Two capacitors C_1 = 3.3 mu F, C_2 = 16.1 mu F are charged individually to V_1 =

Question

Two capacitors C_1 = 3.3 mu F, C_2 = 16.1 mu F are charged individually to V_1 = 17.9 V, V_2 = 4.1 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. Calculate the amount of charge (absolute value) that flows from one capacitor to the other when the capacitors are connected together. 3.78 times 10^-5 C 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

Energy stored in capacitor is given by, E = 0.5 (C V2)

E1 = 0.5 x (3.3 x 10-6 x 17.92) = 5.287 x 10-4 J

E2 = 0.5 x (16.1 x 10-6 x 4.12) = 1.35 x 10-4 J

Total energy separately, E = 6.64 x 10-4 J

When connected parallel:

E' = 0.5 x (19.4 x 10-6 x 6.4472) = 4.03 x 10-4 J

So, reduction in stored energy, delta E = 2.61 x 10-4 J

(Please rate my answer if you find it helpful. Goodluck...)

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.