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Capacitors in Series and Parallel

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Capacitors in Series and Parallel
CHAPTER 19 | ELECTRIC POTENTIAL AND ELECTRIC FIELD
We will find in Atomic Physics that the orbits of electrons are more properly viewed as electron clouds with the density of the cloud related to the
probability of finding an electron in that location (as opposed to the definite locations and paths of planets in their orbits around the Sun). This cloud is
shifted by the Coulomb force so that the atom on average has a separation of charge. Although the atom remains neutral, it can now be the source of
a Coulomb force, since a charge brought near the atom will be closer to one type of charge than the other.
Some molecules, such as those of water, have an inherent separation of charge and are thus called polar molecules. Figure 19.19 illustrates the
⎛
⎞
separation of charge in a water molecule, which has two hydrogen atoms and one oxygen atom ⎝H 2 O⎠ . The water molecule is not symmetric—the
hydrogen atoms are repelled to one side, giving the molecule a boomerang shape. The electrons in a water molecule are more concentrated around
the more highly charged oxygen nucleus than around the hydrogen nuclei. This makes the oxygen end of the molecule slightly negative and leaves
the hydrogen ends slightly positive. The inherent separation of charge in polar molecules makes it easier to align them with external fields and
charges. Polar molecules therefore exhibit greater polarization effects and have greater dielectric constants. Those who study chemistry will find that
the polar nature of water has many effects. For example, water molecules gather ions much more effectively because they have an electric field and
a separation of charge to attract charges of both signs. Also, as brought out in the previous chapter, polar water provides a shield or screening of the
electric fields in the highly charged molecules of interest in biological systems.
Figure 19.19 Artist’s conception of a water molecule. There is an inherent separation of charge, and so water is a polar molecule. Electrons in the molecule are attracted to the
oxygen nucleus and leave an excess of positive charge near the two hydrogen nuclei. (Note that the schematic on the right is a rough illustration of the distribution of electrons
in the water molecule. It does not show the actual numbers of protons and electrons involved in the structure.)
PhET Explorations: Capacitor Lab
Explore how a capacitor works! Change the size of the plates and add a dielectric to see the effect on capacitance. Change the voltage and see
charges built up on the plates. Observe the electric field in the capacitor. Measure the voltage and the electric field.
Figure 19.20 Capacitor Lab (http://cnx.org/content/m42333/1.4/capacitor-lab_en.jar)
19.6 Capacitors in Series and Parallel
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 19.21(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=
Q
.
V
Note in Figure 19.21 that opposite charges of magnitude
Q flow to either side of the originally uncharged combination of capacitors when the
voltage 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 19.21(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.
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684
CHAPTER 19 | ELECTRIC POTENTIAL AND ELECTRIC FIELD
Figure 19.21 (a) Capacitors connected in series. The magnitude of the charge on each plate is
Q . (b) An equivalent capacitor has a larger plate separation 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 19.21. Solving
C=
Q
Q
Q
Q
Q
for V gives V =
. The voltages across the individual capacitors are thus V 1 =
, V2 =
, and V 3 =
. The total voltage is
V
C
C1
C2
C3
the sum of the individual voltages:
V = V 1 + V 2 + V 3.
Now, calling the total capacitance
C S for series capacitance, consider that
V=
Entering the expressions for
Q
= V1 + V2 + V3 .
CS
(19.61)
V 1 , V 2 , and V 3 , we get
Q
Q
Q
Q
=
+
+ .
CS C1 C2 C3
Canceling the
(19.60)
(19.62)
Q s, we obtain the equation for the total capacitance in series C S to be
1 = 1 + 1 + 1 + ...,
CS C1 C2 C3
(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 C S that is less than any of the individual capacitances C 1 , C 2 , ..., as the next example illustrates.
Total Capacitance in Series,
Total capacitance in series:
Cs
1 = 1 + 1 + 1 + ...
CS C1 C2 C3
This content is available for free at http://cnx.org/content/col11406/1.7
CHAPTER 19 | ELECTRIC POTENTIAL AND ELECTRIC FIELD
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
µF .
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
1 gives 1 = 1 + 1 + 1 .
CS
CS C1 C2 C3
1 =
1
1
1
+
+
= 1.325
µF
C S 1.000 µF 5.000 µF 8.000 µF
Inverting to find
C S yields C S =
(19.64)
µF
= 0.755 µF .
1.325
Discussion
C 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,
The total series capacitance
1 = 40 + 8 + 5 = 53 ,
C S 40 µF 40 µF 40 µF 40 µF
(19.65)
so that
CS =
40 µF
= 0.755 µF.
53
(19.66)
Capacitors in Parallel
Figure 19.22(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 C p , we first note that the voltage across each capacitor is 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 Q is the sum of the individual charges:
Q = Q 1 + Q 2 + Q 3.
(19.67)
Figure 19.22 (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 = CV , we see that the total charge is Q = C pV , and the individual charges are Q 1 = C 1V , Q 2 = C 2V , and
Q 3 = C 3V . Entering these into the previous equation gives
C p V = C 1V + C 2V + C 3V.
Canceling
(19.68)
V from the equation, we obtain the equation for the total capacitance in parallel C p :
C p = C 1 + C 2 + C 3 + ....
(19.69)
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