Q.
Two capacitors $C_{1} = 2\mu\ F$ and $C_{2} = 6\mu\ F$ in series, are connected in parallel to a third capacitor $C_{3} = 4\mu\ F.$ This arrangement is then connected to a battery of $e.m.f. = 2V$, as shown in the figure. How much energy is lost by the battery in charging the capacitors
A) $22 \times 10^{- 6}\ J$
B) $11 \times 10^{- 6}\ J$
C) $\left( \frac{32}{3} \right) \times 10^{- 6}\ J$
D) $\left( \frac{16}{3} \right) \times 10^{- 6}\ J$
Explanation
(b) $C_{eq} = \frac{C_{1}C_{2}}{C_{1} + C_{2}} + C_{3} = \frac{2 \times 6}{2 + 6} + 4 = 5.5\ \mu\ F$
Energy supplied $(E) = QV = CV^{2} = 22 \times 10^{- 6}\ J$
P.E. stored $(U) = \frac{1}{2}C_{eq}V^{2} = \frac{1}{2} \times 5.5 \times (2)^{2} = 11 \times 10^{- 6}\ J$
$\Rightarrow$ Energy lost $= E - U = 11 \times 10^{- 6}\ J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.