Two capacitors C1 = 4.1 C2 17.6 F are charged individually to V1 = 15.1 V, V2 =
ID: 1874378 • Letter: T
Question
Two capacitors C1 = 4.1 C2 17.6 F are charged individually to V1 = 15.1 V, V2 = 4.7 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 6.665 V You are correct. Your red 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 know 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 Submit Answer Incorrect. Tries 1/99 Previous Tries By how much (absolute value) is the total stored energy reduced when the two capacitors are connected? Submit Answer Tries 0/99Explanation / Answer
Solution:
Initial Amount of charge:
Q1 = C1 x V1
=> Q1 = 4.1 x 15.1
=> Q1 = 61.91 microCoulomb
Q2 = C2 x V2
=> Q2 = 17.6 x 4.7
=> Q2 = 82.72 microCoulomb
After capacitors are connected in parallel.
Q1 = C1 x V1
=> Q1 = 4.1 x 6.665
=> Q1 = 27.3265 microC
Charge transfered = 61.91 - 27.3265
=> Charge transfered = 34.5835 microC
Energy, Using E = 0.5 x C x V^2
Initial energy
=> E1 = 0.5 x 4.1 x 15.1^2
=> E1 = 467.42 x 10^-6 J
and E2 = 0.5 x 17.6 x 4.7
=> E2 = 194.392 x 10^-6 J
Total energy => 661.812 x 10^-6 J
when capactiors are connected in parallel,
E = 481.981 x 10^-6 J
reduction in energy = 179.831 x 10^-6 J
=> Reduction in energy = 17.9 x 10^-5 J
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