College Physics for AP® Courses

# 19.6Capacitors in Series and Parallel

College Physics for AP® Courses19.6 Capacitors in Series and Parallel

### Learning Objectives

By the end of this section, you will be able to:

• Derive expressions for total capacitance in series and in parallel.
• Identify series and parallel parts in the combination of connection of capacitors.
• Calculate the effective capacitance in series and parallel given individual capacitances.

Several capacitors may be connected together in a variety of applications. Multiple connections of capacitors act like a single equivalent capacitor. The total capacitance of this equivalent single capacitor depends both on the individual capacitors and how they are connected. There are two simple and common types of connections, called series and parallel, for which we can easily calculate the total capacitance. Certain more complicated connections can also be related to combinations of series and parallel.

### Capacitance in Series

Figure 19.26(a) shows a series connection of three capacitors with a voltage applied. As for any capacitor, the capacitance of the combination is related to charge and voltage by $C=QVC=QV size 12{C= { {Q} over {V} } } {}$.

Note in Figure 19.26 that opposite charges of magnitude $QQ size 12{Q} {}$ flow to either side of the originally uncharged combination of capacitors when the voltage $VV size 12{V} {}$ is applied. Conservation of charge requires that equal-magnitude charges be created on the plates of the individual capacitors, since charge is only being separated in these originally neutral devices. The end result is that the combination resembles a single capacitor with an effective plate separation greater than that of the individual capacitors alone. (See Figure 19.26(b).) Larger plate separation means smaller capacitance. It is a general feature of series connections of capacitors that the total capacitance is less than any of the individual capacitances.

Figure 19.26 (a) Capacitors connected in series. The magnitude of the charge on each plate is $Q Q$. (b) An equivalent capacitor has a larger plate separation $d d size 12{d} {}$. Series connections produce a total capacitance that is less than that of any of the individual capacitors.

We can find an expression for the total capacitance by considering the voltage across the individual capacitors shown in Figure 19.26. Solving $C=QVC=QV size 12{C= { {Q} over {V} } } {}$ for $VV size 12{V} {}$ gives $V=QCV=QC size 12{V= { {Q} over {C} } } {}$. The voltages across the individual capacitors are thus $V1=QC1V1=QC1 size 12{ {V} rSub { size 8{1} } = { {Q} over { {C} rSub { size 8{1} } } } } {}$, $V2=QC2V2=QC2 size 12{ {V} rSub { size 8{2} } = { {Q} over { {C} rSub { size 8{2} } } } } {}$, and $V3=QC3V3=QC3 size 12{ {V} rSub { size 8{3} } = { {Q} over { {C} rSub { size 8{3} } } } } {}$. The total voltage is the sum of the individual voltages:

$V=V1+V2+V3.V=V1+V2+V3. size 12{V= {V} rSub { size 8{1} } + {V} rSub { size 8{2} } + {V} rSub { size 8{3} } } {}$
19.60

Now, calling the total capacitance $CSCS size 12{C rSub { size 8{S} } } {}$ for series capacitance, consider that

$V = Q C S = V 1 + V 2 + V 3 . V = Q C S = V 1 + V 2 + V 3 . size 12{V= { {Q} over { {C} rSub { size 8{S} } } } = {V} rSub { size 8{1} } + {V} rSub { size 8{2} } + {V} rSub { size 8{3} } } {}$
19.61

Entering the expressions for $V1V1 size 12{V rSub { size 8{1} } } {}$, $V2V2 size 12{V rSub { size 8{2} } } {}$, and $V3V3 size 12{V rSub { size 8{3} } } {}$, we get

$QCS=QC1+QC2+QC3.QCS=QC1+QC2+QC3. size 12{ { {Q} over { {C} rSub { size 8{S} } } } = { {Q} over { {C} rSub { size 8{1} } } } + { {Q} over { {C} rSub { size 8{2} } } } + { {Q} over { {C} rSub { size 8{3} } } } } {}$
19.62

Canceling the $QQ size 12{Q} {}$s, we obtain the equation for the total capacitance in series $CSCS size 12{ {C} rSub { size 8{S} } } {}$ to be

$1CS=1C1+1C2+1C3+...,1CS=1C1+1C2+1C3+..., size 12{ { {1} over { {C} rSub { size 8{S} } } } = { {1} over { {C} rSub { size 8{1} } } } + { {1} over { {C} rSub { size 8{2} } } } + { {1} over { {C} rSub { size 8{3} } } } + "." "." "." } {}$
19.63

where “...” indicates that the expression is valid for any number of capacitors connected in series. An expression of this form always results in a total capacitance $CSCS size 12{ {C} rSub { size 8{S} } } {}$ that is less than any of the individual capacitances $C1C1 size 12{ {C} rSub { size 8{1} } } {}$, $C2C2 size 12{ {C} rSub { size 8{2} } } {}$, ..., as the next example illustrates.

### Total Capacitance in Series, $C s C s size 12{ {C} rSub { size 8{S} } } {}$

Total capacitance in series: $1CS=1C1+1C2+1C3+...1CS=1C1+1C2+1C3+... size 12{ { {1} over { {C} rSub { size 8{S} } } } = { {1} over { {C} rSub { size 8{1} } } } + { {1} over { {C} rSub { size 8{2} } } } + { {1} over { {C} rSub { size 8{3} } } } + "." "." "." } {}$

### Example 19.9

#### What Is the Series Capacitance?

Find the total capacitance for three capacitors connected in series, given their individual capacitances are 1.000, 5.000, and 8.000 $µFµF size 12{mF} {}$.

#### Strategy

With the given information, the total capacitance can be found using the equation for capacitance in series.

#### Solution

Entering the given capacitances into the expression for $1CS1CS size 12{ { {1} over { {C} rSub { size 8{S} } } } } {}$ gives $1CS=1C1+1C2+1C31CS=1C1+1C2+1C3 size 12{ { {1} over { {C} rSub { size 8{S} } } } = { {1} over { {C} rSub { size 8{1} } } } + { {1} over { {C} rSub { size 8{2} } } } + { {1} over { {C} rSub { size 8{3} } } } } {}$.

$1 C S = 1 1 . 000 µF + 1 5 . 000 µF + 1 8 . 000 µF = 1 . 325 µF 1 C S = 1 1 . 000 µF + 1 5 . 000 µF + 1 8 . 000 µF = 1 . 325 µF size 12{ { {1} over { {C} rSub { size 8{S} } } } = { {1} over {1 "." "00" mF} } + { {1} over {5 "." "00" mF} } + { {1} over {8 "." "00" mF} } = { {1 "." "325"} over {mF} } } {}$
19.64

Inverting to find $CSCS size 12{C rSub { size 8{S} } } {}$ yields ${}$$CS=µF1.325=0.755 µFCS=µF1.325=0.755 µF size 12{ {C} rSub { size 8{S} } = { {mF} over {1 "." "325"} } =0 "." "755" mF} {}$.

#### Discussion

The total series capacitance $CsCs size 12{ {C} rSub { size 8{S} } } {}$ is less than the smallest individual capacitance, as promised. In series connections of capacitors, the sum is less than the parts. In fact, it is less than any individual. Note that it is sometimes possible, and more convenient, to solve an equation like the above by finding the least common denominator, which in this case (showing only whole-number calculations) is 40. Thus,

$1CS=4040 µF+840 µF+540 µF=5340 µF,1CS=4040 µF+840 µF+540 µF=5340 µF, size 12{ { {1} over { {C} rSub { size 8{S} } } } = { {"40"} over {"40" mF} } + { {8} over {"40" mF} } + { {5} over {"40" mF} } = { {"53"} over {"40" mF} } } {}$
19.65

so that

$CS=40 µF53=0.755 µF.CS=40 µF53=0.755 µF. size 12{ {C} rSub { size 8{S} } = { {"40" µF} over {"53"} } =0 "." "755" µF} {}$
19.66

### Capacitors in Parallel

Figure 19.27(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:

$Q=Q1+Q2+Q3.Q=Q1+Q2+Q3. size 12{Q= {Q} rSub { size 8{1} } + {Q} rSub { size 8{2} } + {Q} rSub { size 8{3} } } {}$
19.67
Figure 19.27 (a) Capacitors in parallel. Each is connected directly to the voltage source just as if it were all alone, and so the total capacitance in parallel is just the sum of the individual capacitances. (b) The equivalent capacitor has a larger plate area and can therefore hold more charge than the individual capacitors.

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

$CpV=C1V+C2V+C3V.CpV=C1V+C2V+C3V. size 12{ {C} rSub { size 8{p} } V= {C} rSub { size 8{1} } V+ {C} rSub { size 8{2} } V+ {C} rSub { size 8{3} } V} {}$
19.68

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

$Cp=C1+C2+C3+....Cp=C1+C2+C3+.... size 12{ {C} rSub { size 8{p} } = {C} rSub { size 8{1} } + {C} rSub { size 8{2} } + {C} rSub { size 8{3} } + "." "." "." } {}$
19.69

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

$Cp=1.000 µF+5.000 µF+8.000 µF=14.000 µF.Cp=1.000 µF+5.000 µF+8.000 µF=14.000 µF. size 12{ {C} rSub { size 8{p} } =1 "." "00" µF+5 "." "00" µF+8 "." "00" µF="14" "." 0 µF} {}$
19.70

The equivalent capacitor for a parallel connection has an effectively larger plate area and, thus, a larger capacitance, as illustrated in Figure 19.27(b).

### Total Capacitance in Parallel, $C p C p size 12{C rSub { size 8{p} } } {}$

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 19.28.) To find the total capacitance of such combinations, we identify series and parallel parts, compute their capacitances, and then find the total.

Figure 19.28 (a) This circuit contains both series and parallel connections of capacitors. See Example 19.10 for the calculation of the overall capacitance of the circuit. (b) $C 1 C 1 size 12{ {C} rSub { size 8{1} } } {}$ and $C 2 C 2 size 12{ {C} rSub { size 8{2} } } {}$ are in series; their equivalent capacitance $C S C S size 12{ {C} rSub { size 8{S} } } {}$ is less than either of them. (c) Note that $C S C S size 12{ {C} rSub { size 8{S} } } {}$ is in parallel with $C 3 C 3 size 12{ {C} rSub { size 8{3} } } {}$. The total capacitance is, thus, the sum of $C S C S size 12{ {C} rSub { size 8{S} } } {}$ and $C 3 C 3 size 12{ {C} rSub { size 8{3} } } {}$.

### Example 19.10

#### A Mixture of Series and Parallel Capacitance

Find the total capacitance of the combination of capacitors shown in Figure 19.28. Assume the capacitances in Figure 19.28 are known to three decimal places ( $C1=1.000 µFC1=1.000 µF$, $C2=5.000 µFC2=5.000 µF$, 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

$1 C S = 1 C 1 + 1 C 2 = 1 1 . 000 μF + 1 5 . 000 μF = 1 . 200 μF . 1 C S = 1 C 1 + 1 C 2 = 1 1 . 000 μF + 1 5 . 000 μF = 1 . 200 μF . size 12{ { {1} over { {C} rSub { size 8{S} } } } = { {1} over { {C} rSub { size 8{1} } } } + { {1} over { {C} rSub { size 8{2} } } } = { {1} over {1 "." "000"" μF"} } + { {1} over {5 "." "000"" μF"} } = { {1 "." "200"} over {"μF"} } } {}$
19.71

Inverting gives

$CS=0.833 µF.CS=0.833 µF. size 12{ {C} rSub { size 8{S} } =0 "." "833" µF} {}$
19.72

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" {} } } {}
19.73

#### 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.

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