Two capacitors C_1=6.8 mu F, C_2=18.0 mu F are charged individually to V_1=16.3

Question

Two capacitors C_1=6.8 mu F, C_2=18.0 mu F are charged individually to V_1=16.3 V, V_2=4.1 V. The two capacitors are then connected to in parallel with the positive plates together and the negative together. Calculate the final potential difference across the plates of the capacitors once they are Connected. Calculate the amount of charge (absolute value) that flows from one capacitor to the other when the capacitors are connected together. Pick one capacitor, you know the charge on this capacitor before they were connected. Now that you now the potential difference after they are connected remember the potential drop is the same for both of them you can calculate the charge: Q=CV. You have to calculate the difference of Q before and after. Tries By how much (absolute value) 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 on capacitor= Q= CV = 6.8 x 10^-6 ( 16.3) = 110.84 x 10^-6 C

charge on capacitor after connceted in parallel = ( C1+C2) V = ( 6.8+18) x 10^-6 x 7.445 = 184.636 x 10^-6 C

Charge flow= 73.796 x 10^-6 C

b) Energy = 0.5 cv^2

Energy before capacitors are connected = 0.5 ( 6.8 x 10^-6 ) ( 16.3)^2 + 0.5 ( 18 x 10^-6) ( 4.1)^2 = 1054.636 x 10^-6 J

Energy in parallel = 0.5 ( 24.8 x 10^-6) ( 7.445)^2= 687.307 x 10^-6 J

Reduced energy = 1054.636 x 10^-6 - 687.307 x 10^-6 J= 367.328 x 10^-6 J apprx

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