<< Chapter < Page Chapter >> Page >
  • 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

[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 = Q V 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.

When capacitors are connected in series, an equivalent capacitor would have a plate separation that is greater than that of any individual capacitor. Hence the series connections produce a resultant capacitance less than that of the individual capacitors.
(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 = Q V size 12{C= { {Q} over {V} } } {} for V size 12{V} {} gives V = Q C size 12{V= { {Q} over {C} } } {} . The voltages across the individual capacitors are thus V 1 = Q C 1 size 12{ {V} rSub { size 8{1} } = { {Q} over { {C} rSub { size 8{1} } } } } {} , V 2 = Q C 2 size 12{ {V} rSub { size 8{2} } = { {Q} over { {C} rSub { size 8{2} } } } } {} , and V 3 = Q C 3 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 = V 1 + V 2 + V 3 . size 12{V= {V} rSub { size 8{1} } + {V} rSub { size 8{2} } + {V} rSub { size 8{3} } } {}

Now, calling the total capacitance C S 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 V 1 size 12{V rSub { size 8{1} } } {} , V 2 size 12{V rSub { size 8{2} } } {} , and V 3 size 12{V rSub { size 8{3} } } {} , we get

Q C S = Q C 1 + Q C 2 + Q C 3 . 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 C S size 12{ {C} rSub { size 8{S} } } {} to be

1 C S = 1 C 1 + 1 C 2 + 1 C 3 + . . . , 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 C S size 12{ {C} rSub { size 8{S} } } {} that is less than any of the individual capacitances C 1 size 12{ {C} rSub { size 8{1} } } {} , C 2 size 12{ {C} rSub { size 8{2} } } {} , ..., as the next example illustrates.

Total capacitance in series, C s size 12{ {C} rSub { size 8{S} } } {}

Total capacitance in series: 1 C S = 1 C 1 + 1 C 2 + 1 C 3 + . . . 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} } } } + "." "." "." } {}

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 1 C S size 12{ { {1} over { {C} rSub { size 8{S} } } } } {} gives 1 C S = 1 C 1 + 1 C 2 + 1 C 3 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 C S size 12{C rSub { size 8{S} } } {} yields {} C S = µF 1 . 325 = 0 . 755 µF size 12{ {C} rSub { size 8{S} } = { {mF} over {1 "." "325"} } =0 "." "755" mF} {} .

Discussion

The total series capacitance C s 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,

1 C S = 40 40 µF + 8 40 µF + 5 40 µF = 53 40 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

C S = 40 µF 53 = 0 . 755 µF . size 12{ {C} rSub { size 8{S} } = { {"40" µF} over {"53"} } =0 "." "755" µF} {}
Got questions? Get instant answers now!

Questions & Answers

material that allows electric current to pass through
Deng Reply
material which don't allow electric current is called
Deng
insulators
Covenant
how to study physic and understand
Ewa Reply
what is conservative force with examples
Moses
what is work
Fredrick Reply
the transfer of energy by a force that causes an object to be displaced; the product of the component of the force in the direction of the displacement and the magnitude of the displacement
AI-Robot
why is it from light to gravity
Esther Reply
difference between model and theory
Esther
Is the ship moving at a constant velocity?
Kamogelo Reply
The full note of modern physics
aluet Reply
introduction to applications of nuclear physics
aluet Reply
the explanation is not in full details
Moses Reply
I need more explanation or all about kinematics
Moses
yes
zephaniah
I need more explanation or all about nuclear physics
aluet
Show that the equal masses particles emarge from collision at right angle by making explicit used of fact that momentum is a vector quantity
Muhammad Reply
yh
Isaac
A wave is described by the function D(x,t)=(1.6cm) sin[(1.2cm^-1(x+6.8cm/st] what are:a.Amplitude b. wavelength c. wave number d. frequency e. period f. velocity of speed.
Majok Reply
what is frontier of physics
Somto Reply
A body is projected upward at an angle 45° 18minutes with the horizontal with an initial speed of 40km per second. In hoe many seconds will the body reach the ground then how far from the point of projection will it strike. At what angle will the horizontal will strike
Gufraan Reply
Suppose hydrogen and oxygen are diffusing through air. A small amount of each is released simultaneously. How much time passes before the hydrogen is 1.00 s ahead of the oxygen? Such differences in arrival times are used as an analytical tool in gas chromatography.
Ezekiel Reply
please explain
Samuel
what's the definition of physics
Mobolaji Reply
what is physics
Nangun Reply
the science concerned with describing the interactions of energy, matter, space, and time; it is especially interested in what fundamental mechanisms underlie every phenomenon
AI-Robot

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play Download on the App Store Now




Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'College physics' conversation and receive update notifications?

Ask