Two capacitors C1 = 5.5 F, C2 = 16.7 F are charged individually to V1 = 18.5 V,

Question

Two capacitors C1 = 5.5 F, C2 = 16.7 F are charged individually to V1 = 18.5 V, V2 = 3.0 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. A) 6.840 V

(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

Energy Stored in Capacitor is given by, U = 1/2*cv^2

Charge Stored in Individual Capacitor,
Q1 = C1*V1
Q2 = C2*V2

When Connected in parallel,
Ceq = C1 + C2
Ceq = 5.5 F + 16.7 F = 22.2 F

Total Charge, Q = Q1 + Q2
Q = C1*V1 + C2*V2
Q = 5.5*18.5 + 16.7*3.0 C
Q = 151.85 C

Now,
Final Potential Difference, V = Q/C
V = 151.85/22.2
Final Potential Difference, V = 6.840 V

(B)
As they are in parallel, Voltage remains same across both the capacitors.

V = Q2/C2 = Q1/C1
Q2/16.7 = Q1/5.5
Q2 = 3.04 * Q1

Q1 + Q2 = 151.85
Q1 + 3.04 * Q1 = 151.85
Q1 = 37.58 C

Initially Charge was, 101.75 C

So Amount of charge that flows from one capacitor to the other , = 101.75 C - 37.58 C
Amount of charge that flows from one capacitor to the other , = 64.2 C

(C)
Total stored energy reduced = Ui - Uf
Total stored energy reduced = 1/2 * c1v1^2 + 1/2*c2v2^2 - 1/2*cv^2
Total stored energy reduced = 1/2 * 5.5*18.5 + 1/2 * 16.7*3.0^2 - 1/2 * 22.2 * 6.84^2
Total stored energy reduced = 497 * 10^-6 J

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