# μ μ μ μ μ μ μ μ μ μC μF volts μJ QCVU 1 1203.376 163 7.382674

```PHYS-1150(9)(Kaldon-40203)
WMU - Fall 2014
Quiz 5 - 30,000 points
Name ________________________________
Section:
Rev. 09/30/14.r1
9a
9b
9c
9d
9e
28f
A to D
E to M
N to S
T to Z
Q to S
T to Z
1150
Show all work/Circle your final answer: [TAKE-HOME Due Fri. 3 Oct. 2014]
5.) Reduce this capacitor network to a single capacitor. Then use this
information, along with the rules about how charge (Q) and voltage
difference (V) are split or shared for series and parallel capacitors, as
well as the potential energy (U) to completely fill in the table. All five
capacitors are different.
Attach all work to this sheet – don’t worry if it’s not pretty.
For full credit you must draw all the intermediate circuits.
Q5 Solution
I could do C45 right at the start, but I
don’t have to.
C2 and C3 are in Series.
1
1
1


C23 C2 C3
1
1

263  F 363  F
C23 =152.5  F  &quot;0.006557  F&quot;

C1 and C23 are in Parallel.
C123  C1  C23  163  F+152.5  F
=315.5  F
C123, C4 and C5 are in Series.
1
C12345

1
1
1


C123 C4 C5
1
1
1


315.5  F 463  F 563  F
C12345 =140.7  F  &quot;0.007106  F&quot;

Q
=
C
C1
163 F
C2
263 F
C3
363 F
C4
463 F
C5
563 F
Ceq
V
U
C
F
Q
C
1
1203.376
163
2
1125.706
263
3
1125.706
363
4
2329.082
463
5
2329.082
563
Total 2329.082 140.73
16.55 V
(1)
(2)
(3)
(4)
volts
V
7.382674
4.280249
3.101117
5.030414
4.136912
16.55
Have Ceq and V, can find Q0.
Q4 = Q5 = Q0. Can now find V4 and V5.
V1 = 16.55 volts - V4 - V5.
Can now finish the table.
J
U
4442.066
2409.15
1745.473
5858.122
4817.603 19272.41
19273.15
```