Two capacitors C_1 = 3.8 mu F, C_2 = 16.5 mu F are charged individually to V_1 =

Question

Two capacitors C_1 = 3.8 mu F, C_2 = 16.5 mu F are charged individually to V_1 = 18.9 V, V_2 = 7.6 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. Tries 0/99 Calculate the amount of charge (absolute value) that flows from one capacitor to the other when the capacitors are connected together. Tries 0/99 By how much (absolute value) is the total stored energy reduced when the two capacitors are connected? Tries 0/99

Explanation / Answer

1) The charges on each are
Q1 = CV = 3.8µF x 18.9v = 71.82µC
Q2 = CV = 16.5µF x 7.6v = 12.4µC

when they are connected together the total charge is the sum, 197.22µC
Total C is the sum, 20.3µF
V = Q/C = 197.22µC / 20.3µF = 9.71 volts

2)  use Q=Ctot x V, where Ctot =C1+C2

Q = 20.3µF x 9.71 = 197.22µC

3) use E=1/2CV^2 ( or E=1/2QV, since Q=C/V) for each individually then for both in parallel.

E = (1/2)*20.3µF*9.71^2

= 9.569*10-4 J

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