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Capacitors in Series and Parallel

Science and Technology

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

[link](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=QV size 12{C= { {Q} over {V} } } {}.

Note in [link] that opposite charges of magnitude Q size 12{Q} {} flow to either side of the originally uncharged combination of capacitors when the voltage V 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 [link](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.

(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 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 [link]. Solving C=QV size 12{C= { {Q} over {V} } } {} for V size 12{V} {} gives V=QC size 12{V= { {Q} over {C} } } {}. The voltages across the individual capacitors are thus V1=QC1 size 12{ {V} rSub { size 8{1} } = { {Q} over { {C} rSub { size 8{1} } } } } {}, V2=QC2 size 12{ {V} rSub { size 8{2} } = { {Q} over { {C} rSub { size 8{2} } } } } {}, and V3=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. size 12{V= {V} rSub { size 8{1} } + {V} rSub { size 8{2} } + {V} rSub { size 8{3} } } {}

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

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} } } {}

Entering the expressions for V1 size 12{V rSub { size 8{1} } } {}, V2 size 12{V rSub { size 8{2} } } {}, and V3 size 12{V rSub { size 8{3} } } {}, we get

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} } } } } {}

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

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} } } } + "." "." "." } {}

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 CS size 12{ {C} rSub { size 8{S} } } {} that is less than any of the individual capacitances C1 size 12{ {C} rSub { size 8{1} } } {}, C2 size 12{ {C} rSub { size 8{2} } } {}, ..., as the next example illustrates.

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 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 1CS size 12{ { {1} over { {C} rSub { size 8{S} } } } } {} gives 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} } } } } {}.

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} } } {}

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

Discussion

The total series capacitance Cs 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, 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} } } {}

so that

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

Capacitors in Parallel

[link](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 Cp size 12{ {C} rSub { size 8{p} } } {}, we first note that the voltage across each capacitor is V 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 Q size 12{Q} {} is the sum of the individual charges:

Q=Q1+Q2+Q3. size 12{Q= {Q} rSub { size 8{1} } + {Q} rSub { size 8{2} } + {Q} rSub { size 8{3} } } {}
(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 size 12{Q= ital "CV"} {}, we see that the total charge is Q=CpV size 12{Q= {C} rSub { size 8{p} } V} {}, and the individual charges are Q1=C1V size 12{ {Q} rSub { size 8{1} } = {C} rSub { size 8{1} } V} {}, Q2=C2V size 12{ {Q} rSub { size 8{2} } = {C} rSub { size 8{2} } V} {}, and Q3=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. 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} {}

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

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} } + "." "." "." } {}

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. size 12{ {C} rSub { size 8{p} } =1 "." "00" µF+5 "." "00" µF+8 "." "00" µF="14" "." 0 µF} {}

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

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

(a) This circuit contains both series and parallel connections of capacitors. See [link] for the calculation of the overall capacitance of the circuit. (b) C1 size 12{ {C} rSub { size 8{1} } } {} and C2 size 12{ {C} rSub { size 8{2} } } {} are in series; their equivalent capacitance CS size 12{ {C} rSub { size 8{S} } } {} is less than either of them. (c) Note that CS size 12{ {C} rSub { size 8{S} } } {} is in parallel with C3 size 12{ {C} rSub { size 8{3} } } {}. The total capacitance is, thus, the sum of CS size 12{ {C} rSub { size 8{S} } } {} and C3 size 12{ {C} rSub { size 8{3} } } {}.
A Mixture of Series and Parallel Capacitance

Find the total capacitance of the combination of capacitors shown in [link]. Assume the capacitances in [link] are known to three decimal places ( C1=1.000 µF, C2=3.000 µF, and C3=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 C1 size 12{ {C} rSub { size 8{1} } } {} and C2 size 12{ {C} rSub { size 8{2} } } {} are in series. Their combination, labeled CS size 12{ {C} rSub { size 8{S} } } {} in the figure, is in parallel with C3 size 12{ {C} rSub { size 8{3} } } {}.

Solution

Since C1 size 12{ {C} rSub { size 8{1} } } {} and C2 size 12{ {C} rSub { size 8{2} } } {} are in series, their total capacitance is given by 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} } } } } {}. 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 . 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"} } } {}

Inverting gives

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

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. 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" {} } } {}

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.

Section Summary

  • Total capacitance in series 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} } } } + "." "." "." } {}
  • Total capacitance in parallel 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} } + "." "." "." } {}
  • If a circuit contains a combination of capacitors in series and parallel, identify series and parallel parts, compute their capacitances, and then find the total.

Conceptual Questions

If you wish to store a large amount of energy in a capacitor bank, would you connect capacitors in series or parallel? Explain.

Problems & Exercises

Find the total capacitance of the combination of capacitors in [link].

A combination of series and parallel connections of capacitors.

0.293 μF

Suppose you want a capacitor bank with a total capacitance of 0.750 F and you possess numerous 1.50 mF capacitors. What is the smallest number you could hook together to achieve your goal, and how would you connect them?

What total capacitances can you make by connecting a 5.00 µF size 12{8 "." "00" mF} {} and an 8.00 µF size 12{8 "." "00" mF} {} capacitor together?

3.08 µF size 12{3 "." "08" µF } {} in series combination, 13.0 µF size 12{"13" "." "0 "µF} {} in parallel combination

Find the total capacitance of the combination of capacitors shown in [link].

A combination of series and parallel connections of capacitors.

2 . 79 µF size 12{2 "." "79"" µF"} {}

Find the total capacitance of the combination of capacitors shown in [link].

A combination of series and parallel connections of capacitors.

Unreasonable Results

(a) An 8.00 µF size 12{8 "." "00" mF} {} capacitor is connected in parallel to another capacitor, producing a total capacitance of 5.00 µF size 12{5 "." "00" mF} {}. What is the capacitance of the second capacitor? (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent?

(a) –3.00 µF size 12{8 "." "00" mF} {}

(b) You cannot have a negative value of capacitance.

(c) The assumption that the capacitors were hooked up in parallel, rather than in series, was incorrect. A parallel connection always produces a greater capacitance, while here a smaller capacitance was assumed. This could happen only if the capacitors are connected in series.

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