} = \frac{2}{2 \times 10 \times 10} = \frac{1}{100}$