Answer:
(a) V1 = 9V, V2= 3V
(b)E = 324µJ, E1 = 243µJ, E2 = 81µJ.
Explanation:
Given C1 = 6 µF, C2 = 18 µF
V = 12V
Ceq = C1×C2/(C1 + C2) series connection
Ceq = 6×18/(6 + 18) = 108/24 = 4.5µF
Q = Ceq×V
Q = 4.5×12×10-⁶ = 54µC
The same amount of charge is on both capacitors as they are connected in series.
When the capacitors are reconnected,
V1 = voltage across capacitor 1
V2 = voltage across capacitor 2
V1 = Q/C1 = 54µC/ 6µF = 9V
V2 = Q/C2 = 54µC/ 18µF = 3V
(b)
Before Disconnection
E = 1/2×Ceq×V² = 1/2×4.5×10-⁶× 12² = 324µJ
After the reconnection,
E1 = energy stored in capacitor 1
E2 = energy stored in capacitor 2
E1 = 1/2C1V1² = 0.5× 6µF×9² = 243µJ
E2 = 1/2C2V2² = 0.5×18µF×3² = 81µJ
E = E1 +E2 = (243 + 81)µJ = 324µJ.
