A 19.0-μF capacitor and a 44.0-μF capacitor are charged by being connected across separate 45.0-V batteries.

a. Determine the resulting charge on each capacitor. (Give your answer to at least three significant figures.)

b. The capacitors are then disconnected from their batteries and connected to each other, with each negative plate connected to the other positive plate. What is the final charge of each capacitor?

c. What is the final potential difference across the 44.0-μF capacitor?

Question

Physics

Tate

26 April, 21:44

A 19.0-μF capacitor and a 44.0-μF capacitor are charged by being connected across separate 45.0-V batteries.

a. Determine the resulting charge on each capacitor. (Give your answer to at least three significant figures.)

b. The capacitors are then disconnected from their batteries and connected to each other, with each negative plate connected to the other positive plate. What is the final charge of each capacitor?

c. What is the final potential difference across the 44.0-μF capacitor?

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Answers (1)

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Maia · Answer 1

C1 = 19 micro farad

C2 = 44 micro farad

V = 45 V

A.

The relation between the charge and the potential is

q = C x V

where, c is the capacitance and V be the potential difference

q1 = C1 x V = 19 x 45 = 855 micro Coulomb

q2 = C2 x V = 44 x 45 = 1980 micro Coulomb

B.

Now the charge is shared, the charge on both the capacitors = (q1 + q2) / 2

q = (855 + 1980) / 2 = 1417.5 micro Coulomb

C.

Potentia across 44 micro farad

V' = q/C2 = 1417.5 / 44 = 32.2 V