Higher 2016 J. A. Hargreaves Lockerbie Academy March 2016

ELECTRICITY Part 1

Content

Content

Content	2

Data Sheet	5

Relationships Required For Higher Physics	6

CHAPTER 1: SUMMARY ELECTRICITY CONTENT	7

Summary of electricity content	7

CHAPTER 2: BACKGROUND TO ELECTRICITY	9

Prefixes	9

Definitions	9

Measuring Current and Voltage	10

Series and Parallel: Circuit Rules	10

Resistance in series and parallel	11

Ohm’s Law	12

Practical 1- the Van de Graaff	13

Practical 2 – Resistors in Series Practical	13

Practical 3- Resistance in Parallel	14

Revision Tutorials	14

Tutorial 1: Electric Fields	14

Tutorial 2 Circuits	15

Tutorial 3- Resistance	15

Tutorial Answers	16

Worked Answers for tutorials	16

CHAPTER 3: MONITORING AND MEASURING A.C	18

Summary of content	18

Measuring the Frequency	18

Peak Voltage and R.M.S.	19

Peak Voltage and R.M.S.	20

Practical 1: Comparing bulb brightness with an a.c and d.c supply.	22

Practical 2: Checking the equation for RMS current:	23

Tutorial 1: alternating current & voltage	24

Practical 3 – electron gun	24

Practical 4- energy is measured in joules per coloumb	24

Tutorial 2- Past Paper Questions	25

Alternating current & voltage/ tutorial answers	26

Solutions to Past paper questions	26

CHAPTER 4: CURRENT, POTENTIAL DIFFERENCE, POWER AND RESISTANCE	28

Summary of Content	28

Power	28

Potential Divider Circuits	28

Potential Divider examples	31

Movement of Charge	33

Tutorial 1: Current, voltage, power and resistance	33

Tutorial 2- Wheatstone bridges	38

Tutorial 1 Solutions: Current, voltage, power and resistance :	39

Tutorial 2 Solutions	39

Worked answers for tutorial 2	39

CHAPTER 5: ELECTRICAL SOURCES AND INTERNAL RESISTANCE	40

Summary of Content	40

EMF-electromotive force	40

Cells	40

Practical 1 – A dry Cell	41

Practical 2: Measuring the emf and internal resistance of a 1.5v cell.	42

EMF Questions from graphs	44

GRAPH 1- V against I	44

GRAPH 2 – R against 1/I	45

GRAPH 3 - graph of lost volts against R	46

Power matching	47

[application of internal resistance]	47

Tutorial 1- EMF	50

Tutorial 2 More EMF questions	50

Tutorial 3- Past paper questions	52

Tutorial Answers	56

Tutorial Answers SQA Solutions	57

CHAPTER 6 CAPACITORS	58

Summary of content	58

Background to Capacitors	58

What is a capacitor?	58

Capacitor circuit symbol	58

So how do Capacitors work?	59

Capacitance	61

Charging and Discharging	62

Measuring the capacitance of a capacitor.	64

Charging a capacitor on a d.c. supply.	66

Discharging a capacitor	66

Energy stored in a Capacitor	67

Energy Stored	67

Capacitors and a.c.	68

Charging on a.c.	68

Resistance and Frequency	69

Part 1 Resistance and Frequency	69

Reactance of capacitors in a.c. circuits	70

Relationship Between Current and Frequency in an A.C capacitor Circuit	71

Blocking and Smoothing	72

Blocking	72

Smoothing	72

Tutorial 1: capacitance	73

Tutorial 2: capacitance	73

Tutorial 3: capacitance	73

Tutorial 4: CAPACITANCE	74

Tutorial 5: Exam Questions	76

Tutorial Answers Capacitance	79

Tutorial Work Answers	81

Capacitance/ tutorial answers	81

Tutorial Answers: Exam question soutions	82

Data Sheet

Relationships Required For Higher Physics

CHAPTER 1: SUMMARY ELECTRICITY CONTENT

CHAPTER 1: SUMMARY ELECTRICITY CONTENT

Summary of electricity content

MONITORING AND MEASURING A.C. (chapter 3)

Knowledge that a.c. is a current which changes direction and instantaneous value with time.

Use of appropriate relationships to solve problems involving peak and r.m.s. values.

Determination of frequency, peak voltage and r.m.s. values from graphical data.

Current, potential difference, power and resistance (chapter 4)

Use of appropriate relationships to solve problems involving potential difference, current, resistance and power. Solutions may involve several steps.

Use of appropriate relationships to solve problems involving potential divider circuits.

Use relationships involving potential difference, current, resistance and power to analyse circuits. Calculations may involve several steps.

Calculations involving potential dividers circuits.

Electrical sources and internal resistance (chapter 5)

Knowledge of the terms electromotive force (e.m.f.), internal resistance and terminal potential difference (t.p.d.), ideal supplies, short circuit and open circuit.

Use of an appropriate relationship to solve problems involving e.m.f., t.p.d., current and internal resistance.

Determination of internal resistance and e.m.f. using graphical analysis.

4	Capacitors (chapter 6)

Definition of capacitance.

Use of an appropriate relationship to solve problems involving capacitance, charge and potential difference.

Knowledge that the total energy stored in a charged capacitor is the area under the charge against potential difference graph.

Use of data from a charge against potential difference graph.

Use of appropriate relationships to solve problems involving energy, charge, capacitance and potential difference.

Awareness of the variation of current and potential difference with time for both charging and discharging cycles of a capacitor in a CR circuit (charging and discharging curves).

Awareness of the effect of resistance and capacitance on charging and discharging curves in a CR circuit.

5 Conductors, semiconductors and insulators (chapter 7)

Knowledge that solids can be categorised into conductors, semiconductors or insulators by their ability to conduct electricity.

Awareness of the terms conduction band and valance band.

Qualitative explanation of the electrical properties of conductors, insulators and semiconductors using the electron population of the conduction and valance bands and the energy difference between the conduction and valance bands.

6 p-n junctions (chapter 8)

Awareness that, during manufacture, the conductivity of semiconductors can be controlled, resulting in two types: p-type and n-type.

Knowledge that, when p-type and n-type materials are joined, a layer is formed at the junction. The electrical properties of this layer are used in a number of devices.

Awareness of the terms forward bias and reverse bias.

Knowledge that solar cells are p-n junctions designed so that a potential difference is produced when photons enter the layer. (This is known as the photovoltaic effect.)

Knowledge that LEDs are forward biased p-n junction diodes that emit photons when electrons ‘fall’ from the conduction band into the valence band of the p-type semiconductor.

CHAPTER 2: BACKGROUND TO ELECTRICITY

Prefixes

tera	T	1012	x 1,000,000,000,000
giga	G	109	x 1,000,000,000
mega	M	106	x 1,000,000
kilo	K	103	x 1,000
centi	C	10-2	/100
milli	M	10-3	/1,000
micro	µ	10-6	/1,000,000
nano	N	10-9	/1,000,000,000
pico	P	10-12	/1,000,000,000,000

Definitions

Current

Current is the rate of flow of charge in a circuit and can be calculated using

Where Q =quantity of charge measured in Coulombs (C), I = current (A), t=time (s)

In a complete circuit containing a cell, switch and a bulb the free electrons in the conductor will experience a force which will cause them to move drifting away from the negatively charged end towards the positively charged end

Electrons are negatively charged, 1 electron has a charge of 1610-19 C.

Potential Difference

If one joule of work is done in moving one Coulomb of charge between two points then the potential difference (p.d.) between the two points is 1 volt. (This means that work is done when moving a charge in an electric field)

Where Ew = the work done in moving quantity of charge between two points in an electric field (J), Q= quantity of charge (C ), V= potential difference between two points in an electric field in Joules per Coloumb (JC-1) or volts (V)

Measuring Current and Voltage

Component Name	Circuit Symbol	Function
Voltmeter	Measures potential difference. Must be placed in parallel to measure the difference in electrical potential between two points.
Ammeter	Measures current. Must be placed in series to measure the current flowing in a circuit.
Ohmmeter	Measures resistance. Must be placed in parallel with the component(s) which are to be measured.

They can be used in simple, or more complex circuits in order to investigate the size of any of these quantities.

The diagram also shows how the voltmeter and ammeter must be connected. The voltmeter must be connected in parallel because it is measuring the difference in electrical potential between two points. The ammeter must be connected in series because it is measuring the flow of current THROUGH a component or circuit.
In most cases, resistance should only be measured when a circuit is not ‘on’ or fully connected. To use an ohmmeter, you would place it ‘across’ a component – in parallel, just like a voltmeter.

Series and Parallel: Circuit Rules

Measurement	Series Circuits	Parallel Circuits
Potential Difference
Current
Resistance

Resistance in series and parallel

Worked examples

Calculate the resistance of the following circuit

60 Ω 35 Ω 22 Ω

Rtotal R1 R2 R3	= ? = 60 Ω = 35 Ω = 22 Ω	Rtotal Rtotal Rtotal	= R1 + R2 + R3 = 60 + 35 + 22 = 117 Ω

The total resistance of this circuit is 25 kΩ. Calculate the value of Resistor 2

12 kΩ R2 500 Ω

Rtotal R1 R2 R3	= 25 kΩ = 12 kΩ = ? = 500 Ω	Rtotal 25000 R2 R2 R2	= R1 + R2 + R3 = 12000 + R2 + 500 = 25000 – 12500 = 12500 = 12.5 kΩ

Calculate the resistance of the circuit shown below

Rtotal = ? R1 = 100 Ω R2 = 200 Ω R3 = 400 Ω	1/Rtotal 1/Rtotal 1/Rtotal 1/Rtotal Rtotal Rtotal	= 1/R1 + 1/R2 + 1/R3 = 1/100 + 1/200 + 1/400 = 4/400 + 2/400 + 1/400 = 7/400 = = 400/7 = 57 Ω

100 Ω 200 Ω 400 Ω

The total resistance of this circuit is 100Ω. Calculate the value of Resistor 1

Rtotal = 100 Ω R1 = ? R2 = 200 Ω	1/Rtotal 1/100 1/R1 1/R1 1/R1 R1	= 1/R1 + 1/R2 = 1/R1 + 1/200 = 1/100 - 1/200 = 2/200 - 1/200 = 1/200 = 200 Ω

200 Ω R1

Parallel Section first:

Now have 3 resistors in series

Simplify the parallel branches

Branch 1 Branch 2

Calculate resistance of parallel section

Now find total

Ohm’s Law

Ohm's law forms the basis for understanding how electrons or charge flows through circuits. Ohm's Law deals with the relationship between voltage and current in an ideal conductor.

This relationship states that:

For a constant temperature for a given resistor:

V α I or V / I = constant

Where V= potential difference in Volts, I= current (A), R= resistance ()

There are components which do not have a constant resistance as the current through them is altered – eg a light bulb. Graphs of potential difference against current for this type of component will not be a straight line.

Practical 1- the Van de Graaff

Aim: To familiarise yourself with the Van de Graaff generator.

Watch and participate in the experiments demonstrating the Van de Graaff generator.

Ensure that you know the following:

How the Van de Graaff creates the charge.

How the Van de Graaff stores charge.

The size and shape of the field pattern and force create around the VdG.

How the charge can be distributed.

Write up your experiences including the Physics behind the Van de Graaff.

Practical 2 – Resistors in Series Practical

Aim:	To check the formula for resistors in series against our experience, set up the following circuit:-

Put 10cm of resistance wire into the holder and note the current and voltage readings.

Feed out more wire so that 20cm of the same wire lies between the terminals of the holder.

Adjust the variable resistor until the current is the same as for the 10cm length of wire.

Note the voltmeter reading.

Predict the voltage reading after you have adjusted the current when 30cm of wire stretches between the holder's terminals.

Discuss why the voltmeter reading increases as the resistance wire is lengthened.

Practical 3- Resistance in Parallel

Using the same basic apparatus as in Practical 2 put 10cm of resistance wire into the holder.

Note the current and voltage readings.

Now put a second 10cm length of resistance wire into the holder along with the first piece.

Use the variable resistor to adjust the voltage to the same as the previous value, then take note of the current.

PREDICT: What the current reading is after you have put 3 separate 10cm lengths of wire into the holder and adjusted the voltage to its previous value.

DISCUSS: Why does the current reading increase as the experiment progresses.

Revision Tutorials

Tutorial 1: Electric Fields

1.

Explain with the aid of a diagram, what effect we get when a positively charged object is brought near the right hand edge of these rods:-

2. What is the size of angle X?

3.	An oil drop with a charge of 3 pC and a mass of 1.22 x 10-8 kg is placed in a uniform electric field of 4 x 104 NC-1. If the oil drop floats between the charged plates, what is its weight?

4.	A small polystyrene sphere with a positive charge of 5pC is placed in a uniform electric field at the 10 V level. If the sphere falls downwards to a place where the voltage level is 150 V, how much work does the gravitational field do?

Tutorial 2 Circuits

1.	A lamp is rated at 48W 12V. When running normally,

a)	What is the current through the lamp?

b)	What is the lamp's resistance while it is operating?

2.	If a 2kΩ resistor has a p.d. of 9.6 V across it, what is the current in it?

3.	A model power transmission line is 3 m long and carries a current of 4A. What is the p.d. across it if the wire used has a resistance of 125 milliohms per metre?

Tutorial 3- Resistance

1.	What is the total resistance of this arrangement?

900 Ω 900 Ω 900 Ω 2.	If the equivalent resistance of the following arrangement is 250Ω, what is the value of the unknown resistor?

3.	What is the equivalent resistance of this arrangement of resistors?

500 If the equivalent resistance of the following arrangement is 250Ω, what is the value of the unknown resistor?

R

5.	What is the equivalent resistance of 200Ω, 300Ω and 600Ω when they are all connected in parallel?

6.	What is the equivalent resistance of three resistors, 4Ω, 4Ω and 2Ω in parallel?

7.	What is the value of a resistor which gives an equivalent resistance of 18.75Ω when connected in parallel with a 75Ω resistor?

Tutorial Answers

Tutorial 1

1.

2.	X = 15o since the field is uniform.

3.	Since the drop is floating, the electric force, 1.2x10-7N, is the same as its weight.

4.	Since the motion is caused by the sphere's weight, the work done by the gravitational field is 7x10-10J.

Tutorial 2

1.	a)	The current in the lamp is 4A

b)	The lamp's resistance is 3Ω

2.	The current in the resistor is 4.8mA

3.	The wire has a p.d. of 1.5V across it.

Tutorial 3

1.	Total resistance =1210Ω

2.	The unknown resistance is 40Ω

3.	The equivalent resistance is 300Ω

4.	The unknown resistance is 500Ω

5.	The equivalent resistance is 100Ω

6.	The equivalent resistance is 1Ω

7.	The unknown resistance is 25Ω

Worked Answers for tutorials

WORKED ANSWERS FOR TUTORIAL 2

1.	a)	P=VI gives 48=12xI

Thus I = 4

The current in the lamp is 4A

b)

The lamp's resistance is 3Ω

2.	The current in the resistor is 4.8mA

3.

The wire has a p.d. of 1.5V across it.

WORKED ANSWERS FOR TUTORIAL 3

1.

The total resistance is 1210Ω

2.

The unknown resistance is 40Ω

3.

The equivalent resistance is 300Ω

4.

The unknown resistance is 500Ω

5.

The equivalent resistance is 100Ω

6.

The equivalent resistance is 1Ω

7.

The unknown resistance is 25Ω

CHAPTER 3: MONITORING AND MEASURING A.C

Summary of content

MONITORING AND MEASURING A.C. (CHAPTER 3)

A.C. as a current which changes direction and instantaneous value with time.

Calculations involving peak and r.m.s. values.

Determination of frequency, peak voltage and r.m.s. values from graphical data.

Use of a multimeter as an ammeter, voltmeter and ohmmeter.

Use of an oscilloscope as a voltmeter and waveform monitor.

Use of an oscilloscope to monitor a.c. signals,including the measurement of frequency and peak/r.m.s. values.

Measuring the Frequency

At Higher it is not enough to find the frequency by inspection (this means comparing your known trace to your unknown trace using a double beam oscilloscope and a signal generator) At Higher you need to be able to use the timebase! We first need to work out the period, T where T is the period or time for one wave.

Set up the oscilloscope to give a stationary trace.

Adjust the trace so that the time period can be found.

Time period = time for one wave

Count the number of squares for 1 wave

Multiply the number of squares for 1 wave by the time base setting

Find

(Time setting will be in ms or s. DON’T FORGET TO CHANGE IT!!)

KNOW HOW TO MEASURE FREQUENCY USING AN OSCILLOSCOPE

e.g.

An oscilloscope is connected to the output terminals of a signal generator.

The trace displayed on the screen is shown below.

The timebase of the oscilloscope is set at 30 ms/div.

The frequency of the output signal from the signal generator is

A 4∙2 × 10−3 Hz

B 8∙3 × 10−3 Hz

C 0∙28 Hz

D 4∙2 Hz

E 8∙3 Hz .

Answer

1 wave is 4 boxes across in the trace above. Each box is 30ms. T=430=120 ms =1.210-1s

An alternating voltage is displayed on an oscilloscope screen. The Y-gain and the timebase settings are shown.

Which row in the table gives the values for the peak voltage and frequency of the signal?

Peak voltage/V	Frequency/Hz
A	10	100
B	10	250
C	20	250
D	10	500
E	20	1000

Peak Voltage and R.M.S.

An a.c is one where the magnitude and direction of the current changes, usually many times per second. The mains is an example of an a.c.

V t But what do we take as the voltage to quote?

With an a.c trace the value keeps changing.

V AverageVoltage t The average voltage is not a good indication of the effect of the voltage as the average voltage is always zero.

V Peak Voltage t Peak voltage isn’t a good indication of the voltage to a device. It also cannot be compared with a d.c. supply. It will over- estimate the power dissipated in the circuit

The solution is to do with the fact that half the time the charges move in one direction and half the time in the opposite direction. We can get rid of this problem by squaring everything.

The pattern of the graph now changes and it is possible to work out an average for the new graph.

It turns out that the average is . To get back to the original we now need to take the square root of this value which will be .This is called the r.m.s. value =

Peak Voltage and R.M.S.

The answer is the r.m.s (root mean square) value. Now a.c. and d.c. are comparable!

R.M.S. V

NB.	i)	Peak is always the biggest	ii)

THE SAME APPLIES FOR CURRENT

a.c. is one in which electrons flow first in one direction then in the other.

In a resistor the rate at which heat is produced is given by P = I2R

With a.c. what value do you take as I?

average I = 0 A

peak only occurs for a very short period therefore it is an overestimate.

We therefore want something in between.

The clue is in I2

Squaring the current and plotting it on a graph shows that I2 is always positive. The graph is symmetrical and the average is half its maximum.

average of

Practical 1: Comparing brightness

Comparing bulb brightness with an a.c and d.c supply.

Set up the circuit as shown below and take your own readings to shown that the rms of an a.c trace provides the same power to a circuit as a d.c trace.

B A Two-way switch oscilloscope photo cell & meter

A

Power on d.c	Power on a.c.
Current = 1.42A Voltage=12.35V P=IV	No. of squares from top of the trace to the bottom =6.6 Each square is 5V/div
Power on d.c.=17.5W	Therefore peak voltage = 3.35V=16.5V r.m.s value=
Current= 1.464A Voltage= 11.7V Power on a.c=17.1W

We can now calculate the average power delivered by this a.c. to the resistor.

where IP = peak current.

Compare with d.c. which provides a steady current:

If we arrange the equipment so that the same power is delivered by a.c. and d.c. supply then we can compare I and IP.

This can also be written as:

REMEMBER THAT IP IS ALWAYS > Irms

In a circuit containing a resistor the current is completely independent of frequency. i.e. you change the frequency and the current stays the same. This would not be the same for a capacitor.

Practical 2: Checking the equation for RMS current:

Check the equation above by finding the connection between the peak value of an a.c. supply and the value of a direct current which lights a bulb with the same brightness. The circuit needed is shown here: -

With the two-way switch at A, the bulb lights on a.c.

With the two-way switch at B, the bulb lights on d.c. whose value can be adjusted by using the variable resistor as a voltage divider.

Adjust the d.c. output until movement of the switch between A and B shows no difference in the brightness of the bulb. This ensures that both the a.c. and d.c. supplies are providing the same power.

Record the value of the direct current.

Use the C.R.O. screen and Ohm's law to find the alternating current's peak value.

Adjust the a.c. power supply, repeat the d.c. adjustments and get another pair of alternating and direct current values.

Use your table of results to plot a graph of Peak a.c. value against d.c. value

Use this graph to find the average of the ratio

Tutorial 1: alternating current & voltage

1.	What is the peak value of our 230V mains supply?

2.	An oscilloscope shows the peak voltage from a laboratory power supply as 141 Volts. What is the r.m.s. voltage it supplies?

3.	A motor for use with an alternating current supply is rated at 250V, 500W. What is the peak voltage and current the motor is designed to handle?

4.	What is the frequency of the signal shown on this oscilloscope screen?

5.	A mass of 10kg falls steadily through a height of 5m while attached by a wire to an electric generator. If the mass takes 10 seconds to fall and all of its energy goes into the generator,

a)	What power does the generator give out?

b)	What is the peak voltage output if the generator produces a steady 3.5A?

6.	A mains electric clock stores enough energy in 10s to run itself for 10 hours. If the current during the storage phase has a peak value of 10.6A, what is the power used by the clock mechanism?

7.	An electric heater has a heating element of resistance 50 with an alternating current of peak value 8.5A in it. By how much does this raise the temperature of 9.7kg of water in 15minutes? (Take the specific heat capacity of water as 4.19103 Jkg –1oC –1 .

Practical 3 – electron gun

Examine the electron gun from a cathode ray oscilloscope and note the subtle shapes employed to alter the uniformity of the fields that guide electrons into the desired beam shape.

Practical 4- energy is measured in joules per coloumb

Aim:	To show that the energy delivered by the current is measured in joules per coulomb (volts).

(i)	Use the ammeter and stopwatch together (Q=It) to calculate the number of coulombs passed through the immersion heater.

(ii)	Use the relationship Eh = mc∆T to calculate the number of joules delivered by the heater.

(iii)	Use the answers to parts (i) and (ii) to calculate the number of joules delivered by each coulomb and compare it with the reading shown on the voltmeter.

Tutorial 2- Past Paper Questions

1997 Paper 1 Question 34.

The output from a signal generator is connected to the input terminals of an oscilloscope. A trace is obtained on the oscilloscope screen. The oscilloscope control settings and the trace on the oscilloscope screen are shown in the diagram below.

Calculate the frequency of the output from the signal generator

The frequency and amplitude of the output from the signal generator are kept constant. The time base control setting is changed to 5 ms/division. What will be the effect on the trace shown on the oscilloscope?

The diagram shows the screen and the settings of an oscilloscope, which is being used to measure the output frequency of a signal generator.

What is the frequency of the signal applied to the input of the oscilloscope?

A	2.5 Hz

B	12.5 Hz

C	40 Hz

D	250 Hz

E	500 Hz

1992 P1Q17

1993 Paper 1 Q34

An oscilloscope is connected across a resistor in a circuit. The trace obtained is shown below.

The peak voltage shown on the oscilloscope is 10 volts and the time base setting is 0.2 ms cm-1. Calculate

The r.m.s voltage across the resistor

The frequency of the a.c. voltage

A.c & voltage/ tutorial answers

Tutorial 1

1.	The peak value is 325V

2.	The r.m.s. voltage is 100V

3.	The motor is designed to handle a maximum voltage of 354V and a maximum current of 2.82A

4.	Frequency = 1/0.001 = 1000Hz

5.	a)	The power given out is 49W

b)	The peak voltage output is 19.8V

6.	The power used by the clock is 0.48W

7.	The water temperature is raised by 40oC.

Solutions to Past paper questions

1997 34.a.	Wave period(T) = 4x2.5ms

T = 10x10-3s

f=1/T

f=1/10x10-3s

f=100Hz

The amplitude of the waves displayed on the oscilloscope will be unchanged, but, five complete waves will now appear on the screen.

1992 P1 Q14 D 250 Hz

1992 P1 Q17 B 10/

1993 P1 34.a.

VRMS = 10/1.414

VRMS = 7.07V

b.	f = 1/T

T = 8x0.2ms = 1.6ms

T = 1.6x10-3

f = 1/1.6x10-3

f = 625Hz

CHAPTER 4: I, V, P & R

Summary of Content

Current, potential difference, power and resistance

Use relationships involving potential difference, current, resistance and power to analyse circuits. Calculations may involve several steps.

Calculations involving potential dividers circuits.

Investigating a.c. or d.c. circuits with switches and resistive components.

Use of potential dividers in circuits to set and control voltages in electronic circuits.

Power

The power of a circuit component (such as a resistor) tells us how much electrical potential energy the component transforms (changes into other forms of energy) every second: The following formulae are also used to calculate power (P):

Where P= Power (W),

E=energy (J), t=time (s),	V=voltage (V),	I= current (A),	R=resistance ()

+5V 0V Potential Divider Circuits

Any circuit that contains more than one component can be described as a potential divider circuit. In its simplest form a potential divider is 2 resistors connected across a power supply. If another component is placed in parallel with a part of the potential divider circuit, the operating potential difference of this component can be controlled.

Formula for a series circuit

R3 R1 R2 V3 V2 V1 VS I In a SERIES circuit the current through each resistor is the same. To find the current use the formula:

We already know that IT is the same as the current going through R1, R2, R3 etc. So to find V1, V2, V3, use:

As IT is the same

To find the voltage across resistors you do not need to work out the current.

V2 V1 12 V 90 30

Either work out by ratios (quick if you can do it but costly if it goes wrong!)

EITHER: Summarise VS = 12 V, R1 = 90 , R2 = 30 V1 = ?, V2 = ? Find RT Find RT = R1 + R2 RT = 90 + 30 = 120 Find V1 Find V2	OR: Summarise VS = 12 V, R1 = 90 , R2 = 30 V1 = ?, V2 = ? .Find RT Find RT = R1 + R2 RT = 90 + 30 = 120 Find IT Find V1 Find V2
Check Vs = V1 + V2 = 9 + 3 = 12 V ☺	6.Check Vs = V1 + V2 = 9 + 3 = 12 V ☺

OR USE

Potential Divider examples

15 7 The voltage across the branch of the circuit must be equal to the terminal p.d (potential difference). NB I am not using in this case, although the examiners could be horrible and combine the question!

or use

e.g.

If part of the circuit has a resistor in parallel with one of the resistors then this section has to be worked out first. Don’t forget that if the two resistors are identical then the total resistance in the parallel section will be half the resistance of one of the resistors in parallel.

10 15 15 For example keeping the 12V supply a 10 resistor is placed in series with two 15 resistors in parallel, what is the voltage across each part of the circuit?

Well the 15 resistors in parallel would have a total resistance of 7.5, if you don’t believe me work it out!

Then work out the voltage divider part as previously, but using the total of the two parallel resistors instead of the top resistor.

You are likely to feel quite confident about this after practice; after all it was in the National 5 Physics Course.

15 15 Now consider adding another branch to the circuit. The voltage drop across the bottom branch of the circuit must be equal to the voltage drop across the top branch.

5 10 The voltage across each of the branches of the circuit must be equal to the terminal p.d (potential difference) as voltages in parallel are equal to the terminal p.d

The voltage drop across the resistors in the top branch is completely independent of the voltage dropped across each resistor in the bottom branch. If you don’t believe me try it! Imagine that the top resistors are both 15, if the voltage supply is 12V the voltage across each of the top resistors is 6V. If the resistors in the bottom were 5 and 10, then 4V will be dropped across the 5 resistor and 8V will be dropped across the 10 resistor. Or if the top branch still contains our two 15 resistors and our bottom branch now has two 150 resistors then the voltage across each resistor will be 6V. However, the current in the top branch will now be 10 times that of the bottom branch. Use V=IR to prove this using V as the terminal p.d and R as the total resistance of each branch.

But there is more….

B A B If we put a wire connecting these two sets of resistors charge can flow up or down the wire if there is a p.d. between A and B

Movement of Charge

Which way does the current flow? Here is the information that you need.

The +ve terminal of a power supply has a high potential.

The -ve terminal of a power supply has a low potential.

Conventional current suggests that +ve charge moves from areas of HIGH potential to LOW potential.

conventional current	electron flow

Electron flow says that electrons (-ve charge) moves from areas of LOW potential to HIGH potential.

Tutorial 1: Current, voltage, power and resistance

1.	There is a current of 40 mA in a lamp for 16 s.

Calculate the quantity of charge that passes any point in the circuit in this time.

2.	A flash of lightning lasts for 1 ms. The charge transferred between the cloud and the ground in this time is 5 C.

Calculate the value of the average current in this flash of lightning.

3.	The current in a circuit is 2 5 × 10–2 A. How long does it take for 500 C of charge to pass any given point in the circuit?

4.	There is a current of 3 mA in a 2 kΩ resistor. Calculate the p.d. across the resistor.

A 10 V + – 70 10 20 (a) A 12 V + – 15 9 V (b) A + – 3 5 V 6 V (c) 5.	Calculate the values of the readings on the meters in the following circuits.

.	Calculate the unknown values R of the resistors in the following circuits.

+ 2 A – 10  + –

40 V A A

40 V

R = ? R = ? (b) 4  (a) 5 

10 

V

20 V

(b) 10  X Y 10  10  10  7.	Calculate the total resistance between X and Y for the following combinations of resistors.

(a) 10  X Y 20  10 

8  (f) 12  X Y 5  6  3  (d) 25  X Y 5  20  10  (c) 20  X 10  4  10  5  1 

10  (e) 10  X Y 10  5  25 

In the following circuit the reading on the ammeter is 2 mA. Calculate the reading on the voltmeter

+ 5 k A V 3 k –

9.	Calculate the power in each of the following situations.

(a)	A 12 V battery is connected to a motor. There is a current of 5 A in the motor.

(b)	A heater of resistance 60 Ω connected across a 140 V supply.

(c)	A current of 5 A in a heater coil of resistance 20 Ω.

10.	The heating element in an electric kettle has a resistance of 30 Ω.

(a)	What is the current in the heating element when it is connected to a 230 V supply?

(b)	Calculate the power rating of the element in the kettle.

11.	A 15 V supply produces a current of 2 A in a lamp for 5 minutes. Calculate the energy supplied in this time.

12.	Calculate the readings on the ammeter and the voltmeter in the circuit shown below.

A V 6  6  6  6 V + –

13.	Each of the four cells in the circuit shown is identical.

A 20 5 2 6 V Calculate

(a)	the reading on the ammeter

(b)	the current in the 20 Ω resistor

(c)	the voltage across the 2 Ω resistor.

14.	A voltage of 12 V is applied across a resistor. The current in the resistor is 50 mA. Calculate the resistance of the resistor.

15.	The LED in the circuit below is to emit light.

A B R

(a)	What is the required polarity of A and B when connected to a 5 V supply so that the LED emits light?

(b)	What is the purpose of the resistor R in the circuit?

(c)	The LED rating is 20 mA at 1·5 V. Calculate the value of resistor R.

16.	Write down the rules which connect the (a) potential differences and (b) the currents in series and parallel circuits.

17.	What is the name given to the circuit shown?

Write down the relationship between V1, V2, R1 and R2.

V1 V2 R1 R2 10V

18.	Calculate the values of V1 and V2 of the circuit in question 17 when:

(a)	R1 = 1 kΩ	R2 = 49 kΩ

(b)	R1 = 5 kΩ	R2 = 15 kΩ

R1 10V V1 V2 19.	The light dependent resistor in the circuit is in darkness. Light is now shone on the LDR. Explain what happens to the readings on V1 and V2.¶

20.	Calculate the p.d. across resistor R2 in each of the following circuits.

(a)	(b)	(c)

21.	Calculate the p.d. across AB (voltmeter reading) in each of the following circuits.

B B +12 V 0 V 3 k 3 k V 9 k 3 k A +10 V 0V 3 k 2 k V 6 k 4 k A B +5 V 0 0V 5 k 2 k V 10 k 8 k A (a) (b) (c)

22.	A circuit consisting of two potential dividers is set up as shown.

+9 V 0 V 6 k 3 k V 9 k 6 k A B

(a)	Calculate the reading on the voltmeter.

(b)	(i)	Suggest a value of a resistor to replace the 9 kΩ resistor that would give a reading of 0 V on the voltmeter.

(ii)	Suggest a value of resistor to replace the 3 kΩ resistor that would give a reading of 0 V on the voltmeter.

0 V 12 V 15 4K 12k vVB Y 0 V 12 V 9 120 120 vVB X 23.	In the circuits shown the reading on the voltmeters is zero. Calculate the value of the unknown resistors X and Y in each of the circuits.

Tutorial 2- Wheatstone bridges

1.	What is the value of the unknown resistor shown in this circuit when the micro-ammeter reads zero?

2.	What value must R have in the following circuit for the micro-ammeter to read zero?

3.	Is there a p.d. between A and B in the following circuit?

Tutorial 1 Solutions: Current, voltage, power and resistance :

1.	0.64 C

2.	5 × 103 A

3.	2·0 × 104 s

4.	6 V

5.	(a)	I = 0·1 A	(b)	I = 0·5 A, V = 4·5 V	(c)	I = 2 A, V = 10 V

6.	(a)	5	(b)	6

7.	(a)	25 (b)	25 (c)	24·2 (d)	13·3 (e)	22·9

(f)	14·7 .

8.	3·75 × 10–3 V

9.	(a)	60 W	(b)	327 W	(c)	500 W

10.	(a)	7·7 A	(b)	1763 W

11.	9000 J

12.	I = 0·67 A, V = 4 V

13.	(a)	0·67 A	(b)	0·13 A	(c)	1·34 V

14.	240

15.	(c)	175

18.	(a)	V1 = 0·2 V, V2 = 9·8 V	(b)	V1 = 2·5 V, V2 = 7·5 V

20.	(a)	4 V	(b)	1 V	(c)	3 V

21.	(a)	3 V	(b)	-0·8 V	(c)	0 V

22.	(a)	0·6 V	(b)	(i)	12 k	(ii)	4 k

23.	X = 9 , Y = 45

Tutorial 2 Solutions

1.	The unknown resistance is 266Ω

2.	R has a value of 150Ω

3.	Yes; the resistors do not have the same ratio in each branch and so the micro-ammeter registers a current which requires a p.d. across it. If you care to calculate the actual values, you find that A is at 8V and B at 6V. This gives a 2V p.d. between B and A.

Worked answers for tutorial 2

1.

The unknown resistance is 266Ω

2.

R has a value of 150Ω

3.	Yes; the resistors do not have the same ratio in each branch and so the micro-ammeter registers a current which requires a p.d. across it. If you care to calculate the actual values, you find that A is at 8V and B at 6V. This gives a 2V p.d. between B and A.

CHAPTER 5: ELECTRICAL SOURCES & INTERNAL RESISTANCE

Summary of Content

Electrical sources and internal resistance

Electromotive force, internal resistance and terminal potential difference. Ideal supplies, short circuits and open circuits.

Determining internal resistance and electromotive force using graphical analysis.

Investigating internal resistance of low voltage power supplies.

Load matching. Maximum power is transferred when internal and external resistances are equal.

Investigate the reduction in t.p.d. when additional components are added in parallel.

EMF-electromotive force

Definition- the emf of a source is the electrical potential energy supplied to each coulomb of charge as it passes through the source.

E is measured in volts

Q is measured in coulombs and

Ew is measured in joules

Don’t confuse E when thinking about EMF with E which is energy or E which is electric field strength. In this case E is EMF and EW is the electrical potential energy.

Cells

Cells and other sources of electrical energy are not 100% efficient. There is a resistance inside the source. Energy is used as the charge passes through the resistance within the cell (source).

This can be represented as a resistor in series with the source.

Therefore an electrical source is equivalent to a source with an e.m.f with a resistor in series (internal resistance)

We use r to represent the internal resistance.

So according to Ohm’s Law, V=Ir. This V is equal to the lost volts as the charge passes through the source.

E is the “push” shoving the charges around the circuit or the energy supplied to each coulomb of charge.

For each coulomb of charge

Energy supplied = Energy used in the circuit.

E	=	V	+	Vlost

energy supplied =	voltage dropped +	lost volts

by the source	in the external circuit

but V=IR

E = IR+ Ir

E = V + Vlost

E = IR + Vlost

E = V + Ir

E = I(R+r)

Practical 1 – A dry Cell

Use a dry cell and measure: -

(i)	Its e.m.f. using a voltmeter. (i.e. place a voltmeter directly across the terminals of the 1.5V cell without a load.)

(ii)	Now add a bulb or resistor across the cell and measure the new voltage across the cells. This is the cells terminal p.d. and the corresponding current when an external resistor is connected to it.

(iii) Add an Ammeter in the circuit and find the internal resistance of the cell with this load.

Calculate the internal resistance of the cell using E=V+Ir.

EMF

An open circuit has a no complete circuit.

In an open circuit no charge flows

E = V + Ir

If	I =0A

E = V + 0

E = V

Therefore for an open circuit the p.d. across the terminals is equal to the e.m.f.

Hmm! How come you can measure this with a voltmeter across the terminal? Doesn’t this lead to a flow of charge?

Answer: The resistance inside a voltmeter is very large. If R is very large then I is extremely small (from V=IR), in fact it is just about negligible. AHH!

Practical 2: Finding emf & r.

A V A Set up the experiment shown in the diagram using the Alba interface and emf board.

Alter the variable resistor or the resistance of the external circuit so that I changes

Take readings of I and V from the ammeter and voltmeter.

Repeat for other values of I and V

Plot a graph of V against I

Use the graph to find and r

A 12V car battery has an internal resistance of 0.001. What is its “short” circuit current?	What is the short circuit current of a 1.5V dry cell of internal resistance 1.25?

EMF (graphs)

GRAPH 1- V against I

For these Questions

E= intercept of y axis measured in VOLTS

-gradient = r (internal resistance of the cell)

Or	m=-r

Short circuit current occurs when the external resistance is

zero (R=O) which occurs where the line passes the x-intercept. Short circuit current

E Imax -r

GRAPH 2 – R against 1/I

E = I(R + r)

For these Q

fits into the equation for a straight line or

m=E

+c=-r so the y intercept is -r

Short circuit current occurs when the external resistance is zero (R=O) which occurs where the line passes the x-intercept. However remember that the x-axis is 1/I so to calculate I take the inverse of the intercept.

1/Imax -r E

GRAPH 3 - graph of lost volts against R

For these Q

Lost volts =Ir

when the external resistance is zero (R=O) which occurs where the line passes the y-intercept then the E=Ir or E=lost volts So the y-intercept is equal to the emf.

The voltage available to the circuit or terminal p.d will equal the y intercept subtract the lost volts for a particular external resistance. The resistance in this graph is the external resistance so

E(y-intercept)= I R (value from x axis)-Lost volts (y value for x-axis value)

From this I can be found

Then LOST VOLTS=Ir

E Terminal p.d Or E=I(R+r)

e.g. EMF = 6V When the external resistance of the circuit is 2 ohms

then the terminal p.d = 6V – 1.2 V (read from the y axis) = 3.8V

A cell has an internal resistance of 3 and an e.m.f. of 2V and another cell has an internal resistance of 1 and e.m.f. of 1.5V.

Calculate the resistance of the wire which, when connected to either cell, will produce the same current.

In this question there are 2 different cells, however it is one piece of wire which tells us the external resistance is constant and the question tells us that for these conditions of r and E the current remains constant so

Power matching

[application of internal resistance]

In order to have the most efficient transfer of energy between a source and an output the resistances of both must be considered. It is possible to calculate the optimum ratio of resistance to give maximum output. This is shown below, fortunately you do not need to be able to reproduce this explanation

Maximum Power in a simple d.c. circuit.

We know that

(1)

If we have a simple circuit consisting of a power supply of e.m.f., E, internal resistance, r, passing current through a resistor, R, then:

Substituting for I in equation (1) gives:

Dividing top and bottom of the r.h.s. by R gives:

Thus the power dissipated in the resistor, R, will be at a maximum when the denominator is at a minimum. Remember that r is effectively a constant. Thus if we differentiate the denominator with respect to R we can find the turning points. i.e.

Since resistances cannot be negative we conclude that for the maximum power to be produced we must set R = r. You will test this in an experimental activity.

Here are some results for you to try out. Find the total power in the external circuit (load) using I2R and then add this to the power lost in the internal circuit using I2r

emf E (V)	12
Internal res r ()	10

Load res R ()	0.00	2.00	4.00	6.00	8.00	10.00	12.00	14.00	16.00	18.00	20.00
Current I (A)	1.20	1.00	0.86	0.75	0.67	0.60	0.55	0.50	0.46	0.43	0.40
Power PL in load (W)
Power lost in internal circuit (W)
Total Power (W)

The maximum power transfer theorem states that "the maximum amount of power will be dissipated in the load resistance if it is equal in value to the source resistance of the network supplying the power".

In other words, the load resistance resulting in greatest power dissipation must be equal in value to the equivalent source resistance, then rl = rs but if the load resistance is lower or higher in value than the source resistance of the network, its dissipated power will be less than maximum.

Tutorial 1- EMF

1.	a)	If 15C of charge moves between two electrodes with 1200V between them, how much work is done?

b)	If the current in this device lasted for 600s, what is its power consumption?

c)	What is the current in this device?

d)	Using 'volts = joules per coulomb’ and 'amps = coulombs per second', find what quantity we get by multiplying volts by amps.

2.	A 12V 24W bulb is used in series with a 4 Ω resistor so that the bulb can run normally form a 20V supply. What is the resistance of the bulb?

3.	A battery of e.m.f. 12V has a terminal p.d. of 9V when connected to an external circuit drawing 3A. What is the internal resistance of the battery?

4.	a)	A 12V car battery has an internal resistance of 0.001Ω. What is its 'short circuit' current?

b)	What is the short circuit current of a 1.5V dry cell of internal resistance 1.25Ω?

5.	A power source has a terminal p.d. of 5.7V when its external circuit is receiving 1.5A from it. When the external circuit is changed so that the current drawn from the source is 2A, the terminal p.d. measurers 4.6V	What is the source's e.m.f. and internal resistance?

Tutorial 2 More EMF

What is the e.m.f. of a cell of resistance 3 which can produce a current of 0.2A in a wire of resistance 6?

A cell of e.m.f. 1.2V and internal resistance 0.4 maintains a current in an external resistance of 2. Find the p.d. between the terminals of the cell.

A voltmeter gives a reading of 2V when connected open circuit to an accumulator. When it lights a lamp of resistance 3.5 the reading on the meter falls to 1.4V. Find the internal resistance of the accumulator.

A cell has a resistance of 3 and an e.m.f. of 2V and another cell has a resistance of 1 and e.m.f. of 1.5V.

Calculate the resistance of the wire which, when connected to either cell, will produce the same current.

If the e.m.f. of the cell is 4V calculate

The t.p.d.

The lost volts.

What would happen to these values if another 10 resistance was added in series?

If the e.m.f. is 16V and the current going through the circuit is 10 mA find the internal resistance.

. Calculate the e.m.f. of the cell if the current is 0.4A.

3.	A battery of e.m.f. 12V has a terminal p.d. of 9V when connected to an external circuit drawing 3A. What is the internal resistance of the battery?

4.	a)	A 12V car battery has an internal resistance of 0.001Ω. What is its 'short circuit' current?

b)	What is the short circuit current of a 1.5V dry cell of internal resistance 1.25Ω?

5.	A power source has a terminal p.d. of 5.7V when its external circuit is receiving 1.5A from it. When the external circuit is changed so that the current drawn from the source is 2A, the terminal p.d. measurers 4.6V

What is the source's e.m.f. and internal resistance?

What is the e.m.f. of a cell of resistance 3 which can produce a current of 0.2A in a wire of resistance 6?

A cell of e.m.f. 1.2V and internal resistance 0.4 maintains a current in an external resistance of 2. Find the p.d. between the terminals of the cell.

A voltmeter gives a reading of 2V when connected open circuit to an accumulator. When it lights a lamp of resistance 3.5 the reading on the meter falls to 1.4V. Find the internal resistance of the accumulator.

A cell has a resistance of 3 and an e.m.f. of 2V and another cell has a resistance of 1 and e.m.f. of 1.5V.

Calculate the resistance of the wire which, when connected to either cell, will produce the same current.

Internal Resistance.

If the e.m.f. of the cell is 4V calculate

The t.p.d.

The lost volts.

What would happen to these values if another 10 resistance was added in series?

If the e.m.f. is 16V and the current going through the circuit is 10 mA find the internal resistance.

3. Calculate the e.m.f. of the cell if the current is 0.4A.

Tutorial 3- Past papers

1.

A rechargeable cell is rated at 0.5 A h (ampere hour). This means that, for example, it can supply a constant current of 0.50 A for a period of 1 hour. The cell then requires to be recharged.

What charge, in coulombs, is available from a fully charged cell?

A fully charged cell is connected to a load resistor and left until the cell requires recharging. During this time the p.d. across the terminals of the cell remains constant at 1.2 V. Calculate the electrical energy supplied to the load resistor in this case 3

State what is meant by the e.m.f. of a cell

The circuit shown below is used in an experiment to find the e.m.f. and internal resistnce of the rechargeable cell.

The voltmeter and ammeter readings for a range of settings of the variable resistor are used to produce the graph below.

Use the graph (next page) to find the values for the e.m.f. and internal resistance of the cell. 4

(a) A cell of e.m.f. 1.5 V and internal resistance 0.75 is connected as shown in the following circuit.

Calculate the value of the reading on the voltmeter.

What is the value of the “lost volts” in this circuit?	3

A battery of e.m.f. 6 V and an internal resistance, r, is connected to a variable resistor R as shown in the following circuit diagram.

The graph below shows how the “lost volts” of this battery changes as the resistance of R increases.

Use information from the graph to calculate the p.d. across the terminals of the battery (t.p.d.) when the resistance of R is 1 .

Calculate the internal resistance, r, of the battery.	4

The Circuit below is used to determine the internal resistance r of a battery of e.m.f. E.

The variable resistor provides known values of resistance R.

For each value of resistance R, the switch S is closed and the current I is noted.

For each current, the value of is calculated.

In one such experiment, the following graph of R against is obtained.

Conservation of energy applied to the complete circuit gives the following relationship.

E = I(R + r) 1

Show that this relationship can be written in the form

Use information from the graph to find:

the internal resistance of the battery;

the e.m.f. of the battery. 3

The battery is accidentally short-circuited. Calculate the current in the battery when this happens.	2

Tutorial Answers

Tutorial 1

1.	a)	There are 18000J of work done.

b)	It is a 30W device.

c)	The current in the device is 25mA

d)	The product of volts and amps is watts.

2.	The bulb has a resistance of 6Ω

3.	The battery's internal resistance is 1Ω

4.	a)	The short circuit current is 12000A

b)	The short circuit current of a dry cell is 1.2A

5.	The e.m.f. of the power supply is 9V and its internal resistance is 2.2Ω.

Tutorial 2

Worked ANSWERS FOR TUTORIAL 1

1.	a)

There are 18000J of work done.

b)	It is a 30W device.

c)	The current in the device is 25mA

d)

The product of volts and amps is watts.

2.

The bulb has a resistance of 6Ω

3.	The battery's internal resistance is 1Ω

4.	a)	E = IR + Ir

A 'short circuit' means that R = zero

Thus the equation above gives 12 = 0 + Ix0.001 or I = 1.2x104

The short circuit current is 12000A

b)	1.5 = 0 + Ix1.25

Thus I = 1.2. The short circuit current of a dry cell is 1.2A

5.

Substitute this value of r in equation (1) to get

E= 5.7 + 1.5x2.2

=5.7 + 3.3

= 9

The e.m.f. of the power supply is 9V and its internal resistance is 2.2Ω.

Tutorial Answers

1.	(a) (i)	Q = It	= 0.5x3600	= 1800C

(ii)	E = Pt but	P = IV	=>E=ItV

E = 0.5x3600x1.2 = 2160J

OR use E = QV (same answer)

(b) (i)	The emf is the energy the cell supplies to each coulomb of charge	passing through it.

(ii)	The emf = 1.4V

The internal resistance is equal to the negative of the gradient of the line given.

m = (y1-y2)/(x1-x2)	m = (1.0-0.6)/(1.0-2.0)	m = -4

r = 4Ω

3.	(a) E = I(R+r)	E/I = R+r	R = E/I-r	As required

(b) (i)	The above equation can be written as:	R = E(1/I) - r

Compare this to: y = mx + c

The y-intercept(c) is equal to -r.

The gradient(m) is equal to E.

c = -2.5 =>r = 2.5Ω

(ii)	m = (y2-y1)/(x2-x1)

m = (6-0)/(0.5-0.15)

m = 6/0.35

m = 17.1	=>E = 17.1V

(c)	I = E/r	= 17.1/2.5 = 6.8A

CHAPTER 6 CAPACITORS

Summary of content

6	Capacitors

Capacitors and the relationship between capacitance, charge and potential difference.

The total energy stored in a charged capacitor is the area under the charge against potential difference graph. Use the relationships between energy, charge, capacitance and potential difference.

Variation of current and potential difference against time for both charging and discharging.

The effect of resistance and capacitance on charging and discharging curves.

Carry out investigations into capacitor charging/discharging using data loggers or otherwise

Energy storage. Flash photography.

Smoothing and suppressing. Capacitance-based touch screens

Background to Capacitors

What is a capacitor?

A capacitor is a useful device for storing charge, and hence energy. They have numerous uses and so obviously lots of potential for exam questions; many of which are quite tricky, so it is quite important that you have a sound understanding of how they work. They can also be used for an integration question where capacitors are linked with another part of the course.

Capacitor circuit symbol

Capacitance is measured in FARADS.

1 Farad = 1 Coulomb per volt. A capacitor of 10 pF can store 10 pF of charge at a voltage of 1 Volt.

N.B. 1 F is a very large unit. C would normally be expressed as F or nF.

Capacitance

Units of capacitance are therefore C V-1 or Farads (F)

Here is some information on capacitors that is for interest and help in understanding.

Capacitors are analogous with a beaker:

The size of beaker corresponds to the maximum capacitance.

Amount of liquid in the beaker corresponds to the charge on the capacitor.

Show and describe

metal plate insulator another metal plate in front of this to compete a “sandwich” Most capacitors can be connected either way around, but ELECTROLYTIC capacitors must be connected correctly according to the + and – labels.

If the rated p.d. across a capacitor is exceeded it will break down (puncturing the insulation).

As an aside:

Capacitors in parallel:	Capacitors in series:

4)	Factors affecting capacitance:

area of plates;

distance between plates;

material between plates.

So how do Capacitors work?

The capacitor stores electric charge - note that this is entirely different to current & voltage. Once charged, as in the circuit below - the capacitor will store the charge. This charge will not go anywhere (actually, it will slowly leak away - but at an insignificant rate) & can be used to power a circuit for a very short while - lighting a small torch lightbulb for 5 seconds is an impressive feat for a capacitor. They are usually used in a delay circuit. Most capacitors are polarity sensitive - ie. They won't work properly if you insert them the wrong way round.

Note that these little dudes will quite happily blow up if you put too much current through it or voltage across it, or if you connect it the wrong way round (although that only applies to the electrolytic capacitor). They are quite difficult to blow up without trying, as long as you take care when designing your circuits.

No one bothered to explain to me when I was at school how a capacitor worked. It is easy to confuse them with the work that was completed on electric fields and moving charge between two plates. THIS IS NOT THE SAME. With a capacitor charge is NOT moved from one plate to the other plate.

- - - - - - + + + + + + Electrons are transferred from the cell to this plate Electrons are repelled from this plate by the electrons on the other plate. The first thing to realise is that the electrons do not pass between the capacitor plates. Electrons travel from the power supply to one of the capacitor plates. Because the plates are in such close proximity the electrons on that plate repel electrons from the other plate which pass around the circuit and back to the power supply. The first electron is very easy to put on the plate and the capacitor offers no resistance. As the charge builds up on one plate it becomes more difficult to add charge and so the reactance (resistance) of the capacitor increases. Eventually the voltage on the capacitor equals the supply voltage. There is then no potential difference between the plates and the source so no more charge can be added. As the reactance (resistance) of the capacitor increases the current in the circuit decreases. In this circuit we have also placed a resistor as this is used to control the maximum current passing through the capacitor. Too much current can destroy a capacitor.

Charging the capacitor requires a potential difference to be placed across it. Work is done transferring charge (Q) onto the plates. The rate at which the charge is transferred is controlled by the capacity of the capacitor and the value of the resistor.

Capacitance

VS VC RR I

constants

RR = RR

When the switch is open and

the capacitor holds no charge:

I = 0 A

Vc = 0 V

Qc = 0 C

VR = 0 V

Rc = 0

The instant the switch is closed

Vc = 0 V

Qc = 0 C

VR = Vs

Rc = 0

thereafter

I decreases as charge builds up on the capacitor

Vc increases

Qc increases

VR decreases

one plate of the capacitor becomes positive

one plate of the capacitor becomes negative

When the capacitor is fully charged

Vc = Vs

VR = 0 V

I = 0 A

During the charging process the voltage across the capacitor gradually increases until it reaches a maximum which is equal to the supply voltage Vs. The rate of change in voltage across the capacitor is greatest at the beginning. Vs remains constant (ignoring lost volts so there would be a difficult question!) therefore as Vc increases, VR must decrease.

The total of these two voltages (Vc and VR) must remain equal to the supply voltage so if you know the supply voltage at any time and one of the voltages Vc or VR then the other can be found.

To find VR use the current given for that time multiplied by RR.

Charging and Discharging

If we charge the capacitor by closing the switch at time A then arrange the circuit so that at point B the capacitor is DISCHARGED through the same resistor we would observe the following voltage and current graphs.

The area under the current-time graph (= I x t) is the total charge, Q, stored on the capacitor. The area under the graph in the charging phase must be equal to the area under the graph in the discharging phase (assuming the capacitor is COMPLETELY discharged). In the above example the areas are obviously the same since the capacitor is charged and discharged through the same resistor. However, if the capacitor is discharged through a smaller value resistor it will discharge more quickly and have a larger initial current. BUT THE AREA UNDER THE DISCHARGE CURVE WILL BE EXACTLY THE SAME AS BEFORE.

When discharging a capacitor, if the load resistor is halved in value then the initial current will be doubled and therefore the time for discharging must be halved, as the total charge is the same.

If values can be put on the graph then make sure you do put them on!

Measuring the capacitance of a capacitor.

mA V constant current supply + - C R 0:00:00

Discharge C.

Connect up. Start clock and take regular values of V and I at set time intervals.

Find the charge that has accumulated on the capacitor at each time interval according to Q = I x t.

Plot a graph of Q against V. Since the gradient of the graph will be the capacitance

Therefore the capacitor has a value of 3000 μF (that is quite a large capacitor!)

Q and V are directly proportional.

1 Farad = 1 Coulomb per volt.

N.B. 1 F is a very large unit. C would normally be expressed as F or nF.

Using your knowledge work out what the area under this graph would represent. We will come back to it later.

V Q 0

Charging a capacitor on a d.c. supply.

Discharging a capacitor.

The current and voltage fall to zero, but be aware that the current flows in the opposite

direction to the original charging current.

Energy stored in a Capacitor

When a charge is moved between 2 charged, parallel plates, work is done. If we move a charge from one plate to the other against the uniform field then work done is given by:

work done = QV

When a capacitor is being charged, work is done moving charges against opposing forces. Electrons are pushed onto the already negative plate. This repels them. They are also removed from the positive plate which tends to attract them.

Energy Stored

The voltage across the capacitor is, however, constantly changing so:

E QV

We average the voltage over the charging period so that

i.e. it is the area under a V-Q graph for a capacitor.

In summary, and using Q = CV,

V Q 0

Capacitors and a.c.

NB.	Remember that “REACTANCE” is used to describe the “RESISTANCE” of a capacitor.

Charging on a.c.

With a.c. a capacitor will begin to charge up, then current starts flowing in the opposite direction. The capacitor therefore discharges and then begins to charge up again.

Hopefully you can imagine that the storage of charge will be greatly affected by how quickly this change of direction is occurring.

Low frequency a.c.

When the frequency of the supply is low the capacitor is fully charged up (voltage is high across the capacitor and the current zero) before the current flows in the opposite direction.

High frequency a.c.

The capacitor does not get time to fully charge up or discharge before the current direction is changed.

This shows us that capacitors block d.c. but allows a.c. through.

Resistance and Frequency

Part 1 Resistance and Frequency

A Signal Generator V R Set up the above circuit and use it to see how the voltages across and the current in R change as you vary the frequency of the supply.

Examine your results to see if there is any connection between an increase in frequency and the resisting effect of R.

Write up your experiment briefly.

Part 2 Resistance and frequency with an oscilloscope

Replace the voltmeter with an oscilloscope and explain how to measure the frequency.

Write up this experiment INCLUDING ALL YOUR WORKING.

NB.	THE RESISTANCE OF A RESISTOR IS UNAFFECTED BY FREQUENCY.

This can therefore be a method of determining whether a component is a capacitor or a resistor.

Reactance of capacitors in a.c. circuits

Resistance (properly called reactance when used with capacitors) of a capacitor can be found out in the usual way.

With a LOW FREQUENCY supply the average VOLTAGE is LARGE and the average CURRENT is SMALL.

Reactance is large.

With a HIGH FREQUENCY supply the average VOLTAGE is SMALL and the average CURRENT is LARGE.

Reactance is small.

We tend therefore to say that capacitors allow current to flow in an a.c. circuit but not in a d.c. circuit.

Read up on BLOCKING and SMOOTHING.

Relationship Between Current and Frequency in an A.C capacitor Circuit.

Blocking and Smoothing

The change in reactance of a capacitor with frequency can be exploited as follows.

Blocking

An electrical signal consists of a steady d.c. voltage with an a.c. voltage superimposed on it.

If this signal is now fed into the following circuit:

then the d.c. component of the signal is removed (or blocked).

Smoothing

A capacitor can also be used in to smooth an a.c. voltage as follows.

A normal a.c. signal is fed into the following circuit.

In the absence of the capacitor the four diodes would rectify this signal thus:

But with the capacitor present this signal is smoothed by the repeated charging and discharging of the capacitor. This output is known as the ripple voltage.

Tutorial 1: capacitance

1.	What is the capacitance of a capacitor storing 0.005C when the p.d. between its plates is 50V?

2.	A capacitor has a p.d. of 20V between its plates. If its capacitance is 25µF, what charge is on it?

3.	What is the p.d. across a 50µF capacitor storing 2.5mC?

Tutorial 2: capacitance

1.	How much work is done in charging up a 30µF capacitor which stores 0.001C at 20V?

2.	What is the capacitance of a capacitor storing 0.016J at a p.d. of 40V?

3.	What is the charge on a 30µF capacitor storing 1.35J?

4.	Calculate the p.d. across a 10µF capacitor if 0.0125J is required to charge it.

Tutorial 3: capacitance

1.	What is the reactance of a capacitor when the p.d. across it is 20V and the current in the circuit 5mA?

2.

What is the reactance of the capacitor?

3.	A 2000Ω resistor is connected in series with a capacitor whose reactance measures 1500Ω. If the a.c. supply is quoted as 14V, what is the peak voltage across the capacitor?

Tutorial 4: CAPACITANCE

1.

How do you show that the component in the box is a resistor?

2.	A 24V a.c. power supply sends a 100Hz current through a 480Ω resistor:-

What is the current when the frequency is doubled to 200Hz?

3.

The 2µF capacitor holds 20mC of charge. What is the initial current when S is closed?

4.	If the switch S is moved to A and held for a time, then moved quickly to B, what do the graphs of current against time and p.d. against time look like for the resistor R while the switch is held at B?

5.

How does the brightness of each lamp change when the supply voltage is increased to 25V?

6.	In the diagram below, the bulbs are identical and the three capacitors have the values shown. How do the bulbs compare in brightness?

7.

8.	What is the p.d. of the supply at the instant C is storing 1.125x10-4J in the circuit shown here? : -

Tutorial 5: Exam Questions

In an experiment to measure the capacitance of a capacitor a student sets up the following circuit:

C coulombmeter Y X V VS

When the switch is in position X the capacitor charges up to the supply voltage, VS. When the switch is in position Y the coulombmeter indicates the charge stored by the capacitor.
The student records the following measurements and uncertainties.
Reading on voltmeter	= (2·56 ± 0.01) V
Reading on coulombmeter	= (32 ± 1) μC
Calculate the value of the capacitance and the percentage uncertainty in this value. You must give your answer in the form

value ± percentage uncertainty. 3

V + - 12 V S relay 2200 F 3·3 k The student designs the circuit shown below to switch off a lamp after a certain time.
The 12 V battery has negligible internal resistance.
The relay contacts are normally open. When there is a current in the relay coil the

contacts close and complete the lamp circuit. Switch S is initially closed and the lamp is on.

What is the maximum energy stored in the capacitor?

Switch S is now opened. Explain why the lamp stays lit for a few seconds.

The 2200 F capacitor is replaced with a 1000 F capacitor. Describe and explain the effect of this change on the operation of the circuit. 6 (9)

a) 6·0 V 470 F 1·5 k V A S The following diagram shows a circuit that is used to investigate the charging of a capacitor.
The capacitor is initially uncharged. It has a capacitance of 470 F and the resistor has a resistance of 1.5 k. The battery has an e.m.f of 6 V and negligible internal resistance.

Switch S is now closed. What is the initial current in the circuit?

How much energy is stored in the capacitor when it is fully charged?

What change could be made to this circuit to ensure that the same capacitor stores more energy?

A capacitor is used to provide the energy for an electronic flash in a camera.
When the flash is fired, 6·35 x 10-3 J of the stored energy is emitted as light.
The mean value of the frequency of photons of light from the flash is 5·80 x 1014 Hz.
Calculate the number of photons emitted in each flash of light.

R S V A 2000 F + - The circuit below is used to investigate the charging of a 2000 F capacitor. The d.c. supply has negligible internal resistance.

The graphs below show how the potential difference, VR across the resistor and the current, I, in the circuit vary with time from the instant switch S is closed.
I)	What is the potential difference across the capacitor when it is fully charged?

II)	Calculate the energy stored in the capacitor when it is fully charged.

III)	Calculate the resistance of R in the circuit above. 5

The circuit below is used to investigate the charging and discharging of a capacitor.
The graph below shows how the power supply voltage varies with time after switch S is closed.
The capacitor is initially uncharged.
The capacitor charges fully in 0·3 s and discharges fully in 0·3 s.
Sketch a graph of the reading on the voltmeter for the first 2·5 s after switch S is closed.
The axes on your graph must have the same numerical values as those in the above graph 2

R S V A C + - power supply

A technician carries out an experiment to measure the capacitance of a capacitor C.
The capacitor, initially uncharged, is charged up to 2 V using the circuit below.

The charging current is kept constant at 0·20 mA during the charging process by adjusting the resistance of R. The capacitor is fully charged in 10 seconds.

Explain whether the resistance of R is increased or decreased during the charging period.	1

What is the charge supplied to the capacitor?

Calculate the capacitance of the capacitor.	2

The capacitor is then used in a circuit where it is connected across a 10 V supply. Calculate:

the charge stored on the capacitor

the energy stored on the capacitor.	(9)

An audio engineer obtains the results shown on the graph below. The graph shows how the current in a circuit containing an 8 F capacitor varies with frequency. The output of the electrical supply is 2 V r.m.s.

Describe an experiment to obtain such a graph. Your answer should include the following:

a circuit diagram of the apparatus required

a statement of the variables measured and controlled

a description of how the measurements were taken

conclusions which can be drawn from the graph.

An engineer has two loudspeakers, LS1 and LS2, to connect to an audio amplifier. One of the speakers is designed so that it is able to produce low frequency sounds. The engineer connects the loudspeakers to the amplifier using the circuit shown below.

Which of the loudspeakers, LS1 or LS2, is intended to emit low frequency sounds. You must explain your answer. (7)

Tutorial Answers Capacitance

Tutorial 1

1.	The capacitance is 1x10-4F (100µF)

2.	The capacitor is storing 0.0005C

3.	The p.d. across the capacitor is 50V

Tutorial 2

1.	The work done during charging is 0.01J

2.	The capacitance is 2x10-5F (20µF)

3.	The stored charge is 0.009C

.	The p.d. across the capacitor is 50V

Tutorial 3

1.	The reactance of the capacitor is 4000Ω

2.	Since the resistance of the whole circuit is 2500Ω, the reactance of the capacitor must be 1000Ω

3.	The peak voltage must = 8V

Tutorial 4

1.	Note the voltage and current at some frequency. Use them to calculate the resistance of the component. Repeat the measurements for different supply frequencies. If the resistance remains constant, then the component is a resistor.

2.	Since the resistance is independent of the frequency, the current remains at 0.005A when the frequency doubles to 200Hz.

3.	The initial current is 1A

4.

5.	Each is 2.5 times brighter since neither the resistance nor the capacitive reactance depend on the supply voltage.

6.	The bulbs increase in brightness from (1) to(3) since the capacitive reactance becomes less with increasing capacitance.

7.	Bulb (1) remains at constant brightness while bulb (2) increases in brightness as the capacitive reactance decreases with the increase in supply frequency.

8.	The supply voltage is 8V.

Tutorial Work Answers

Capacitance/ tutorial answers

Answers for tutorial 1

1.

The capacitance is 1x10-4F (100µF)

2.	The capacitor is storing 0.0005C

3.	The p.d. across the capacitor is 50V

ANSWERS FOR TUTORIAL 2

1. The work done during charging is 0.01J

2. The capacitance is 2x10-5 F (20µF)

3.

The stored charge is 0.009C

4.

The p.d. across the capacitor is 50V

ANSWERS FOR TUTORIAL 3

1.

The reactance of the capacitor is 4000Ω

2.	Apply V=RI to the whole circuit to get:-

Since the resistance of the whole circuit is 2500Ω, the reactance of the capacitor must be 1000Ω.

3.	Since the circuit is behaving as a voltage divider, the p.d. across the capacitor is This is the r.m.s. voltage across the capacitor so the peak voltage must by 6x√2 = 8V

ANSWERS FOR TUTORIAL 4

1.	Note the voltage and current at some frequency. Use them to calculate the resistance of the component. Repeat the measurements for different supply frequencies. If the resistance remains constant, then the component is a resistor.

2. Since the resistance is independent of the frequency, the current remains at 0.005A when the frequency doubles to 200Hz.

3.	The initial current depends only on the p.d. and the resistance.

Since the p.d. is 10000V,

The initial current is 1A

4.

5.	Each is 2.5 times brighter since neither the resistance nor the capacitive reactance depends on the supply voltage.

6.	The bulbs increase in brightness from (1) to (3) since the capacitive reactance becomes less with increasing capacitance.

7.	Bulb (1) remains at constant brightness while bulb (2) increases in brightness as the capacitive reactance decreases with the increase in supply frequency.

8.

The p.d. across the capacitor is 3V.

V=RI applied to the resistor gives V = 2000 x 2.5x10-3 = 5 volts.

Since the supply voltage equals the sum of the individual p.d.'s round the circuit, its value is 8V.

Tutorial Answers: Exam question soutions

2000

24.a.	To calculate the capacitance uses the mean values of charge and voltage.

C = Q/V

C = (32/2.56)F

C=12.5F

The percentage error in capacitance value can be taken to be equal

to that of largest individual percentage error.

% error in voltage =	(0.01/2.56)x100 = 0.39%

% error in charge =	(1/32)x100	= 3.125%

C=12.5F+-3.125%

OR

3.125% of 12.5F = 0.39F =>	C = (12.5+-0.4)F

NB/	Only quote the error to the same number of decimal places as the capacitance value.

b.i.The maximum energy is stored in the capacitor when the voltage across the capacitor is equal to the supply voltage.

Vc= 12V

C = 2200x10-6

E = ?

E = 1/2(QVc)

Q = CVc

=>E = 1/2(CVc2)

E = 0.5(2200x10-6x122)

E = 0.1584J

b.ii.(A)	When the switch is opened the capacitor discharges through the resistor and relay coil. The discharge current magnetises the coil closing the switch in the lamp circuit, causing the lamp	to light. As the discharge current gradually falls the coil loses its magnetism and the switch in the lamp circuit opens. When this happens the lamp goes off.

(B)	Increasing the value of the capacitor increases the discharge time. The energy stored in the capacitor is also greater. This means that the lamp will stay lit for longer.

2001

25.a.i.	The initial charging current(Imax) occurs when all of the supply voltage(Vsupply) is across the 1.5k resistor(R).

Imax = Vsupply/R

Imax = 6/1500

Imax = 4x10-3A

a.ii.	When fully charged the voltage across the supply voltage is equal to the voltage across the capacitor.

Vsupply = Vc = 6V

Ecapacitor = QVc/2

Q = CVc

=>Ecapacitor = CVc2/2

Ecapacitor = 470x10-6x62/2

Ecapacitor = 8.46x10-3J

a.iii.	Increasing the supply voltage would increase the energy storing capacity of the capacitor. This is because the final voltage, across the fully charged capacitor, would be higher.

25.b.	Etotal = 6.35x10-3J

fphoton = 5.80x1014Hz

h = 6.63x10-34Js

Ephoton = ?

Ephoton = hfphoton

Ephoton = 6.63x10-34 x 5.80x1014

Ephoton = 3.84x10-19J

Etotal = NEphoton

N = Etotal/Ephoton

N = 6.35x10-3/3.84x10-19

N = 1.65x1016

2002

25.a.i.	Initially all the supply voltage is across the resistor.

VR = Vsupply = 6V

When the capacitor is fully charged: Vsupply = Vcapacitor

Vcapacitor = 6V

a.ii. E = CV2/2

E = (2000x10-6x62)/2

E = 0.072/2

E = 0.036J

a.iii.	Imax = Vsupply/R

R = Vsupply/Imax

R = 6/7.5x10-3

R = 800