Question

A capacitor of capacitance $3\, \mu F$ is first charged by connecting it across a $10\, V$ battery by closing key $K_{1}$, then it is allowed to get discharged through $2\, \Omega$ and $4 \, \Omega$ resistors by opening $K_{1}$ and closing the key $K_{2}$. The total energy dissipated in the $2\, \Omega$ resistor is equal to

• A. 0.5 mJ
• B. 0.05 mJ
• C. 0.15 mJ
• D. none of these

Answer 1

Answer: (D)

Initial energy stored on capacitor:
$U = \frac{1}{2} \times 3 \times 10^{-6} \times (10)^2 $
$ = 150 \times 10^{-6} \,J$
When capacitor is allowed to discharge, the whole energy stored on capacitor will be dissipated as heat in $2\,\Omega$ and $4\,\Omega$ resistors. Heat dissipated in either resistor will be directly proportional to the value of that resistor. So, heat dissipated in $2\,\Omega$ resistor is
$H = \frac{2U}{ 2 + 4} = \frac{U}{3} = 50 \times 10^{-6} J$
$\Rightarrow H = 0.050\,mJ$