JEE Main20202 Sep 2020Evening ShiftPhysicsCapacitanceActual
A 10 μF capacitor is fully charged to a potential difference of 50 V After removing the source voltage it is connected to an uncharged capacitor in parallel. Now the potential difference across them becomes 20 V . The capacitance of the second capacitor is :
Options
- A15   μF
- B30   μF
- C20   μF
- D10   μF
Correct answer
A. 15   μF
Step-by-step solution
V = C 1 V 1 + C 2 V 2 C 1 + C 2 20 = 10 × 50 + 0 20 + C C = 15 μ F