A $4\,\,\mu F$ conductor is charged to ${\rm{400}}\,\,{\rm{V}}$ and then its plates are joined through a resistance of ${\rm{1}}\,\,{\rm{K\Omega }}$. The heat produced in the resistance is

$\begin{array}{*{20}{l}} {{\rm{0}}{\rm{.32}}\,\,{\rm{J}}} \end{array}$

$\begin{array}{*{20}{l}} {{\rm{0}}{\rm{.18}}\,\,{\rm{J}}} \end{array}$

$\begin{array}{*{20}{l}} {{\rm{0}}{\rm{.21}}\,\,{\rm{J}}} \end{array}$

$\begin{array}{*{20}{l}} {{\rm{0}}{\rm{.25}}\,\,{\rm{J}}} \end{array}$

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JEE PHYSICS MCQ: ELECTRIC POTENTIAL AND CAPACITANCE

1 Medium

1 of 1

Application +4 | -1

4\,\,\mu F

conductor is charged to

\mathrm{400}\,\,\mathrm{V}

and then its plates are joined through a resistance of

\mathrm{1}\,\,\mathrm{K\Omega }

. The heat produced in the resistance is

View answer and explanation

Correct option: D

\begin{array}{llllllllllllllllllll} {\mathrm{0}\mathrm{.32}\,\,\mathrm{J}} \end{array}

Option D (0.32 J) is correct because the energy stored in the capacitor is converted into heat in the resistor, and the energy stored is calculated by

\frac{1}{2}CV^2

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