Two capacitors C_1 = 7.3 mu F, C_2 = 14.7 mu F are charged individually to V_1 =
ID: 1626936 • Letter: T
Question
Two capacitors C_1 = 7.3 mu F, C_2 = 14.7 mu F are charged individually to V_1 = 15.8 V, V_2 = 5.0 V. The two capacitors are then connected together in parallel with the positive plates together and the negative pates together: Calculate the final potential difference across the plates of the capacitors once they are connected. Calculate the amount of charge that flows from one capacitor to the other when the capacitors are connected together. 5.27 times 10^-5 C By how much is the total stored energy reduced when the two capacitors are connected? Calculate the stored energy for the two capacitors individually before they were connected and add them up, and compare that to the energy of the combined system.Explanation / Answer
Charge in capacitor C1 = Q1 =C1V1
= 7.3 x 15.8 x 10-6 = 0.00011534 Coulomb
Charge in capacitor C2 = Q2 =C2V2
= 14.7 x 5 x 10-6 = 0.0000735 Coulomb
Total charge in system = Q=Q1+Q2 = 0.00018884 Coulomb
When the capacitors are connected in parallel, the capacitance of circuit = C=C1+C2
= (7.3+14.7)x10-6 = 22 x 10-6 F
Final voltage of system = V= Q/C
= 0.00018884/22 x 10-6 = 8.58 V
Final charge in C1 = C1V = 7.3 x 8.58 x 10-6 = 0.000062634 Coulomb
Final charge in C2 = C2V = 14.7 x 8.58 x 10-6 =0.000126126 Coulomb
Total initial energy of system = (1/2)(C1V12+C2V22) = 0.001094936 J
Total final energy of system = (1/2)(C1+C2)V2 = 0.0008097804 J
Reduction in energy = 0.001094936-0.0008097804 = 0.0002851556 J
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