Q.
A $2\mu F$ capacitor is charged as shown in figure. The percentage of its stored energy dissipated after the switch S is turned to position 2 is
A) 0%
B) 20%
C) 75%
D) 80%
Explanation
(d) [image:1_image337.emf]
[image:1_image338.emf]
$U_{i} = \frac{1}{2}(2)V^{2},\ V_{common} = \frac{V}{5}$
$U_{f} = \frac{1}{2}(2 + 8)\left( \frac{V}{5} \right)^{2}$
$\frac{U_{i} - U_{f}}{U_{i}} \times 100$
$= \frac{V^{2} - \frac{V^{2}}{5}}{V^{2}} \times 100$
$\frac{4}{5} \times 100 = 80\ \%$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.