Discussion Question 5B
P212, Week 5
Analysis of Capacitor Networks
You’re ready to analyze a circuit without any help! A battery of voltage V0 is again hooked up to a
network of 5 capacitors, but in a different arrangement. You are asked four questions:
1. What is the voltage
VA at the point A?
2. What is the voltage
VB at the point B?
3. What is the charge
Q2 on capacitor
C2?
4. What is the energy
U5 stored in
capacitor C5?
Begin by simplifying the circuit using equivalent capacitors.
12
45
5813
310 13
1111 4 534
13 13 15
54 419
0
.
Each capacitor gets a charge of
Q=V .
eq
eq
eq
CCCC
CCCC
CF
C
CC
μμ μ
μμμ
μ
μ
=+=
=+ =
=++→ =
=
()
3
12 0 12
6
12 2 12
54 419 3 628
15
54 419 4 186 12 4 186 7 814
13
8 10 4 186 33 5
3B
12
22
.
The voltage across C is V .
.
Voltage across C is . . .
Voltage across C is . Hence Q . .
Volt
B
A
C
VV V
F
C
VVVVVV
F
VCV VC
μ
μ
μ
μ
μ
−
=Δ = =
Δ= = =−Δ=− =
Δ=Δ=×=
()
12
6
512
10 10 4 186 41 86
45 45 12
5
age across C is also since C C
Hence Q . .
V
CV V C
μ
−
Δ=
=Δ =× =
Finally, from Q5 we calculate the energy stored on C5 as U5 = Q52/(2C5) = 87.6 μJ
This could of course also be obtained from U5 = C5 ∆V122 / 2
Remember, the voltages VA and VB must be determined relative to ground …
V0 = 12 V
C1 = 5 μF
C2 = 8 μF
C3 = 15 μF
C4 = 3 μF
C5 = 10 μF
C3
C2
C1
C4 C5
V0
A
B
C3
C45
A
B
C12
Ceq
V0