The Capacitor

A capacitor is an electronic device which can store charge.

The symbol for a capacitor is:

Capacitance is measured in Farads (F) or, more usually, mF and F.

means x10 -6

Notes p.15

Switch Open

Switch Closed

0V

0V

After a While!

6V

6V

As the capacitor charges up, the voltage across it increases so the voltage across the resistor decreases to 0V - and the current decreases to 0A.

RELATIONSHIP BETWEEN CHARGE AND APPLIED P.D.AIM To show that the charge stored on a capacitor is proportional to the applied voltage.MethodThe following circuit was set up

VC

1. With switch in position A the capacitor charges. The charging voltage is recorded from the voltmeter.2. Switch is moved to position B and the coulomb meter records the charge stored on the capacitor.3. This procedure is repeated as the voltage is increased in regular steps.

ResultsVoltage (V) Charge (C) Charge (C)

Voltage (V)

Graph of Charge (C) against Voltage (V) CapacitanceFrom the graph we can see that charge is directly proportional to voltage.

Q VSo … Q = C V

This constant is “Capacitance” and is measured in Farads. One Farad is a big value so it is normally measured in milli Farads (mF) or micro Farads (F).

Q

C V

Q

V

Memorise – mF = x 10-3

F = x 10-6

Q = charge in coulombs (C)V = voltage in volts (V)C = capacitance in farads (F)

Q

C V

Calculations Using Q = CVIn Q = CV calculations you will also often have to use the SG equation …

Q = I t

Problems p. 19 Q. 1 - 6

OLD Problems p.66, Q. 1-67.a) 0.005 C b) 1.25 A8. 5 x 10-7 F = 0.5 F9. a) 40 V b) 16.7 % c) 46.7 V (or 48V using actual C = (30 – 5) F = 25

F)10. 22.5 C = 2.25 x 10-5 C11.a) 1 x 10-6 F = 1 F b) 8 x 10-7 C = 0.8

C12. 50 F = 5 x 10-5 F

Energy Stored in a Capacitor

The electrical energy stored in a capacitor is given by …

E = ½ Q V

(Recall for an electric field at constant voltage … W = QV)

This is because the voltage steadily increases from 0V to the supply.Q

V

(Area = ½ QV)

Energy Stored in a Capacitor (cont…)

By substituting Q = CV into this equation we can obtain 2 other equations that relate energy stored to the capacitance…

Why Must Work Be Done to Charge a Capacitor?

As electrons collect on one plate of the capacitor they repel any more electrons moving on to the plate.This repulsion force must be overcome for more electrons to collect on the plate so “work has to be done”.This “Work Done” is the energy stored in the capacitor ….. E = ½ Q V …. Area under Q vs V graph.

Observe the virtual demonstration of “Energy Stored in a Capacitor”.

Over to you – Problems p.20 Q. 7 – 11

Capacitor – Resistor Circuits (“CR Circuits”)

** During charging Vc + VR = Vsupply **Charging

Vc

t

Vsupply

VR

t

Vsupply

I

t

Imax

Vsupply

I

VC

VR

** Imax = Vsupply / R

** During discharging the capacitor acts like a battery with a decreasing voltage and the charges move in the opposite (negative!) direction.**

Discharging

+++ +++

- - - - - -Vc

VR

So, as the voltage across the capacitor decreases to zero, the voltage across the resistor does the same.

+++ +++

- - - - - -Vc

VR

Discharging (cont.)t

Vc

t

Vsupply

VR

t

Vsupply I

Imax

0

0 0

All four graphs can be summarised as follows -

Now read typed notes pages 14, 15 and 16.

Example calculation on capacitor/resistor circuits

During charging the voltages across each component aremonitored

VR8 K

1000 F VC

12 V

8k

1000F

The graph below shows how the voltage across the capacitor, VC, varies with time from the instant switch S is closed.

a) Draw the graph showing how VR varies over the time 0s –

7s. Include values on both axes.

1 2 3 4 65 7 time (s)

p.dacrossthe capacitor(V)

0123456789101112

b) Calculate the initial charging current.

c) Calculate the current after 7s.

d) When the capacitor is fully charged it is removed from the circuit and connected across a 15 resistor to discharge.How much energy is dissipated in the resistor.

VR = 12 V R = 8000

I = VR R = 12 8000 = 0.0015 A = 1.5 mA

VR = 2 V R = 8000

I = VR R = 2 8000 = 2.5 x 10-4 A = 0.25 mA

Vc = 12 V C = 1000 x 10-6 F E = ?

E = ½ CV2

= ½ 1000 x 10-6 x 122

= 0.072 JNB The value of R is irrelevant

Frequency Effects

In a RESISTIVE CIRCUIT (with no capacitor) the frequency of an a.c. supply has NO EFFECT on current.

In a CAPACITIVE CIRCUIT the current is directly proportional to the frequency of the a.c. supply.

Frequency of ac supplyCurrent

Frequency of ac supplyCurrent

Uses of CapacitorsRead pages 194 & 195 of “Higher Physics for CfE”Some uses of capacitors are:

•Storing charge (eg flash lamps)•Smoothing ac to dc

•Blocking dc (but allowing ac)

ProblemsAll problems to complete up to p.26, Q. 17.

OLD Problems 14-2314. 0.6J 15. a) initially 1.2mA then drops to zero b) ammmeter reads zero c) range 0 – 2mA d) no effect on Vmax but takes longer to charge and

discharge.16. (i) The capacitor can charge up from 0 to 100V while the bulb

does not conduct. At 100V the bulb conducts so the capacitor discharges

through it making it light. This makes bthe pd across C drop again.

When it reaches 80V the bulb stops conducting (goes off) so C charges up again to 100V and process repeats so bulb flashes.

(ii) To decrease flash rate INCREASE value of C or R.17. a) 0.005 F b) -- 5.26mF 5.00mF 4.84mF 4.94mF c) C = (5.01 ± 0.11) mF18. a) (i) position 1 (ii) position 2 b) see jotter for graphs

OLD Problems 14-23 (cont.)17. a) 0.005 F b) -- 5.26mF 5.00mF 4.84mF 4.94mF c) C = (5.01 ± 0.11) mF18. a) (i) position 1 (ii) position 2 b) see jotter for graphs c ) (i) 40 A or 4 x 10-5 A (ii) 400 C or 4 x 10-4 C 19. a) (i) VC graph rising in curve from 0V to 3V

(ii) VR graph decreasing in curve from 3V to 0Vb) final voltage across C = 3V …final charge stored = 9C or 9 x 10-6 C

20. (i) Circuit 1 no effect - Circuit 2 current increases (ii) Circuit 1 no effect - Circuit 2 current decreases