Two capacitors C1 = 4.9 C2 = 17.1 F are charged individually to V1 = 15.8 V, V2

Question

Two capacitors C1 = 4.9 C2 = 17.1 F are charged individually to V1 = 15.8 V, V2 = 6.5 V. The two capacitors are then connected together in parallel with the positive plates together and the negative plates together. Calculate the final potential difference across the plates of the capacitors once they are connected. Submit Answer Tries 0/10 Calculate the amount of charge (absolute value) that flows from one capacitor to the other when the capacitors are connected together. Submit Answer Tries 0/10 By how much (absolute value) is the total stored energy reduced when the two capacitors are connected? Submit Answer Tries 0/10

Explanation / Answer

(1) Chare on first capacitor
Q1 = C1V1 = 4.9*10-6 *(15.8) = 77.42*10-6 C
Q2 = C2V2 = 17.1*10-6 *(6.5) = 111.15*10-6 C
Total charge on the capacitor when they are connected = 188.56*10-6 C
and the combined parallel capacitance = 22*10-6 F
hence the potential difference = q/C =188.56*10-6 /22*10-6 = 8.57 volt
(b) Now charge on the capacitor 1
Q1 = C1V = 4.9*10-6*(8.57) = 41.993*10-6 C
Since charge has become less on capacitor 1 by = 77.42 - 41.993 uC = 35.427*10-6 C
has been transferred to the second capacitor.
(c) Initial energy on capacitor 1 = (1/2)C1V12 = 611.62*10-6 J
Initial energy on the capacitor 2 = (1/2)C2V22 = 361.237*10-6 J
Total energy initially = 972.857*10-6 J
Now
energy on the capacitor 1 = (1/2)C1V2 = 179.94*10-6 J
Energy on the capacitor 2 = (1/2)C2V2 = 627.954*10-6 J
Total energy on the capacitor = 807.894*10-6 J
Energy reduction = (972.857 - 807.894)*10-6 = 164.963*10-6 J

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