1. Home
  2. College physics
  3. Electric potential and electric
  4. Capacitors in series and parallel

19.6 Capacitors in series and parallel  (Page 2/2)

<< Chapter < Page
  College physics     Page 2 / 2
Chapter >> Page >

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 C p 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 = Q 1 + Q 2 + Q 3 . size 12{Q= {Q} rSub { size 8{1} } + {Q} rSub { size 8{2} } + {Q} rSub { size 8{3} } } {}
Part a of the figure shows three capacitors connected in parallel to each other and to the applied voltage. The total capacitance when they are connected in parallel is simply the sum of the individual capacitances. Part b of the figure shows the larger equivalent plate area of the capacitors connected in parallel, which in turn can hold more charge than the individual capacitors.
(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 = C p V size 12{Q= {C} rSub { size 8{p} } V} {} , and the individual charges are Q 1 = C 1 V size 12{ {Q} rSub { size 8{1} } = {C} rSub { size 8{1} } V} {} , Q 2 = C 2 V size 12{ {Q} rSub { size 8{2} } = {C} rSub { size 8{2} } V} {} , and Q 3 = C 3 V size 12{ {Q} rSub { size 8{3} } = {C} rSub { size 8{3} } V} {} . Entering these into the previous equation gives

C p V = C 1 V + C 2 V + C 3 V . 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 C p size 12{C rSub { size 8{p} } } {} :

C p = C 1 + C 2 + C 3 + . . . . 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

C p = 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).

Total capacitance in parallel, C p size 12{C rSub { size 8{p} } } {}

Total capacitance in parallel C p = C 1 + C 2 + C 3 + . . . 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 [link] .) To find the total capacitance of such combinations, we identify series and parallel parts, compute their capacitances, and then find the total.

The first figure has two capacitors, C sub1 and C sub2 in series and the third capacitor C sub 3 is parallel to C sub 1 and C sub 2. The second figure shows C sub S, the equivalent capacitance of C sub 1 and C sub 2, in parallel to C sub 3. The third figure represents the total capacitance of C sub S and C sub 3.
(a) This circuit contains both series and parallel connections of capacitors. See [link] for the calculation of the overall capacitance of the circuit. (b) C 1 size 12{ {C} rSub { size 8{1} } } {} and C 2 size 12{ {C} rSub { size 8{2} } } {} are in series; their equivalent capacitance C S size 12{ {C} rSub { size 8{S} } } {} is less than either of them. (c) Note that C S size 12{ {C} rSub { size 8{S} } } {} is in parallel with C 3 size 12{ {C} rSub { size 8{3} } } {} . The total capacitance is, thus, the sum of C S size 12{ {C} rSub { size 8{S} } } {} and C 3 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 ( C 1 = 1.000 µF , C 2 = 5.000 µF , and C 3 = 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 C 1 size 12{ {C} rSub { size 8{1} } } {} and C 2 size 12{ {C} rSub { size 8{2} } } {} are in series. Their combination, labeled C S size 12{ {C} rSub { size 8{S} } } {} in the figure, is in parallel with C 3 size 12{ {C} rSub { size 8{3} } } {} .

Solution

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

C S = 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.

Got questions? Get instant answers now!

Section summary

  • 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} } } } + "." "." "." } {}
  • Total capacitance in parallel C p = C 1 + C 2 + C 3 + . . . 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.

Got questions? Get instant answers now!

Problems&Exercises

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

A circuit is shown with three capacitors. Two capacitors, of ten microfarad and two point five microfarad capacitance, are in parallel to each other, and their combination is in series with a zero point three zero microfarad capacitor.
A combination of series and parallel connections of capacitors.

0.293 μF

Got questions? Get instant answers now!

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?

Got questions? Get instant answers now!

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

Got questions? Get instant answers now!

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

The circuit includes three capacitors. A zero point three zero microfarad capacitor and a ten microfarad capacitor are connected in series, and together they are connected in parallel with a two point five microfarad capacitor.
A combination of series and parallel connections of capacitors.

2 . 79 µF size 12{2 "." "79"" µF"} {}

Got questions? Get instant answers now!

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

The figure shows a circuit that is a combination of series and parallel connections of capacitors. On the left of the circuit is a five point zero microfarad capacitor in series with a three point five microfarad capacitor. In the middle is an eight point zero microfarad capacitor. On the right, a zero point seven five microfarad capacitor is in parallel with a fifteen microfarad capacitor, and together they are in series with a one point five microfarad capacitor. Altogether, the system of capacitors on the left, the capacitor in the middle, and the system of capacitors on the right are connected in parallel.
A combination of series and parallel connections of capacitors.
Got questions? Get instant answers now!

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.

Got questions? Get instant answers now!

Questions & Answers

it is the relatively stable flow of income
Chidubem Reply
what is circular flow of income
Divine Reply
branches of macroeconomics
SHEDRACK Reply
what is Flexible exchang rate?
poudel Reply
is gdp a reliable measurement of wealth
Atega Reply
introduction to econometrics
Husseini Reply
Hi
mostafa
hi
LEMLEM
hello
Sammol
hi
Mahesh
bi
Ruqayat
hi
Ruqayat
Hi fellas
Nyawa
hey
Sammol
hi
God
hello
Jahara
Good morning
Jorge
hi
abubakar
hi
Nmesoma
hi
Mahesh
Hi
Tom
Why is unemployment rate never zero at full employment?
Priyanka Reply
bcoz of existence of frictional unemployment in our economy.
Umashankar
what is flexible exchang rate?
poudel
due to existence of the pple with disabilities
Abdulraufu
the demand of a good rises, causing the demand for another good to fall
Rushawn Reply
is it possible to leave every good at the same level
Joseph
I don't think so. because check it, if the demand for chicken increases, people will no longer consume fish like they used to causing a fall in the demand for fish
Anuolu
is not really possible to let the value of a goods to be same at the same time.....
Salome
Suppose the inflation rate is 6%, does it mean that all the goods you purchase will cost 6% more than previous year? Provide with reasoning.
Geetha Reply
Not necessarily. To measure the inflation rate economists normally use an averaged price index of a basket of certain goods. So if you purchase goods included in the basket, you will notice that you pay 6% more, otherwise not necessarily.
Waeth
discus major problems of macroeconomics
Alii Reply
what is the problem of macroeconomics
Yoal
Economic growth Stable prices and low unemployment
Ephraim
explain inflationcause and itis degre
Miresa Reply
what is inflation
Getu
increase in general price levels
WEETO
Good day How do I calculate this question: C= 100+5yd G= 2000 T= 2000 I(planned)=200. Suppose the actual output is 3000. What is the level of planned expenditures at this level of output?
Chisomo Reply
how to calculate actual output?
Chisomo
how to calculate the equilibrium income
Beshir
Criteria for determining money supply
Thapase Reply
who we can define macroeconomics in one line
Muhammad
Aggregate demand
Mohammed
C=k100 +9y and i=k50.calculate the equilibrium level of output
Mercy Reply
Hi
Isiaka
Hi
Geli
hy
Man
👋
Bahunda
hy how are you?
Man
ys
Amisha
how are you guys
Sekou
f9 guys
Amisha
how are you guys
Sekou
ys am also fine
Amisha
fine and you guys
Geli
from Nepal
Amisha
nawalparasi district from belatari
Amisha
nd u
Amisha
I am Camara from Guinea west Africa... happy to meet you guys here
Sekou
ma management ho
Amisha
ahile becheclor ho
Amisha
hjr ktm bta ho ani k kaam grnu hunxa tw
Amisha
belatari
Amisha
1st year ho
Amisha
nd u
Amisha
ahh
Amisha
kaha biratnagar
Amisha
ys
Amisha
kina k vo
Amisha
money as unit of account means what?
Kalombe
A unit of account is something that can be used to value goods and services and make calculations
Jim
all of you please speak in English I can't understand you're language
Muhammad
I want to know how can we define macroeconomics in one line
Muhammad
it must be .9 or 0.9 no Mpc is greater than 1 Y=100+.9Y+50 Y-.9Y=150 0.1Y/0.1=150/0.1 Y=1500
Kalombe
Mercy is it clear?😋
Kalombe
hi can someone help me on this question If a negative shocks shifts the IS curve to the left, what type of policy do you suggest so as to stabilize the level of output? discuss your answer using appropriate graph.
Galge Reply
if interest rate is increased this will will reduce the level of income shifting the curve to the left ◀️
Kalombe
Got questions? Join the online conversation and get instant answers!
Jobilize.com Reply
<< Chapter < Page Page > Chapter >>

Read also:

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