W5 Circuits containing Capacitance

Physics 212 S.I. — Fall 2014
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S.I. Leader: K. Oskar
Circuits Containing Capacitors
We first consider a parallel connection.
If a battery supplies a potential difference V to points a and b, this same potential difference,
V = Vab , exists across each of the capacitors. That is, since the left hand plates of all the
capacitors are connected by conductors, they all reach the same potential Va when connected
to the battery; and the right-hand plates each reach potential Vb . Each capacitor plate
acquires a charge given by Q1 = C1 V , Q2 = C2 V , and Q3 = C3 V . The total charge Q that
must leave the battery is then
Q = Q1 + Q2 + Q3
= C1 V + C2 V + C3 V
Let us try to find a single equivalent capacitor that will hold the same charge Q at the same
voltage V = Vab . It will have a capacitance given by Ceq where Q = Ceq V . Thus,
Ceq = C1 V + C2 V + C3 V
= (C1 + C2 + C3 )V
So, Ceq = C1 + C2 + C3 .
We can consider the case where we have n capacitors in parallel with the ith capacitor
has a capacitance of Ci (i.e., the capacitance of the first is C1 for the second is C2 and so
on), and using the procedure as before we would get
Ceq =
n
X
i=1
Ci = C 1 + C2 + · · · + Cn .
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Physics 212 S.I., Fall 2014, S.I. Leader: K. Oskar
Capacitors can also be connected in series: that is, end to end.
A +Q charge flows from the battery to one plate of C1 , and Q flows to one plate of C1 .
The regions A and B between the capacitors were originally neutral, so the net charge there
must still be zero. The +Q on the left plate of C1 attracts a charge of −Q on the opposite
plate. Because region A must have a zero net charge, there is +Q on the left plate of C2 .
The same considerations apply to the other capacitors, so we see that the charge on each
capacitor plate has the same magnitude Q. A single capacitor that could replace these three
in series without affecting the circuit (that is, Q and V the same) would have a capacitance
Ceq where
Q = Ceq V
The total voltage V across the three capacitors in series must equal the sum of the voltages
across each capacitor: V = V1 + V2 + V3 . We also have for each capacitor Q = C1 V1 ,
Q = C2 V2 , and Q = C3 V3 , so we substitute for V1 , V2 , V3 , and V
Q
Q
Q
Q
=
+
+
Ceq
C1 C2 C3
1
1
1
= Q(
+
+
)
C1 C2 C3
So C1eq = C11 + C12 + C13 . Again, if we have n capacitors in series we would get that, using the
same procedure as before,
n
X 1
1
1
1
1
=
=
+
+ ··· +
Ceq
Ci
C1 C2
Cn
i=1
2
Problems
1 Six 4.8 − µF capacitors are connected in parallel. What is the equivalent capacitance?
What is their equivalent capacitance if connected in series?
2 A 3.00µF and a 4.00µF capacitor are connected in series, and this combination is
connected in parallel with a 2.00µF capacitor. What is the net capacitance?
Physics 212 S.I., Fall 2014, S.I. Leader: K. Oskar
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3 If 21.0 V is applied across the whole network,
calculate the voltage across each capacitor and the charge on each capacitor.
4 An electric circuit was accidentally constructed using a 7µF capacitor instead of the
required 16µF value. Without removing the 7.0µF capacitor, what can a technician
add to correct this circuit?
5 Determine the equivalent capacitance between points a and b for the combination of
capacitors
6 Two capacitors connected in parallel produce an equivalent capacitance of 35.0µF
but when connected in series the equivalent capacitance is only 4.6µF . What is the
individual capacitance of each capacitor?
7 A circuit contains a single 250-pF capacitor hooked across a battery. It is desired to
store four times as much energy in a combination of two capacitors by adding a single
capacitor to this one. How would you hook it up, and what would its value be?

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