Three 0.18 mu F capacitors are connected in parallel across a V = 17-V battery,

Question

Three 0.18 mu F capacitors are connected in parallel across a V = 17-V battery, as shown in the figure below. The battery is then disconnected. Next, one capacitor is carefully disconnected so that it doesn't lose any charge and is reconnected backward, that is, with its positively charged side and its negatively charged side reversed. What is the potential difference across the capacitors now? (Express your answer to two significant figures.) V By how much has the stored energy of the combination of capacitors changed in the process? Find the rate U_electric, _i/U_electric, _f. (Express your answer to two significant figures.)

Explanation / Answer

Equivalent capacitance = C + C + C = 3C

since they are connected in parallel.

=> Ceq = 5.4 x 10-7 F

so, total charge = Q = CeqV = 5.4 x 10-7 (17) = 9.18 x 10-6 C.

and charge on each capacitor will be: q = CV = 0.18 x 10-6 (17) = 3.06 x 10-6 C.

now when one of the capacitors are reversed, the new total charge will be:

Q' = 2q - q = q = 3.06 x 10-6 C.

=> q' = Q'/3 = 1.02 x 10-6 C.

while the capacitance remains the same

therefore, Potential difference across the capacitors now will be: V ' = q' / C = 5.667 Volts.

2]

Change in Energy = (1/2)CeqV2 - (1/2)CeqV '2 = (1/2)(5.4 x 10-7)[172 - 5.6672] = 69.36 x 10-6 J

and Ui/Uf = 172/5.6672 = 9.0

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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