Two parallel plate capacitors A and B having capacitance 1µF and 5 µF are charged separately to the same potential 100V. They are then connected such that +ve plate of A is connected to –ve plate of B. Find the charge on each capacitor and total loss of energy in the capacitors
Answers
Answered by
21
The capacitors are connected in series.
Energy U1 + U2 = 1/2 C1 V1²
= 1/2 * 10⁻⁶ * 100² + 1/2 * 5 * 10⁻⁶ * 100²
= 0.03 Joules
The total charge gets redistributed. So there is work done. That becomes the loss.
Effective capacitance of both = C1 C2 / (C1 + C2) = 5/6 μF
Total voltage across both = 100 + 100 = 200 V
Final energy = 1/2 * 5/6 * 10⁻⁶ * 200² J = 0.0167 J
Loss = 0.0133 J
Energy U1 + U2 = 1/2 C1 V1²
= 1/2 * 10⁻⁶ * 100² + 1/2 * 5 * 10⁻⁶ * 100²
= 0.03 Joules
The total charge gets redistributed. So there is work done. That becomes the loss.
Effective capacitance of both = C1 C2 / (C1 + C2) = 5/6 μF
Total voltage across both = 100 + 100 = 200 V
Final energy = 1/2 * 5/6 * 10⁻⁶ * 200² J = 0.0167 J
Loss = 0.0133 J
Similar questions