Question

A capacitor of capacitance $3\mu F$ is first charged by connecting across $10V$ battery, then it is allowed to get discharged through $2\Omega$ and $4\Omega$ resistor by closing the key $K$ as shown in figure. The total energy dissipated in $2\Omega$ resistor is equal to

• A. 0.15 mJ
• B. 0.5 mJ
• C. 0.05 mJ
• D. 1.0 mJ

Answer 1

Answer: (C)

Total energy stored in capacitor, $E_{\text{total }}=\frac{1}{2}C{V}^2$
$=\frac{1}{2}\times 3\times 10^{-6}\times 10^2=1.5\times 10^{-4}J$
Energy dissipated in $2\Omega =\frac{2}{(2+4)}\times E_{\text{total }}$
$=\frac{2}{6}\times 1.5\times 10^{-4}=0.5\times 10^{-4}J=0.05mJ$