Figure 2(a) shows a parallel connection of three capacitors with a voltage applied. Here the total capacitance is easier to find than in the series case. To find the equivalent total capacitance CpCp size 12{ {C} rSub { size 8{p} } } {}, we first note that the voltage across each capacitor is VV size 12{V} {}, the same as that of the source, since they are connected directly to it through a conductor. (Conductors are equipotentials, and so the voltage across the capacitors is the same as that across the voltage source.) Thus the capacitors have the same charges on them as they would have if connected individually to the voltage source. The total charge QQ size 12{Q} {} is the sum of the individual charges:

Q=Q1+Q2+Q3.Q=Q1+Q2+Q3. size 12{Q= {Q} rSub { size 8{1} } + {Q} rSub { size 8{2} } + {Q} rSub { size 8{3} } } {}

(8)Using the relationship Q=CVQ=CV size 12{Q= ital "CV"} {}, we see that the total charge is Q=CpVQ=CpV size 12{Q= {C} rSub { size 8{p} } V} {}, and the individual charges are Q1=C1VQ1=C1V size 12{ {Q} rSub { size 8{1} } = {C} rSub { size 8{1} } V} {}*, *Q2=C2VQ2=C2V size 12{ {Q} rSub { size 8{2} } = {C} rSub { size 8{2} } V} {}*,* and Q3=C3VQ3=C3V size 12{ {Q} rSub { size 8{3} } = {C} rSub { size 8{3} } V} {}. Entering these into the previous equation gives

CpV=C1V+C2V+C3V.CpV=C1V+C2V+C3V. size 12{ {C} rSub { size 8{p} } V= {C} rSub { size 8{1} } V+ {C} rSub { size 8{2} } V+ {C} rSub { size 8{3} } V} {}

(9)Canceling VV size 12{V} {} from the equation, we obtain the equation for the total capacitance in parallel
CpCp size 12{C rSub { size 8{p} } } {}:

Cp=C1+C2+C3+....Cp=C1+C2+C3+.... size 12{ {C} rSub { size 8{p} } = {C} rSub { size 8{1} } + {C} rSub { size 8{2} } + {C} rSub { size 8{3} } + "." "." "." } {}

(10)Total capacitance in parallel is simply the sum of the individual capacitances. (Again the “*...*” indicates the expression is valid for any number of capacitors connected in parallel.) So, for example, if the capacitors in the example above were connected in parallel, their capacitance would be

Cp=1.000 µF+5.000 µF+8.000 µF=14.000 µF.Cp=1.000 µF+5.000 µF+8.000 µF=14.000 µF. size 12{ {C} rSub { size 8{p} } =1 "." "00" µF+5 "." "00" µF+8 "." "00" µF="14" "." 0 µF} {}

(11)The equivalent capacitor for a parallel connection has an effectively larger plate area and, thus, a larger capacitance, as illustrated in Figure 2(b).

Total capacitance in parallel Cp=C1+C2+C3+...Cp=C1+C2+C3+... size 12{ {C} rSub { size 8{p} } = {C} rSub { size 8{1} } + {C} rSub { size 8{2} } + {C} rSub { size 8{3} } + "." "." "." } {}

More complicated connections of capacitors can sometimes be combinations of series and parallel. (See Figure 3.) To find the total capacitance of such combinations, we identify series and parallel parts, compute their capacitances, and then find the total.

Find the total capacitance of the combination of capacitors shown in Figure 3. Assume the capacitances in Figure 3 are known to three decimal places (
C1=1.000 µFC1=1.000 µF, C2=3.000 µFC2=3.000 µF, and C3=8.000 µFC3=8.000 µF), and round your answer to three decimal places.

*Strategy*

To find the total capacitance, we first identify which capacitors are in series and which are in parallel. Capacitors C1C1 size 12{ {C} rSub { size 8{1} } } {} and C2C2 size 12{ {C} rSub { size 8{2} } } {} are in series. Their combination, labeled CSCS size 12{ {C} rSub { size 8{S} } } {} in the figure, is in parallel with C3C3 size 12{ {C} rSub { size 8{3} } } {}.

*Solution*

Since C1C1 size 12{ {C} rSub { size 8{1} } } {} and C2C2 size 12{ {C} rSub { size 8{2} } } {} are in series, their total capacitance is given by 1CS=1C1+1C2+1C31CS=1C1+1C2+1C3 size 12{ { {1} over { {C} rSub { size 8{S} } } } = { {1} over { {C} rSub { size 8{1} } } } + { {1} over { {C} rSub { size 8{2} } } } + { {1} over { {C} rSub { size 8{3} } } } } {}. Entering their values into the equation gives

1
C
S
=
1
C
1
+
1
C
2
=
1
1
.
000
μF
+
1
5
.
000
μF
=
1
.
200
μF
.
1
C
S
=
1
C
1
+
1
C
2
=
1
1
.
000
μF
+
1
5
.
000
μF
=
1
.
200
μF
.
size 12{ { {1} over { {C} rSub { size 8{S} } } } = { {1} over { {C} rSub { size 8{1} } } } + { {1} over { {C} rSub { size 8{2} } } } = { {1} over {1 "." "000"" μF"} } + { {1} over {5 "." "000"" μF"} } = { {1 "." "200"} over {"μF"} } } {}

(12)Inverting gives

CS=0.833 µF.CS=0.833 µF. size 12{ {C} rSub { size 8{S} } =0 "." "833" µF} {}

(13)This equivalent series capacitance is in parallel with the third capacitor; thus, the total is the sum

C
tot
=
C
S
+
C
S
=
0
.
833
μF
+
8
.
000
μF
=
8
.
833
μF.
C
tot
=
C
S
+
C
S
=
0
.
833
μF
+
8
.
000
μF
=
8
.
833
μF.
alignl { stack {
size 12{C rSub { size 8{"tot"} } =C rSub { size 8{S} } +C rSub { size 8{S} } } {} #
=0 "." "833"" μF "+ 8 "." "000"" μF" {} #
=8 "." "833"" μF" {}
} } {}

(14)
*Discussion*

This technique of analyzing the combinations of capacitors piece by piece until a total is obtained can be applied to larger combinations of capacitors.

Comments:"This introductory, algebra-based, two-semester college physics book is grounded with real-world examples, illustrations, and explanations to help students grasp key, fundamental physics concepts. […]"