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Physics-

General

Easy

Question

A 2 $mu F$ capacitor is charged as shown in the figure. The percentage of its stored energy dissipated after the switch $s$ is turned to positions 2 is

0 %
20 %
75%
80%

The correct answer is: 80%

$q subscript i end subscript equals C subscript i end subscript V equals 2 V equals q$ (say)
This charge will remain constant after switch is shifted from position 1 to position 2.
$U subscript i end subscript equals fraction numerator 1 over denominator 2 end fraction fraction numerator q to the power of 2 end exponent over denominator C subscript i end subscript end fraction equals fraction numerator q to the power of 2 end exponent over denominator 2 cross times 2 end fraction equals fraction numerator q to the power of 2 end exponent over denominator 4 end fraction$
$U subscript i end subscript equals fraction numerator 1 over denominator 2 end fraction fraction numerator q to the power of 2 end exponent over denominator C subscript i end subscript end fraction equals fraction numerator q to the power of 2 end exponent over denominator 2 cross times 10 end fraction equals fraction numerator q to the power of 2 end exponent over denominator 20 end fraction$
$therefore$ Energy dissipated $open parentheses fraction numerator q to the power of 2 end exponent over denominator 5 end fraction close parentheses i s blank$ 80% of the initial stored energy $open parentheses equals fraction numerator q to the power of 2 end exponent over denominator 4 end fraction close parentheses$ .

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