Question

A C₁= 5.0 µF capacitor is charged by a 12.0-V battery and then is disconnected from the battery. When this capacitor (C₁) is then connected to a second (initially uncharged) capacitor, C₂, the voltage on the first drops to 5.0 V. What is the value of C₂?

Answer 1

Solution :

The charge in first capacitor before connecting the second capacitor= 12 x 5 = 60 C

so chage stored in second capacitor = 60 C ( as they are connected in series)

voltage drop across second capacitor = (12 - 5) V = 7 V

so capacitance of second capacitor , C2= 60 / 7 = 8.57 F

Answer:8.57 F

Thank You.