You construct a circuit using capacitors of capacitance C1=2.00 F, C2=5.00 F, C3

Question

You construct a circuit using capacitors of capacitance C1=2.00 F, C2=5.00 F, C3=2.00 F. You connect these to a battery that sustains a voltage difference of 60.00 V. The circuit is illustrated below:

What is the equivalent capacitance of the circuit?

Correct, computer gets: 1.56e+00 uF

(1) What is the magnitude of the voltage difference across capacitor C3

(2) What is the charge on the positive plate of capacitor C1?

(3) You remove capacitor C3 from the circuit. What is the change in the electrical energy stored by the circuit?

Please help with numbers 1, 2, & 3??

Explanation / Answer

C1 and C2 are in parallel connection so equivalent of these two capacitor is C'

then C' = C1 + C2 = 7 uF

now C' and C3 are in series connection.

equivalent of these is Ceq then,

1/Ceq = 1/C' +   1/C3   = 1/7 + 1/2

Ceq = 1.56 uF

charge through battery = Ceq V = 1.56 x 60 = 93.33 uC

in series connection charge will be same.

1) Q3 = C3 * V3

93.33 uC = ( 2 uF ) V3

V3 =46.67 volt

2) so voltage across C1 will be :

V1 = 60 - 46.67 =13.33 Volt

Q1 =C1V1 = 2 uF x 13.33 =26.67 uC

3) capacitor 3 is removed.

now Ceq = C' = 7 uF

finally Energy Ef = C V^2 /2 = (7 x 10^-6 x 60^2) /2 = 0.0126 J

initially energy Ei = (1.56 x 10^-6 x 60^2 ) /2 =0.00281 J

change = Ef - Ei = 0.0098 J

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