Question

A capacitor of capacitance $3\mu F$ is first charged by connecting it across a $10V$ 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
$H=\frac{2U}{2+4}=\frac{U}{3}=50\times 10^{-6}J$
$\Rightarrow H=0.050mJ$