Q. $0.2F$ capacitor is charged to $600\ V$ by a battery. On removing the battery, it is connected with another parallel plate condenser of $1F.$ The potential decreases to
A) $100\ volts$
B) $120\ volts$
C) $300\ volts$
D) $600\ volts$
Explanation
(a) By using charge conservation $0.2 \times 600 = (0.2 + 1)V$
$\Rightarrow V = \frac{0.2 \times 600}{1.2} = 100\ V$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.