Problem 2 a Two capacitors (C 6-1 LuF, C 17.9LF) are charged individually to IV1

Question

Problem 2 a Two capacitors (C 6-1 LuF, C 17.9LF) are charged individually to IV1 313.2 V, V 5.8 V) The two capacitors are then co nnected 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. 7.681 V You are correct. Computer's answer now shown above Previous Tries Your receipt is 170-9039 b) Calculate the amount of charge (absolute value) that flows from one capacitor to the other when the capacitors are connected together. 15.61 10A-6C Pick one capacitor, you know the charge on this capacitor before they were connected. Now that you know the potential difference aft 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 Submit Answer Incorrect Tries 4/6 Previous Tries c By how much (absolute value is the total stored energy reduced when the two capacitors are connected? 3066*10A-6J Calculate the stored energy for the the two capacitors individually before they were connected and add them up, and compare that to the energy of the combined system. Submit Answer Incorrect. Tries 16 Previous Tries

Explanation / Answer

We have already got the final potential difference as 7.681V

b) The charge flow on the smaller cap is its capacitance times the final voltage

Q1 = C1 * V1 = 6.1 *10^-6 * 7.681 = 4.68*10^-5 V

Since its charge started out at 8.052*10^-5V , it has lost 3.37*10^-5

Q2 = C2 *V2 = 17.9*10^-6 * 7.681 = 13.7*10^-6 V which is 3.37*10^-5 greater than its initial charge .

c) Total stored energy

Energy is one half times the capacitance times voltage squared

J = (C*V^2) /2

J1 = (6.1*10^-6 * 13.2^2) / 2 = 5.31*10^-4 J

J2 = ( 17.9*10^-6 * 5.8^2) / 2 = 3.01*10^-4 J

Total energy separately = 8.32*10^-4 J

when paralled

J = ( 23.99 *10^-6 *7.681^2) / 2 = 7.07*10^-4J

reduction in stored energy = 1.25*10^-4 J

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