We will now start drawing electric circuits. These drawings are called schematics because they are a representation of the circuit in an electrical sense, not a physical sense. Electric circuits are composed of circuit elements (batteries, capacitors, resistors, etc.) connected together by conductors represented by lines. Symbols representing capacitors, batteries, and resistors are shown in section 16.7 of the text.
We want to consider how to treat combinations of capacitors. We begin by considering cases with two capacitors combined, and from these we can build up any arbitrarily complex combination. There are only two ways to combine two capacitors, we call them parallel combination and series combination. We will derive an expression for a single capacitance that is equivalent to the combination of capacitors, and in the process we will apply what we have learned about charges, potential difference (voltage), and conductors.
(Draw a parallel combination with capacitors C1 and C2 connected to a battery.)
In a parallel combination, the capacitors are usually drawn side by side.
If we imagine them as parallel-plate capacitors with the same gap, snuggling them right up next to each other, the combination seems to become a single capacitor with an area equal to the sum of the areas.
Then from the equation for capacitance of a parallel-plate capacitor, we have
Ceq = e0Aeq/d = e0(A1+A2)/d = e0A1/d + e0A2/d = C1 + C2
Thus, Ceq = C1 + C2.
Let's do this again, but this time thinking in terms of charges and voltage because it will probe useful for the series combination.
When connected in parallel, the voltage across each capacitor is the same: DV1 = DV2 = DV.
The charge on each capacitor is
Q1 = C1DV and Q2 = C2DV
So the total charge is
Q = Q1 + Q2 = C1DV + C2DV = ( C1 + C2 )DV
The equivalent capacitance is defined to be
Ceq = Q/DV = C1 + C2
the same result as above.
It is pretty clear how to extend this result to more than two capacitors in parallel. In that case
(Draw a series combination.) In a series combination, the capacitors are connected head-to-tail. We want to replace the pair by a single equivalent capacitor (draw equivalent capacitor circuit). To do this, we must understand how the charge is distributed on the plates.
Consider the inner pair of plates, one from each capacitor, connected by a conductor. These three objects are electrically isolated from the remainder of the circuit -- they form a single isolated conductor. Since the net charge on the capacitors is zero before the battery is connected, the net charge on the inner pair of plates must also be zero. After the battery is connected, the plates of the capacitors will hold some charge, but the inner pair of plates will still have zero net charge. Therefore, the charges on the inner pair of plates is equal and opposite, and we see that both capacitors will hold the same charge.
We don't add these charges together, as in the parallel case.
The quantity that adds is the voltage across each capacitor
DV = DV1 + DV2
The voltage across the capacitors is related to their charge
DV1 = Q/C1 and DV2 = Q/C2
The definition of the equivalent capacitance is Ceq = Q/DV, or DV = Q/Ceq.
Therefore
Q/Ceq = Q/C1 + Q/C2
We can divide out the common factor of Q to arrive at
It is not too difficult to see that for three or more capacitors in series, the equivalent capacitance is given by:
Consider the combination of capacitors in Figure P16.31. (a) What is the equivalent capacitance of the group? (b) Determine the charge on each capacitor.
(a) This circuit consists of a three parallel branches, the first two branches are single capacitors, and the last branch is a series combination of two capacitors.
Begin by finding the equivalent capacitance of those two, then the equivalent capacitance of the three parallel combination can be found.
1/C3eq = 1/24µF + 1/8.0µF = 1/6.0µF
Therefore, C3eq = 6.0µF.
Remember to take the inverse to get the equivalent capacitance of a series combination.
Now, the parallel combination gives:
Ceq = 4.0µF + 2.0µF + 6.0µF = 12.0µF
(b) The charge on each depends on the voltage across the capacitor.
The voltage across the 4 and 2µF capacitors is the full 36V of the battery.
Q4 = (4.0µF)(36V) = 144µC
Q2 = (2.0µF)(36V) = 72µC
The charge on the 2 series capacitors is the same, and is equal to the charge that would exist on their equivalent value.
Q24 = Q8 = Q3eq = (6.0µF)(36V) = 216µC
To determine how much energy is stored on a capacitor, we imagine starting with a neutral capacitor, and adding a little charge, then a little more, then a little more, each time adding up how much energy is required. The first bit of charge takes no energy, but the next little bit must be added to a capacitor that's now has a potential difference, V = CDQ, and requires an energy VDQ. For the next bit of charge, the potential is a bit higher, and a little more energy is needed. When all the energy is summed up, the total energy required to fully charge a capacitor to a potential difference V is:
Consider the parallel-plate capacitor formed by the Earth and a cloud layer as described in problem 22. Assume this capacitor will discharge (lightning) when the electric field strength between the plates reaches 3.0×106N/C. What is the energy released if the capacitor discharges completely during a lightning strike?
From last lecture, we found the capacitance of the cloud 800m above the Earth to be C = 1.1×10-8 F.
When the capacitor discharges completely, the energy stored is zero.
Therefore, the energy released is equal to the energy stored before it discharges.
Since we previously determined that the charge stored is Q = 26 C, let's use Estored = Q² / 2 C to determine the energy stored:
Estored = (26 C)² / 2(1.1×10-8 F) =
Note that we could also have calculated this result using Estored = ½CV², and V = Ed.
Estored = ½C(Ed)² = ½(1.1×10-8 F)(3.0×106N/C)²(800m)² =
A dielectric is an insulating material. When such an insulating material is placed between the plates of a capacitor, the atoms and molecules tend to polarize, and the capacitance increases. The entire effect is handled by a number called the dielectric constant for the material used, k. A list of dielectric constants is given in Table 16.1 in the text. When a dielectric is used, the capacitance is changed from what it would be with no dielectric material by: