section v 18 capacitance

7/28/2019 Section v 18 Capacitance

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Capacitance

18. Capacitance

Content

18.1 Capacitors and capacitance

18.2 Energy stored in a capacitor

Learning Outcomes

Candidates should be able to:

(a) show an understanding of the function of capacitors in simple circuits.

(b) define capacitance and the farad.

(c) recall and solve problems using C = Q/V.

(d) derive, using the formula C = Q/V, conservation of charge and the addition of p.ds,

formulae for capacitors in series and in parallel.

(e) solve problems using formulae for capacitors in series and in parallel.

* (f) deduce from the area under a potential-charge graph, the equation W = QV and

hence W = CV2 .


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Capacitors

symbol


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Capacitors - another electrical component

A capacitor is any device that is capable of storing charge.

A practical capacitor is normally made up of two parallel metalplates(called aparallel plate capacitor) separated by an insulatorordielectric e.g. air, mica, waxed paper, polythene, oil etc.

An isolated conductor carrying net charge is considered as storingcharges and hence functioning as a capacitor

A dielectric increases the amount of charge that can be stored

Capacitors are used in electrical and electronic devices same asresistors

Capacitors are generally non-polarity sensitive. However polaritysensitive capacitors are known as electrolytic capacitors andmust be connected with the correct polarity or they will bedamaged

Some uses are to store charge, store energy, coupled with an inductorto tune a radio circuit, to power electromagnets in supercolliders, filterout high frequency radio waves, camera flashes, in computers to

function when there is a power failure, to save valuable data


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Breakdown potential in capacitor dielectrics

When an external voltage applied across a capacitor containing agaseous dielectric is increased gradually, there will reach a potentialwhereby the gas molecules are ionized resulting in electricalconduction and hence a drop in the p.d. across the capacitor.

The potential applied that causes this effect is called the breakdown

potential.


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Capacitance

When an isolated spherical conductor is connected to a highvoltage supply, it is found that as the potential is increased, thecharge stored on the sphere also increases i.e. Q V and hence Q= CV where the gradient C is a constant depending on the size ofthe conductor and known as the capacitance

The capacitance C of a capacitor is defined as the ratio of the

charge Q stored on either plate to the potential difference Vbetween the plates.

Thus: C = Q/V

It can also be defined as the charge stored per unit p.d. applied tothe capacitor. The unit ofCis farads (F).

1 farad is 1 coulomb per volt. i.e F = C V-1

Capacitance is a measure of the charge storing ability of aconductor or the extent to which a capacitor can store charge. Thevalue ofC in radio and hi-fi systems are usually from micro-farads(F = 10-6 F) to pico-farads (pF = 10-12 F).


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Factors affecting capacitance

The material used as a dielectric affects the capacitance of a capacitor

Experiment shows that capacitance is directly proportional to the areaof the plates and inversely proportional to the distance between them

i.e. C A/d

For a capacitor with air or a vacuumbetween the plates, the constantof proportionality is the permittivity of free space0whose value is

8.85 x 10-12 C2 N-1 m-2thus C =0A/dand the unit of permittivity is F m

-1

A quantity called relative permittivityr is introduced to account forthe fact that the use of a dielectric increases the capacitance

The relative permittivity is defined as the capacitance of a parallel

plate capacitor with the dielectric between the plates divided by thecapacitance of the same capacitor with a vacuum between the plates

Including the relative permittivity factor, the full expression for thecapacitance of a parallel plate capacitor is

C =0r A/d


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Relative permittivity of some materials

Materials Relative permittivity

Air 1.0005

Polythene 2.3

Sulphur 4

Paraffin oil 4.7

Mica 6Barium titanate 1200


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Example

A parallel plate air-fi;;ed capacitor has square plates of side 30 cm that

are a distance 1.0 mm apart. Given that 0= 8.85 x 10-12 C2 N-1 m-2

calculate the capacitance of the capacitor.

Solution

Using C =

0

r A/d= 8.0 x 10-10

F


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Q-V graph - relationship between charge and

potential

If the charge and the p.d. is measured at various times of the chargingprocess, the following Q - V graph is obtained.

The gradient is equal to a quantity known as capacitance.

Capacitance C = Q/V

Q

V


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Charging and discharging circuit

A

CV

switch

R


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Charging/Discharging of a parallel plate

capacitor In a circuit where a battery or energy source is connected

through a switch to a parallel plate capacitor and an ammeter inseries, a maximum current I

0is observed initially which then

decreases gradually to zero as time increases.

No current flows through the capacitor. The time period duringwhich the current drops from a maximum value I

0to zero is the

charging stage. When charging a capacitor work is done by the battery to move

charge on to the capacitor hence energy is transferred from thepower supply and stored as electric potential energy in thecapacitor

If this battery is then disconnected and the fully chargedcapacitor is now connected to an external resistor, a maximumcurrent of initial value I

0flows through the circuit in the opposite

direction to that of the charging stage.

The current similarly will drop to zero after a certain time. Thistime period is said to be the discharging stage.

This mechanism is known as induction


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I

+ charging C

O

time

- discharging C


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Induction

In the same circuit earlier with the switch is on, electrons from themetal plate of the capacitor and connecting leads will be attracted to the+ ve terminal of the battery.

Electrons in the connecting leads joined to the ve terminal will bepushed to the other metal plate i.e + ve plate

As time goes by, more electrons are deposited on one plate than theother plate hence becoming more positive.

A potential difference is set up across the plates

This p.d. will become constant when it reaches the e.m.f value of thebattery

The induction process will then stop.

This takes only a fraction of a second if there is negligible resistance inthe circuit.

If a resistor is connected in series, it will slow down the entire process.It will then take a longer time to charge the capacitor


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Arrangement of Capacitors in Circuit

Capacitors can be arranged in a series or parallel manner in a circuit

similar to resistors. Such arrangements will give rise to a resultant or

combined capacitance for each that can be calculated


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Capacitors in Parallel - derivation

For capacitors charged in parallel generally, each capacitor stores a different amountof charge as the current could be different in the different paths.

It can be proved that the effective capacitance C of such connections is given by: C= C

1+ C

2+ C

3+ ... + C

n

For 3 capacitors connected in parallel,

VAB = VCD = VEF= V0 , upon complete charging By the law of conservation of charge, total charge Q for the three capacitors is: Q =

Q1

+ Q2

+ Q3

= C1V

0+ C

2V

0+ C

3V

0since Q = CV

If the effective capacitance is C, then

Q = CV0

= C1V

0+ C

2V

0+ C

3V

0 giving

C = C1 + C2+ C3 i.e in a parallel circuit the effective capacitance is the sum of the individual

capacitances (compare this with resistors)


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Example

Calculate the effective capacitance for

a) 10 F and 40 F in parallel

Ans: 50 F

b) 10 F , 40 F and 8 F in parallel

Ans: 58 F


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Capacitors in Series - derivation

When charged in series, each capacitor stores the same amount ofcharge as the current through all of them is the same.

The effective/combined capacitance C for series connections ofcapacitors is: 1/C = 1/C

1+ 1/C

2+ 1/C

3+ ...+ 1/C

n

For 3 capacitors in series,

V0

= VAB

+ VCD

+ VEF

V0

= Q/C1

+ Q/C2

+ Q/C3

since V = Q/C

Q/C = Q/C1

+ Q/C2

+ Q/C3

1/C = 1/C1

+ 1/C2

+ 1/C3

i.e in a series circuit the effective capacitance is the reciprocal of

the sum of the reciprocals of the individual capacitances (againcompare this with resistors)


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Example

Calculate the effective capacitance for

a) 10 F and 40 F in series

Ans: 8 F

b) 10 F , 40 F and 8 F in series

Ans: 4 F


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Summary of capacitors in series and parallel

For capacitors in parallel, the combined capacitance is greater than the

largest capacitor

For capacitors in series, the combined capacitance is less than the

smallest capacitance For 2 identical capacitance in parallel, the combined capacitance is

equal to double that of a single capacitance

For 2 identical capacitors in series, the combined capacitance is equal

to half that of a single capacitance


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Example

Calculate the charge stored in a capacitor connected to a battery of e.m.f10 V if the capacitance is 470 F.

Solution

Q = CV

= 470 x 10-6 F x 10 V since F = C V-1

= 4.7 x 10-3

= 4.7 mC


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Energy stored

In the Q-V graph, the area under the graph represents the energystored.

E = QV = CV2 = Q2/C = Q2/2C

Theory: if work is needed to separate the charges across the gap, then

work done = Fd

W = Fd = Eqd = qd x V/d = qVHence, total work W = integration of q.dV

= area under graphV

V0

0 Q

0 Q0


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Examples

Calculate the energy stored for a 470 uF capacitor at 10 V.

SolutionE = CV2 = x 470 x 10-6 F x (10 V)2

= 0.0235 J

Try this for the units:

(F x V2 = C V-1 x V2 = C V = J since, energy or work done = Fd = C V)

A camera flash lamp uses a 5000 uF capacitor which is charged by a 9

V battery. Calculate the energy transferred when the capacitor is fully

discharged through the lamp

Solution

E = CV2 = x 5000 x 10-6 x (9)2

= 0.203 J


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Capacitors and resistors on printed circuit

boards


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Resuscitator


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Rating

A capacitor is usually rated as valueF, valueV

e.g. 470 F, 16 V

What does the rating mean?

For every 1 V increase in pd, there will be an increase of 470 C ofcharge

16 V is the max safe operating p.d. beyond which the capacitor will

break down.


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Example

A capacitor of capacitance 220 F is connected to a battery of e.m.f. 20 V.After being fully charged, it is disconnected from the battery and isconnected to an empty capacitor of capacitance 470 F .

What is:

(a) the final voltage across the capacitors.

Ans: 6.38 V

(b) the charge on each capacitor ?

Ans : 1.41 mC; 3 mC


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Example

You are provided with capacitors of ratings 48F, 25 V.

Show the combinations of capacitors to produce

(a) a capacitor of rating 96 F, 25V

(b) a capacitor of rating 24 F, 50V

(c) a capacitor of rating 72 F, 50 V

(d) a capacitor of rating 16.0 F, 75V


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Capacitors in A.C. Circuits

Capacitors are widely used in alternating current and radio circuitsbecause they can transmit alternating currents.

Charge cannot flow through the dielectric of the capacitor due to onedirection flow of the d.c.

When an alternating voltage Vof frequencyfis applied to a capacitor,an a.c. current flows in the circuit even though the capacitor has aninsulator between its plates.

Whenfis high, such as 50 or 100 Hz, a steady currentIflows through

the circuit where :I = fQ sinceI = Q/t


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Time constant CR

The quantity CR in the decay equation may be used to give an

indication of whether the decay is fast or slow and is called the timeconstantof the capacitor-resistor circuit

Since C = Q/V and R = V/I, CR = Q/I = t which is in seconds

To find the charge Q on the capacitor plates after a time t = CR we

substitute t = CR in the exponential equation Q = Q0e-t/CR giving

Q = Q0e-CR/CR = Q0e-1 = Q0/e= Q0/2.718=

Hence the time constant is the time for the charge to have dcecreasedto 1/e (or 1/2.718) of its initial charge

This means that in 1 time constant, the charge stored by the capacitor

drops to roughly one-third of its initial value. In the next time constant

it will drop by the same ratio, to about one-ninth of the value at the

beginning of the decay


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Example

A 500 uF capacitor is connected to a 10 V supply, and is then

discharged through a 100 k resistor. Calculate:a) the initial charge stored by the cpacitor

b) the initial discharge current

c) the value of the time constant

d) the charge on the plates after 100 s

e) the time at which the remaining charge is 2.5 x 10-3 C

Solution

a) Q = CV, so Q = 500 x 10-6 x 10 = 5.0 x 10-3 C

b) I = V/R, so I = 10/(100 x 103) = 1.0 x 10-4 A

c) CR = 500 x 10-6

x 100 x 103

= 50 sd) after 50 s, the charge on the plates is Q0/2.718 = 1.8 x 10

-3 C

after another 50 s, the charge is 1.8 x 10-3/2.718 = 6.8 x 10-4 C

e) using Q = Q0e-t/CR , 2.5 x 10-3 = 5.0 x 10-3e-t/50 or 0.50 = e-t/50

Taking natural log on both sides, -0.693 = -t/50 or t = 35 s


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Worked examples

Define the capacitance of a capacitor

The figure below shows a circuit which is used to determine the capacitance of acapacitorC. The switch S causes the capacitor to be alternatively charged by the

battery and completely discharged through the ammeter. The battery has an e.m.f. of

6.0 V and negligible internal resistance.

(a) Describe, in terms of the flow of charged particles, what happens in the circuit

when the capacitor is being charged.

(b) The switch S is set so that the charge/discharge process occurs 50 times persecond and the steady reading on the ammeter is 0.14 mA. Calculate the capacitance

ofC.

(c) Calculate the maximum energy stored by the capacitor.

AC6 V

switch


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Solution

C = Q/V where Q is the charge and Vis the potential difference. (a) Electrons flow away from the -ve battery terminal or towards the

+ve terminal.

(b) Q = It = 1/f (I) = (0.14 x 10-3)/50 = 2.8 x 10-6 C

C = Q/V= (2.8 x 10-6)/6 = 4.7 x 10-7 F = 0.47 F

(c) Energy stored = CV2 = 0.5 x 4.7 x 10-7 x 62 = 8.4 x 10-6 J


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Summary

Capacitors store charge or energy

Charged by the process of induction

Capacitors have a rating of capacitance and maximum safe voltage

Capacitance = charge/p.d. across plates = C = Q/V Energy stored = area under graph = QV

Capacitance increases in parallel connections but decreases in series

connections

Q = Q0e-t/CR , I = I0e

-t/CR , V = V0e-t/CR


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Exercises

1) A capacitor, C1, of 4 F is charged to a potential difference of 50 V. Another

capacitor, C2, of 6F is charged to a potential difference of 100V.(a) Find the energy stored in each capacitor. [ 5x10-3 J; 30x10-3 J ]

(b) If the two capacitors are now joined, with plates of like charge connectedtogether, what is the final common p.d.? [ 80 V ]

(c) What is the energy stored after the connection?[ 32x10-3 J ]

(d) What happens to the difference in energy stored? [ loss energy = 3x10-3 J ]

2) A 2 F capacitor is charged to a potential of 200 V and then isolated. When it isconnected in parallel with a second capacitor which is initially uncharged, thecommon potential becomes 40 V. Find the capacitance of the second capacitor.

[ 8 F ]3) Given a number of capacitors each with a capacitance of 2F and a maximum

safe working potential difference of 10 V, how would you construct capacitors

of(a) 1 F capacitance, suitable for use up to 20 V

(b) 2 F capacitance, suitable for use up to 20 V?

4) A capacitor of capacitance 160 F is charged to a potential difference of 200 Vand then connected across a discharge tube, which conducts until the potentialdifference across it has fallen to 100 V. Calculate the energy dissipated in thetube. [ 2.4 J ]

section v 18 capacitance

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