6a - Capa Remix

8/13/2019 6a - Capa Remix

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Capacitors


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Capacitors

A capacitor is a device used to store electric charge.It consists of two metal plates (conductors) separated

by some insulating material called the dielectric whichis the energy storage.

There are two (2) types of Capacitors

1. Electrolytic (aka. Polarized caps)

2. Non-electrolytic


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Application

Capacitors are used in a variety of applications in theelectronics world. Examples of its usage can be found

in:Tuning radio circuitsSupercapacitors are replacing car batteriesRechargeable batteries

Smoothing (rectification)Cameras (flash)


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Capacitance (C)

The ratio of the charge (Q) on the plates of thecapacitor to the potential difference (V) across

them.

=

S.I Unit: 1 Farad (F) = 1 C/V 1 Farad is the charge required to increase the

potential difference across the plates by 1 V.


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Parallel-plate Capacitors

o= Permittivity of empty space (airo)= 8.854x10

-12C

2/N.m

2= 1/(4k)

= Permittivity of material between plates

A= Surface Area of one of the plates (SI: m2)

d = Separation of the plates (SI: m)


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Table of Dielectric constants ORRelative Permittivity (K)


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Relative Permittivity (r or K)

The relative permittivity of the dielectric is givenby:

=0

; where

C0= Capacitance of capacitor with vacuum between theplates

C = Capacitance with Dielectric between the plates

But C0= 0A/d , C = A/d

Thus, r= /0


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Tips to Note The amount of Capacitance that a Capacitor has

depends on three (3) characteristics:

1. The distance (d) between the plates of theCapacitor (m)

2. The area (A) of the plates of the Capacitor (m2)3. The insulating material (dielectric) placed in

between the plates.


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Batteries vs CapacitorsBatteries Capacitors

The voltage depends on the

electrochemical reactionsbetween the anode andcathode

Eventually die out

The voltage is directly

proportional to the chargestored

Has the ability to be rechargedmany times before it loses itscapacity.


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Pq1

(a) A capacitor of capacitance 5 mF is connectedto a 6 V supply. What charge is stored in thecapacitor?

Ans: Q =30C

(b) A 400 pF capacitor carries a charge of

2.5 x 10-8

C. What is the potential differenceacross the plates of the capacitor?

Ans:V= 62.5V


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Pq2

A capacitor is charged such that there is a chargeof +20 mC on the positive plate. What is thecharge on the negative plate?

Ans: -20mC


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Pq3 A 4700 F capacitor is connected as shown in the circuit diagram.

When it is fully charged:

(a)What is the charge on the positive plate of the capacitor? (b) What is the potential difference across the capacitor?

(c) How many additional electrons are on the negative plate?


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Solution Pq3

C = 4700F V = 4.5V a) When the Capacitor is fully charged, the

voltage across the plates is equal to thevoltage of the source supplying the charge.

Q = CV Q = 4700 x 10-6x 4.5 Q = 0.021C or 21mC


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Solution Pq3

b) the voltage across the plates is 4.5V as thecapacitor is fully charged

c) the charge on an individual electron is1.6x10-19C and the current flowing is 0.021A.

Therefore the number of additional electronsis given by, # of electrons = 0.021A / 1.6 x 10-19C

= 1.3 x 1017electrons


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Why use a Dielectric? The Dielectric is any insulating material that is

placed in between the two plates of the capacitor

that:1. Reduce the potential difference across the

plates2. Reduce the Electric Field strength between the

plates3. Increase the Capacitance of the Capacitor,

which allows for an increase in the storage ofcharge.


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Action of the Dielectric


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Capacitors in Series

Let C1, C2and C3be the Capacitances of threeCapacitors in Series The charge(Q) supplied to each capacitor will be the

samebut the voltage(V) across each is what varies.


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V = V1+ V2+ V3 But V = Q/C from the definition of

Capacitance, Thus, Q/CT= Q/C1 + Q/C2+ Q/C3

Therefore for Capacitors in Series we have,

1/CT= 1/C1+ 1/C2+ 1/C3 CT= total Capacitance


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Capacitors in Parallel

Let C1, C2and C3be the Capacitances of threecapacitors in Parallel. In this setup, the voltage across each Capacitor is

the same, however, the charge supplied to each will

be different


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Q = Q1+Q2+ Q3 But Q=CV from the definition of Capacitance

Thus, CTV = C1V + C2V +C3V

Therefore for Capacitors in Parallel we have,

CT= C1+ C2 + C3


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Energy storage in Capacitors


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Energy Stored in a Capacitor The energy of a charged capacitor is stored in the

electric field between the plates. There are (3)

equations that govern the energy stored in acapacitor:

=

2

= 2

=

2

W = energy stored in capacitor(Joules, J)C = Capacitance of Capacitor, (Farad, F)V = Voltage across capacitor, V (volts, V)Q = Charge stored on the plates of capacitor(Coulomb, C)


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Q-V Graphs

1. The area under a Charge(Q) vs Voltage (V) graph orvice versa, represents theenergy stored by a capacitor.

Area = bh= QV = W


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Tip to Note From W = CV2, it can be seen that if the pd is

doubled, the energy stored goes up by a factor of

four; energy stored is proportional to the square ofthe pd.

The gradient of a charge (Q) against voltage (V) plot

is equal to the Capacitance (C)of the capacitor.

The gradient of V/V against Q gives 1/C.


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Pq The graph shows the relation between the charge on a capacitor

and the pd across it.

Use the graph to find the capacitance of the capacitor.

2. Use the graph to find the energy stored, first when the pd across the capacitor is

4.0 V and again when the pd is 6.0 V.


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Solution 1) The gradient of the graph gives the Capacitance

(C),

Gradient = (25 x 10-30/6 0) = 4.2mF

2) When the pd is 4.0V the charge stored is 17mC,thus the energy stored is,

W = QV = x 17 x10-3x4.0 = 0.034J Likewise, when p.d is 6.0V the charge stored is 25mC,

thus energy = x25x 10-3x6.0 = 0.075J


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Charging & Discharging

Capacitors


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Charging a Capacitor

As the capacitor charges:

charge Q flows onto the plate connected to the negative

terminal of the supply

charge Q flows off the plate connected to the positive terminalof the supply, leaving it with charge +Q

the capacitor plates always have the same quantity of charge, butof the opposite sign

no charge flows between the plates of the capacitor.


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Example: Discharging through a Lamp

With the switch at A, the Capacitor starts to charge until thevoltage across it equals the supply voltage. At this point,that capacitor cannot contain any more charge and thecurrent flowing is zero. Moving the switch to B, Thecapacitor now discharges through the lamp causing it to bebrightly lit at first then grow dimmer over time.


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Graphs of Charge(Q) vs time (t)


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Graphs of voltage(V) against time (t)


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Time Constant (, tau or T) The time taken for a Capacitor to drop to 1/e or 37% of the

supply voltage, charge and current is known as the time constantof the capacitor. It depends on the resistance (R)andCapacitance (C)in the circuit.

= ; or T = time constant(s) R = Resistor value()

C = Capacitor value (F)


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Half-life of a Capacitor (t1/2)

The half-lifeof a RC circuit is the time taken forthe charge or potential difference (p.d) across itor the discharge current through it to drop to

half on any initial value.


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Half-life of capacitor


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Tips to Note All graphs vary exponentiallywith time After:

1 time constant we have 1/e of Q, V and I 2 time constants 1/e2(e-2) 3 time constants 1/e3(e-3) etc. Time constant is just an indication of how quickly the

capacitor discharges. The greater the time constantthe longer it will take to discharge

Time constant can be compared to the concept ofhalf-life in radioactivity as they both undergo anexponential decay( the equations are similar).


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Time Constant comparison


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Equations for Charging & Discharging

CapacitorsCharging Discharging

Q = Q0(1- et/RC)

= (

)

= (

)

Q = Q0et/RC

=

=


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Pq1a) Calculate the current through theresistor when the switch is first closed.

b) What is the current after the switch hasbeen closed for a long time? Explain youranswer.

c) Calculate the current through theresistor when the pd across the capacitor is

2.0 V.


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Solution (Pq1) a) I = V/R = 6/ 470 x 103

I = 1.3 x 10-5 A

b)When the switch is first closed, thecurrent flowing in the circuit is a maximum.As the capacitor charges up to the p.d of the

battery, the current gradually drops to zero.This happens because the p.d in the circuit isnow zero (5V across battery 5V acrosscapacitor = 0V).


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Solution (Pq1) c) when the p.d across the capacitor is 2.0V,

the potential difference across the resistor

is (6V 2V = 4V). So the current flowing atthis time is,

I = V/R

I = 4/ 470 x 103

I = 8.5 x 10-6A


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Pq2 A student sets up the circuit shown in the

diagram.


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Pq2 cont (a)

(i) She moves the switch from X to Y. Explain what happens to thecapacitor.

(ii) Sketch a graph to show how current varies with time from themoment the switch touches Y. Indicate typical values of current andtime on the axes of your graph.

(b) Calculate the maximum energy stored on the capacitor in this

circuit.

(c) The student wants to produce a time delay equal to the time ittakes for the potential difference across the capacitor to fall to 0.07of its maximum value. Calculate this time delay.


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Solution Pq2 a) i) When the switch is moved from X to Y

the capacitor starts to discharge

exponentially. ii)


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Solution Pq2 b) Max energy stored = CV2

= (10 x 10-6) x (10)2= 5.0 x 10-4J

c) V =V0e-t/RC;V = 0.07V0& V0= 10V, RC = 0.05s0.7 = 10 e-t/(0.05)

0.07 = e-t/(0.05)Taking the natural log (ln) of both sides,ln (0.07) = -t / 0.05-2.7 = -t /0.05; t = 0.13s


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Web links http://www.splung.com/content/sid/3/page/ca

pacitors

http://www.physics.sjsu.edu/becker/physics51/capacitors.htm
http://www.splung.com/content/sid/3/page/capacitorshttp://www.splung.com/content/sid/3/page/capacitorshttp://www.physics.sjsu.edu/becker/physics51/capacitors.htmhttp://www.physics.sjsu.edu/becker/physics51/capacitors.htmhttp://www.physics.sjsu.edu/becker/physics51/capacitors.htmhttp://www.physics.sjsu.edu/becker/physics51/capacitors.htmhttp://www.splung.com/content/sid/3/page/capacitorshttp://www.splung.com/content/sid/3/page/capacitorshttp://www.splung.com/content/sid/3/page/capacitors

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