Two capacitors C 1 = 7.9 F, C 2 = 18.9 F are charged individually to V 1 = 14.9

Question

Two capacitors C1 = 7.9 F, C2 = 18.9 F are charged individually to V1 = 14.9 V, V2 = 7.7 V. The two capacitors are then connected together in parallel with the positive plates together and the negative plates together.
a)Calculate the final potential difference across the plates of the capacitors once they are connected.
b)Calculate the amount of charge (absolute value) that flows from one capacitor to the other when the capacitors are connected together.

c)By how much (absolute value) is the total stored energy reduced when the two capacitors are connected?

Explanation / Answer

a) Q1i=C1V1=117.71*10^-6 C ,

Q2i=C2V2 = 145.53*10^-6 c
Qf=Q1i+Q2i
Vf=Qf/(C1+C2) = 9.82 V

b) Q1=C1V1, Q2=C2V2
Q1+Q2=Qeq

q1/C1=q2/C2

Qeq-q2/C1=q2/C2

q2-Q2=amount of charge

c) Q1=C1V1, Q2=C2V2

Qt/(C1+C2)=Veq
C1(Veq)=q1
C2(Veq)=q2
Q1-Veq=f
[1/2(C1)(V1)^2]+[1/2(C2)(V2)^2]=E1 (J)
01/2(C1+C2)(Veq)=E2 (J)
E1-E2=total energy reduced

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