Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

The diagram above shows a circuit composed of an ideal battery, ideal wires, and

ID: 1259364 • Letter: T

Question

The diagram above shows a circuit composed of an ideal battery, ideal wires, and six capacitors. The battery provides a constant voltage of 9 volts. The capacitances of the six capacitors are given below.

C1= 14 nF , C2= 8 nF , C3= 16 nF , C4= 3 nF , C5= 6 nF , C6= 6 nF

a/ Find the voltage accross C4

b/ find the charge on C3

The diagram above shows a circuit composed of an ideal battery, ideal wires, and six capacitors. The battery provides a constant voltage of 9 volts. The capacitances of the six capacitors are given below. C1= 14 nF , C2= 8 nF , C3= 16 nF , C4= 3 nF , C5= 6 nF , C6= 6 nF Find the voltage accross C4 find the charge on C3

Explanation / Answer

Part A)

First C2 and C3 are in parallel

C of that section = 24 nF

That is in series with C6, so in that leg...

1/C = 1/24 + 1/6

C = 4.8 nF

C1 and C5 are in series, so in that leg...

1/C = 1/14 + 1/6

C = 4.2 nF

The 4.2 and the 4.8 are in parallel

The C of the large parallel section is 9.0 nF

That is in series with C4, so

1/C = 1/3 + 1/9

C = 2.25 nF

Now we can find net Q

Q = CV = (2.25 X 10-9)(9) = 2.025 X 10-8 C

Then for the voltage on C4

Q = CV

(2.025 X 10-8) = (3 X 10-9)(V)

V = 6.75 Volts

Part B)

The voltage over the large parallel portion will be 9 - 6.75 = 2.25 Volts

The charge in the right side portion of that is...

Q = CV

Q = (4.8 X 10-9)(2.25)

Q = 1.08 X 10-8 Coulombs

Use that to find the V in the upper parallel section

Q = CV

(1.08 X 10-8) = (24 X 10-9)(V)

V = .45 Volts

Then finally for C3

Q = CV

Q = (16 X 10-9)(.45)

Q = 7.2 X 10-9 Coluombs (That is 7.2 nC)