A capacitor of 2 is charged as shown in the diagram. When the switch S is turned to position 2, the percentage of its stored energy dissipated is [figure]
- 75%
- 80%
- 0%
- 20%
Answer
(2) 80%
Solution
Initial charge on the 2 capacitor: Q = C1V; initial energy Ui = 12C1V2
In position 2 the charge is shared with the 8 capacitor (in parallel):
Common potential V' = C1VC1 + C2 = 2V10 = V5
Final energy: Uf = 12(C1 + C2)V'2 = 12(10) V225 = V25 (C in μF)
Initial energy: Ui = 12(2)V2 = V2
⇒ Energy dissipated = Ui - Uf = 45V2
∴ Percentage dissipated = 4/51 100 = 80\%
Answer: (2)