A 10   μF capacitor is fully charged to a potential difference of 50   V After removing the…

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 :
  1. 15 μF
  2. 30 μF
  3. 20 μF
  4. 10 μF

Solution

V=C1V1+C2V2C1+C2
20=10×50+020+C
C=15μF

Asked in: JEE Main 2020 (02 Sep Shift 2)

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