Two capacitors, one that has a capacitance of 3.90 µF and one that has a capacit

Question

Two capacitors, one that has a capacitance of 3.90 µF and one that has a capacitance of 11.8 µF, are connected in parallel. The parallel combination is then connected across the terminals of a 11.0-V battery. Next, they are carefully disconneted so that tehy are not discharged. They are then reconnected to each other--the positive plate of each capacitor connected to the negative plate of the other.

(a) Find the potential difference across each capacitor after they are reconnected.

3.90 µF capacitor______ v

11.8 µF capacitor______ v

(b) Find the energy stored in the capacitors before they are disconnceted from the battery, and find the energy stored after they are reconnected.

before disconnected______ mJ

after reconnected______ mJ

Explanation / Answer

C1 =3.9µF, C2 = 11.8 µF, V =11 V

(a) After reconnected they are in series combination

C =C1C2/(C1+C2) = (3.9*11.8)/(3.9+11.8)

C = 2.93 µF

Q=VC = 11*2.93 = 32.23 µC

In series combination charge is constant

V1 =Q/C1 = 32.23/3.9 = 8.265 V

V2 =Q/C2 = 32.23/11.8 =2.73 V

(b) Before disnnected they are in parallel combination

C =C1+C2 =(3.9+11.8)

C = 15.7 µF

Ub = (1/2)CV2 =(1/2)(15.7x10-6)(11*11)

Ub = 0.95 mJ

After reconnection

Ua = Q2 /2C =(32.23x10-6)2/(2.92x10-6)

Ua =0.36 mJ

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