Question

A capacitor of capacitance $3\, \mu F$ is first charged by connecting across $10\, V$ 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 $2 \Omega=\frac{2}{(2+4)} \times E_{\text {total }}$
$=\frac{2}{6} \times 1.5 \times 10^{-4}=0.5 \times 10^{-4} J =0.05\, mJ$