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Physics-

General

Easy

Question

Two inductor coils of self inductance 3H and 6H respectively are connected with a resistance 10 $capital omega$ and a battery 10 V 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 mu(tual inductance between $L subscript 1 end subscript$ and $L subscript 2 end subscript$ ):

$\frac{1}{10}$
$\frac{1}{100}$
$\frac{1}{1000}$
1

The correct answer is: $fraction numerator 1 over denominator 100 end fraction$

$L subscript e f f end subscript equals 2 H$
Energy stored in inductor = $blank fraction numerator 1 over denominator 2 end fraction L I to the power of 2 end exponent equals fraction numerator 1 over denominator 2 end fraction cross times left parenthesis B right parenthesis cross times left parenthesis 1 right parenthesis to the power of 2 end exponent equals 1 J$
Energy developed in resistance= $blank F R T equals 1 to the power of 2 end exponent cross times 10 cross times 10$ = 100 J
Hence the required ratio is $fraction numerator 1 over denominator 100 end fraction$

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