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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.

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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.

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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

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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?

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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

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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"} {}

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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.
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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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Questions & Answers

Why is there no 2nd harmonic in the classical electron orbit?
Shree Reply
how to reform magnet after been demagneted
Inuwa Reply
A petrol engine has a output of 20 kilowatts and uses 4.5 kg of fuel for each hour of running. The energy given out when 1 kg of petrol is burnt is 4.8 × 10 to the power of 7 Joules. a) What is the energy output of the engine every hour? b) What is the energy input of the engine every hour?
Morris Reply
what is the error during taking work done of a body..
Aliyu Reply
what kind of error do you think? and work is held by which force?
Daniela
I am now in this group
smart
theory,laws,principles and what-a-view are not defined. why? you
Douglas Reply
A simple pendulum is used in a physics laboratory experiment to obtain an experimental value for the gravitational acceleration, g . A student measures the length of the pendulum to be 0.510 meters, displaces it 10 o from the equilibrium position, and releases it. Using a s
Emmanuel Reply
so what question are you passing across... sir?
Olalekan
Two masses of 2 kg and 4 kg are held with a compressed spring between them. If the masses are released, the spring will push them away from each other. If the smaller mass moves off with a velocity of 6m/s, what is the stored energy in the spring when it is compressed?
Emmanuel Reply
54 joule
babar
how?
rakesh
Reduce that two body problem into one body problem. Apply potential and k. E formula to get total energy of the system
rakesh
i dont think dere is any potential energy... by d virtue of no height present
Olalekan
there is compressed energy,dats only potential energy na?
rakesh
yes.. but... how will u approach that question without The Height in the question?
Olalekan
Can you explain how you get 54J?
Emmanuel
Because mine is 36J
Emmanuel
got 36J too
Douglas
OK the answer is 54J Babar is correct
Emmanuel
Conservation of Momentum
Emmanuel
woow i see.. can you give the formula for this
joshua
Two masses of 2 kg and 4 kg are held with a compressed spring between them. If the masses are released, the spring will push them away from each other. If the smaller mass moves off with a velocity of 6m/s, what is the stored energy in the spring when it is compressed? Asume there is no external force.
Emmanuel Reply
Please help!
Emmanuel
please help find dy/dx 2x-y/x+y
Inuwa
By using the Quotient Rule dy/dx = 3y/(x +y)²
Emmanuel
3y/(x+y)²
Emmanuel
may be by using MC^2=MC^2 and Total energy=kinetic energy +potential energy so 1st find kinetic energy and den find potential energy which is stored energy
rakesh
i think i m correct
rakesh
But how?
Emmanuel
3y/(x+y)²
Douglas
what's the big bang?
kwame Reply
yes what is it?
LamaBbake
it is the explanation of how the universe began
Zainab
yes
Ana
explain
Chinagorom
in
Chinagorom
it is a theory on how the universe began. to understand more I would suggest researching the topic online.
david
thanks guys
kwame
if a force of 12N is applied to load of 200g what us the work done
Joshua Reply
We can seek accelation first
Nancy
we are given f=12 m=200g which is 0.2kg now from 2nd law of newton a= f/m=60m/s*2 work done=force applied x displacement cos (theta) w= 12x60 =720nm/s*2
Mudang
this very interesting question very complicated for me, í need urgent help. 1,two buses A and B travel along the same road in the same direction from Harper city (asume They both started from the same point) to Monrovia. if bus A maintains a Speedy of 60km/h and bus B a Speedy of 75km/h, how many
mohammed
hours Will it take bus B to overtake bus A assuming bus B starts One hour after bus A started. what is the distance travelled by the buses when They meet?.
mohammed
pls í need help
mohammed
4000 work is done
Ana
speed=distance /time distance=speed/time
Ana
now use this formula
Ana
what's the answer then
Julius
great Mudang
Kossi
please Ana explain 4000 ?
babar
hey mudang there is a product of force and acceleration not force and displacement
babar
@Mohammed answer is 0.8hours or 48mins
Douglas
nice
A.d
its not possible
Olalekan
í want the working procedure
mohammed
the answer is given but how Will One arrive at it. the answers are 4hours and 300m.
mohammed
physics is the science that studies the non living nature
isidor Reply
ancient greek language physis = nature
isidor
what is phyacs
technical Reply
if i am going to start studying physics where should i start?
BRIAN Reply
I think from kinematics
Nancy
You can find physics books at the library or online. That's how I started.
Chelsea
And yes, kinematics is usually where you can begin.
Chelsea
study basic algebra and calculus and can start from classical mechanics
Mudang
yes think so but dimension is the best starting point
Obed
3 formula's of equations of motion
benjamin Reply
vf=vi+at........1 s=vit+1/2(at)2 vf2=vi2+2as
Ana
solve the formula's please
benjamin
those are the three .. what you wanna solve ?
Nihrantz
For first equation simply integrate formula of acceleration in the limit v and u
Tripti
For second itegrate velocity formula by ising first equation
Tripti
similarly for 3 one integrate acceleration again by multiplying and dividing term ds
Tripti
any methods can take to solve this eqtions
a=vf-vi/t vf-vi=at vf=vi+at......1
Ana
suppose a body starts with an initial velocity vi and travels with uniform acceleration a for a period of time t.the distance covered by a body in this time is "s" and its final velocity becomes vf
Ana
what is the question dear
Zeeshan
average velocity=(vi+vf)/2 distance travelled=average velocity ×time therefore s=vi+vf/2×t from the first equation of motion ,we have vf =vi+at s=[vi+(vi+at)]/2×t s=(2vi+at)/2×t s=bit+1/2at2
Ana
find the distance
Ana
how
Zeeshan
Two speakers are arranged so that sound waves with the same frequency are produced and radiated through a room. An interference pattern is created. Calculate the distance between the two speakers?
Hayne Reply
How can we calculate without any information?
Amir
I think the formulae used for this question is lambda=(ax)/D
Amir

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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