Using a battery, a 100 pF capacitor is charged to 60 V and then the battery is removed. After that, a second uncharged capacitor is connected to the first capacitor in parallel. If the final voltage across the second capacitor is 20 V, its capacitance is : (in pF)
(A)600
(B)200✓ Correct
(C)400
(D)100
Answer: (B)
Solution
The first capacitor carries Q=C0V0=100×60=6000pC, and once the battery is removed this charge is fixed. Joining the second capacitor in parallel lets the charge share until both reach the common voltage 20V, so 100+C6000=20. Solving, 100+C=300, so the capacitance of the second capacitor is 200pF.
JEE Main · 2025 · 03 Apr 2025 · Shift 2 · Single correct option