# A parallel plate capacitor of capacitance C is connected to a battery and is charged to potential difference V. Another capacitor of capacitance 2C is charged to potential difference 2V .The battery is now disconnected and capacitors are connected in parallel to each other in such a way that positive terminal of one is connected to negative terminal of other. The final energy of configuration is?

Grade:12

## 1 Answers

Arun
25757 Points
4 years ago
Explanation:
Charge stored in the first capacitor = cv
Charge stored in the second capacitor = 2c x 2v = 4cv
When they are connected total charge of the system = 4cv - cv = 3cv
Because capacitance gets added in parallel connection,
Total capacitance of the parallel connection = c + 2c = 3c
we know that ,
charge = voltage x capacitance
q = cv
=> v = q/c
Hence voltage of the system = 3cv/3c = v
Now Energy of a capacitor is given by,
E = ½ c v2
=> E = 1/2 x 3c x v² = 3cv²/2
Hence energy of the system is 3cv²/2

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