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```Example: determine the
capacitance of a single capacitor
that will have the same effect as
the combination shown. Use
C1 = C2 = C3 = C.
C2
C1
C3
I don’t see a series combination of capacitors, but I do see a
parallel combination.
C23 = C2 + C3 = C + C = 2C
Now I see a series combination.
1
1
1
=
+
Ceq C1 C23
1
1
1
2
1
3
= +
=
+
=
Ceq C 2C 2C 2C 2C
C eq
2
= C
3
C23 = 2C
C 1= C
Example: for the capacitor circuit shown, C1 = 3F, C2 = 6F, C3
= 2F, and C4 =4F. (a) Find the equivalent capacitance. (b) if
V=12 V, find the potential difference across C4.
C1
I’ll work this at the blackboard.
C2
C4
C3
V
Homework Hint: each capacitor has associated
with it a Q, C, and V. If you don’t know what to do
next, near each capacitor, write down Q= , C= ,
and V= . Next to the = sign record the known
value or a “?” if you don’t know the value. As soon
as you know any two of Q, C, and V, you can
determine the third. This technique often provides
visual clues about what to do next.
(a) Find Ceq. (b) if V=12 V, find V4.
C1=3F
C2=6F
C4=4F
C3=2F
C1 and C3 are not in parallel. Make
sure you understand why!
C2 and C4 are not in series. Make
sure you understand why!
V=12 V
C1 and C2 are in series. Make sure you use the correct equation!
1
1
1
1 1
2 1
3
1
=
+
=
+
=
+
=
=
C12
C1 C 2
3 6
6 6
6
2
Don’t forget to invert: C12 = 2 F.
(a) Find Ceq. (b) if V=12 V, find V4.
C12=2F
C4=4F
C12 and C4 are not in series. Make
sure you understand why!
C3=2F
V=12 V
C12 and C3 are in parallel. Make sure you use the correct
equation!
C123 = C12 + C3 = 2 + 2 = 4μF
(a) Find Ceq. (b) if V=12 V, find V4.
C123=4F
C4=4F
C123 and C4 are in series. Make
sure you understand why!
Combined, they make give Ceq.
V=12 V
Make sure you use the correct equation!
1
1
1
1 1
2
1
=
+
=
+
=
=
Ceq
C123 C 24
4 4
4
2
Don’t forget to invert: Ceq = 2 F.
(a) Find Ceq. (b) if V=12 V, find V4.
Ceq=2F
Ceq = 2 F.
V=12 V
If you see a capacitor circuit on the test, read the problem first.
Don’t go rushing off to calculate Ceq. Sometimes you are asked
to do other things.
Truth in advertising: there’s a high probability you will need to calculate Ceq at some point in the problem.
(a) Find Ceq. (b) if V=12 V, find V4.
Q1=?
C1=3F
V1=?
Q2=?
C2=6F Q =?
4
V2=? C =4F
4
V4=?
Q3=?
C3=2F
V3=?
Homework Hint: each capacitor has associated
with it a Q, C, and V. If you don’t know what to do
next, near each capacitor, write down Q= , C= ,
and V= . Next to the = sign record the known
value or a “?” if you don’t know the value. As soon
as you know any two of Q, C, and V, you can
determine the third. This technique often provides
visual clues about what to do next.
V=12 V
We know C4 and want to find V4. If we know Q4 we can
calculate V4. Maybe that is a good way to proceed.
(a) Find Ceq. (b) if V=12 V, find V4.
Q123=?
C123=4F
V123=?
Q4=?
C4=4F
V4=?
C4 is in series with C123 and
together they form Ceq.
Therefore Q4 = Q123 = Qeq.
V=12 V
Qeq = Ceq V =
C=
2 12 =
Q
Q
 V=
V
C
24μC = Q4
 V4 =
Q4
24
=
= 6V
C4
4
You really need to know this:
Capacitors in series…
all have the same charge
add the voltages to get the total voltage
Capacitors in parallel…
all have the same voltage
add the charges to get the total charge
(and it would be nice if you could explain why)
Homework Hint!
C1
What does our text mean by Vab?
C2
C4
b
a
Our text’s convention is Vab = Va – Vb.
This is explained on page 759. This is
in contrast to Physics 1135 notation,
where Vab = Vb – Va.
C3
V
In the figure on this slide, if Vab = 100 V then point a is at a
potential 100 volts higher than point b, and Vab = -100 V;
there is a 100 volt drop on going from a to b.
A “toy” to play with…