Q. 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 m J
B) 0.5 m J
C) 0.05 m J
D) 1.0 m J
Explanation
(c) Total energy stored in capacitor, $E_{total} = \frac{1}{2}CV^{2}$
$= \frac{1}{2} \times 3 \times 10^{- 6} \times 10^{2} = 1.5 \times 10^{- 4}$ J
Energy dissipated in $2\Omega = \frac{2}{(2 + 4)} \times E_{total}$
$= \frac{2}{6} \times 1.5 \times 10^{- 4} = 0.5 \times 10^{- 4}$ J = 0.05 mJ
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.