Name: ________________________________________ Code # _________
UCSD Physics 2B
PE2 Supplement
Summer Session 2008
ID: X
Potential, Work & Capacitors
READ ME !!!
Each of these exercises in this set asks for a different value,
but the approach is the same for them all. These are the steps:
1. Simplify the circuit down to one equivalent capacitance in other words,
reduce the circuit one step a a time into a single device
If this what the problem asks for, you’re done. If not, then
2. Use the capacitor equation Q = CV to find the circuit charge
3. Use the capacitor rules to determine the charges on each element
4. Use the capacitor equation again for each element to find the
voltage across each circuit element.
5. If the problem asks for the energy stored in an element, use the voltage
you found along with the capacitor energy equation, OR...
use the charge you found in #3 above using the alternate equation
Note that there are often shortcuts, but this procedure will always work!!!
1. You connect three capacitors as shown in the diagram below.
If C1 = 5.0 μF , C2 = 4.0 μF and C3 = 3.0 μF , what is the equivalent capacitance between the
terminals A and B of this network?
A. 0.44 μF
B. 2.3 μF
C. 3.5 μF
1
D. 5.2 μF
Name: ________________________
ID: A
2. You connect three capacitors as shown in the diagram below.
If C1 = 5.0 μF , C2 = 4.0 μF and C3 = 3.0 μF and a potential of 12 V is applied between terminals A
and B, the energy stored in C3 is approximately
A. 0.16 mJ
B. 0.041 mJ
C. 0.12 mJ
2
D. 0.41 mJ
Name: ________________________
ID: A
3. You connect three capacitors as shown in the diagram below.
C 1 = 2.5 μF , C 2 = 5.0 μF and C 1 = 7.5 μF . A potential difference of 9.0 V is maintained between the
terminals A and B. What is the magnitude of the charge on capacitor C2 ?
A. 3.4 μ C
B. 68 μ C
C. 0.86 μ C
3
D. 22.5 μ C
Name: ________________________
ID: A
4. You connect three capacitors as shown in the diagram below.
C 1 = 2.5μ F , C 2 = 5.0μ F and C 3 = 2.5μ F . A potential difference of 12.0 V is maintained between the
terminals A and B. What is the voltage across capacitor C3 ?
A. 3.0 V
B. 4.0 V
C. 6.0 V
4
D. 9.0 V
ID: A
UCSD Physics 2B
Answer Section
PE2 Supplement
Potential, Work & Capacitors
MULTIPLE CHOICE
1. ANS: B
C1 and C2 are in parallel so the capacitances ADD: C 12 = C 1 + C 2 = 9.0μF .
The network C12 is in series with C3, so
1
1
1
1
1
4
=
+
=
+
=
C series C 12 C 3 9.0μF 3.0μF 9.0μF
Hence the equivalent capacitance C series = 2.25μF .
See if you can learn the fast way on your calculator:
C 1 x −1 ⟨+⟩C 2 x −1 ⟨=⟩ x −1 ⟨=⟩ where
means function key.
If this doesn’t work for you, we’ll try different syntaxes in class
PTS: 1
2. ANS: C
Step one: find equivalent capacitance: (we just did this, so skip it)
C1 and C2 are in parallel so the capacitances ADD: C 12 = C 1 + C 2 = 9.0μF .
The network C12 is in series with C3, so
1
1
1
1
1
4
=
+
=
+
=
C series C 12 C 3 9.0μF 3.0μF 9.0μF
Hence the equivalent capacitance C series = 2.25μF .
Step two: find total charge
Ê
ˆ
Q TOT = C TOT V TOT = ÁÁ 2.25 × 10−6 F ˜˜ (12V ) = 2.7 × 10 −5 C
Ë
¯
Step three: The Capacitor Rules tell us that Q 3 = Q TOT because it’s not in parallel with anything else. Since
we’re asked for the Energy, and we already know the charge, the fastest way to proceed is to use this form:
2
ÁÊÁ 2.7 × 10 −5 C ˜ˆ˜
Q2
Ë
¯
U3 =
=
= 1.22 × 10 −4 J = 0.122mJ
2C 3 2 ÁÊÁ 3.0 × 10−6 F ˜ˆ˜
Ë
¯
PTS: 1
1
ID: A
3. ANS: D
C1 and C2 are in parallel so we reduce this first into one capacitor.
Parallel capacitances ADD: C 12 = C 1 + C 2 = 7.5μF .
The network C12 is in series with C3, so we reduce again
1
1
1
1
1
1
=
+
=
+
=
C series C 12 C 3 7.5μF 7.5μF 3.75μF
Now we’re done simplifying the circuit, and the equivalent capacitance
C TOT = C series = 3.75μF .
Now find the total charge:
Ê
ˆ
Q TOT = C TOT V TOT = ÁÁ 3.75 × 10 −6 F ˜˜ (9.0V ) = 3.38 × 10−5 C
Ë
¯
Since C2 is in parallel with another element, the charge Q2 is shared with capacitor C1. and the two charges
together must equal the total charge. Using a Capacitor rule, the ratio of charge is the ratio capacitance,
Q 1 C 1 2.5
=
=
= 0.5.
Q 2 C 2 5.0
So Q1 holds half the charge of Q2, and the total charge of the pair is sum:
Q TOT = Q 1 + Q 2 = 1.5 × Q 2
Q TOT 3.38 × 10 −5 C
=
= 2.25 × 10 −5 C = 22.5μC
1.5
1.5
SHORT CUT: If you happen to notice that C 12 = C 3 , and see that this splits the circuit capacitances exactly
in half, then the two voltages V 12 = V 3 . Their sum must of course be the circuit sum, so
1
V 12 = V 3 = V TOT = 4.5V . Now in parallel, all voltages are the same (equipotential) therefore you could
2
immediately use the Capacitor formula & deduce that
Ê
ˆ
Q 2 = C 2 V 2 = ÁÁ 5.0 × 10 −6 F ˜˜ (4.5V ) = 2.25 × 10−6 C
Ë
¯
⇒ Q2 =
PTS: 1
4. ANS: D
C1 and C2 are in parallel so the capacitances ADD: C 12 = C 1 + C 2 = 7.5μF .
The network C12 is in series with C3, so
1
1
1
1
1
0.533
=
+
=
+
=
μF
C series C 12 C 3 7.5μF 2.5μF
Hence the equivalent capacitance C series = 1.88μF . Now apply the "Capacitor Formula"
Ê
ˆ
Q TOT = CV = ÁÁ 1.88 × 10−6 F ˜˜ (12.0V ) = 2.25 × 10 −5 C
Ë
¯
C3 is a lone series element and hence has the same charge, so apply the "Capacitor Formula" again, this time
to find the voltage across C3
Q 3 2.25 × 10 −5 C
V3 =
=
= 9.0V
C3
2.5 × 10−6 F
PTS: 1
2