VCE Physics

Unit 1

ELECTRICITY

Unit Outline• Apply the concepts of Charge (Q), Electric Current (I), Potential Difference (V), Energy (E)

and Power (P), in electric circuits.• Analyse electrical circuits using the relationships I = Q/t, V = E/Q, P = EIt = VI, E = VIt. • Model Resistance in Series and Parallel using:, • Potential Difference versus Current (V-I) Graphs • Resistance as the Potential Difference to Current ratio, including V/I = R =

constant for ohmic devices.• Equivalent effective resistance in arrangements in: • series: RT = R1 + R2 + R3 + …..• parallel: 1/RT = 1/R1 + 1/R2 + …..• Model simple electrical circuits such as car and household AC electrical systems as

simple direct current (DC) circuits.• Model household electricity connections as a simple circuit comprising fuses, switches,

circuit breakers, loads and earth.• Identify causes, effects and treatment of electric shock in homes and relate these to

approximate danger thresholds for current and time.• Investigate practically the operation of simple circuits containing resistors, including

variable resistors, diodes and other nonohmic devices.• Convert energy values to kilowatt-hour (kWh)• Identify and apply safe and responsible when conducting investigations involving

electrical equipment and power supplies.

Chapter 1

The Basics

1.0 Electric Charge

Electric Charge is a property of some atomic particles. Which ones ? Protons and Electrons. These two particles carry an equal and opposite electric charge.This charge is the smallest known amount of charge that exists independently.This charge is called the “ELEMENTARY CHARGE”

Atoms consist of a nucleus, containing protons and neutrons, with electrons circulating around it.

The charge carried by the Proton is DEFINED to be POSITIVE.The charge carried by the Electron is DEFINED to be NEGATIVE.

The UNIT of Electric Charge is the COULOMB, Symbol (C).

Protons carry a charge of +1.6 x 10-19 CoulombsElectrons carry a charge of – 1.6 x 10-19 Coulombs

If each electron (or proton) carries such a small charge, a large number would be needed to make up 1 Coulomb of charge.

Charge

1. How many electrons in 1 coulomb of charge ?

1 electron carries 1.6 x 10-19 CSo 1 Coulomb of charge will have 1/(1.6 x 10-19) electrons= 6.25 x 1018 electrons

2. How many electrons in 7.5 C of charge ?

1C has 6.25 x 1018 electrons so 7.5 C will have 7.5 x 6.25 x 1018 = 4.69 x 1019 electrons

1.2 Electric Current When electric charges are made to move or flow an ELECTRIC CURRENT (Symbol I) is said to exist.The SIZE of the Current depends on the number of Coulombs of Charge passing a given point in a given Time.The Unit of Current is the Ampere often shortened to Amp (Symbol A)

Mathematically:I = Q/t

Where: I = Current in AmpsQ = Charge in Coulombst = Time is Seconds

Current flowing along wire

1 Coulomb passing this point in 1 second = 1 amp of current

So, in 1 second 6.25 x 1018 electrons pass this point

Electric current has the property of starting immediately a circuit is complete and stopping immediately a circuit is broken. Once the current is flowing it stays the same all around the circuit.

Current3. Calculate the current flowing if 3.57 Coulomb of charge passes a point in 1.25 sec.

I = Q/t = 3.57/1.25 = 2.86 Amp

4. If 5.62 A of current flows through a wire in 0.68 sec.(a) How much charge has been moved ?

I = Q/t Q = It = (5.62)(0.68) = 3.82 C

If 1 C of charge is carried by a total of 6.25 x 1018 electrons, then 3.82 C is carried by (3.82)( 6.25 x 1018) = 2.39 x 1019 electrons

(b) How many electrons were needed to transport the charge in (a) ?

5. If a current of 125 A resulted from the movement of 225 C of charge, for how long did the current flow ?

I = Q/t t = Q/I = 225/125 = 1.8 sec

1.3 Conventional versus Electron Current

In Direct Current (DC) electric circuits, the current always flows in one direction.On circuit diagrams, it is ALWAYS shown as flowing from the positive to negative terminal of the power source.

It is a quirk of history that the current direction is shown this way.Electric currents were discovered before the electron. It was thought that the charge carriers were positive and the current must flow this way,

Wire connected to a Battery

ElectronCurrentConventional Current

Never shown on circuit diagrams

Always shown on circuit diagrams

BUT, we know that a current is a stream of electrons (negative particles), which must travel in the other direction.What’s going on ?

This means the current carriers must be positively charged because they will be repelled (like charges repel) from the positive terminal, and attracted (unlike charges attract) to the negative terminal. Protons are the Positive particles.

Conventional vs Electron Current6. Currents shown on circuit diagrams;

A: are from the negative to the positive terminal of the power supply and are called electron currentsB: are from the positive to the negative terminal of the power supply and are called conventional currentsC: are from the positive to the negative terminal of the power supply and are called electron currentsD: are from the negative to the positive terminal of the power supply and are called conventional currents

1.4 Potential DifferenceFor a current to flow around a circuit a “driving force” is needed. This driving force is the difference in VOLTAGE between the start and the end of the circuit.The larger the current you want the greater the Potential Difference (Voltage Difference) you require.Potential Difference (P.D.) is best understood using the water analogy:

Low output from the tap = low current.SO A SMALL P.D. CAN ONLY DRIVE A SMALL CURRENT.

Small Drop = Low Pressure

Low output = low

currentA short drop between storage and tap gives low water pressure = a low P.D. Large Drop =

High Pressure

High output = High current

A large drop between storage and tap gives high pressure = Large P.D.

High output at the tap = high currentSO A LARGE P.D. CAN DRIVE A LARGE CURRENT.

Strictly, Potential Difference is DEFINED as a measure of the energy given to the charge carriers (the electrons) for them to complete their job, that is, to travel once around the circuit.

1.5 Potential Difference (2)Mathematically:

V = E/Q

Where:V = P.D. measured in Volts (V)E = Electrical Potential Energy in Joules (J)Q = Electrical Charge in Coulombs (C)

This means that by passing through a P.D. of 1 Volt, 1 Coulomb of charge picks up 1 J of energy, or more simply 1 V = 1 JC-1

12 V

In this case, each coulomb passing through the battery will pick up 12 J of energy.(The energy is used up in lighting the globe)24 V

The Battery P.D. is now increased to 24 V.How many Joules of energy will each coulomb now pick up ?

Answer: 24 J

There are many terms used in texts to describe Voltage, some of these include Potential, Potential Difference, Potential Drop, Voltage Drop, Voltage DifferenceAt this stage of your studies you can take them all to mean the same thing. The preferred term for the VCAA examiners in Potential Difference

One further voltage, EMF (Electro Motive Force), while still measured in volts, is slightly different and cannot be grouped with the other terms.

Voltage7. One coulomb of charge passing through a battery picks up 15 J of energy. What Potential Difference did the charge pass through ?

8. An external circuit is connected to a 24 V battery. If 6.5 C of charge passes through the battery. (a) How much energy does each coulomb of charge pick up in passing through the battery ?

(b) How much energy (in total) has the battery supplied to the charge passing through it

V = E/Q = 15/1 = 15 V

Each Coulomb will pick up 24 J of energy in passing through the battery

V = E/Q E = VQ = (24)(6.5) = 156 J

1.6 Electrical EnergyThe Charge Carriers in a circuit obtain their energy from a power source or power supply.The amount of energy (E) the charge carriers pick up depends upon the size of the voltage difference through which they are forced to travel.Since energy transferred = work done, another way of defining electrical energy (E) is by the work done on the charge (Q) in passing through a Voltage (V)

Mathematically E = VQ

and since Q = ItSubstituting we get E = VItWhere:E = Electrical Energy (J)V = Voltage (V)Q = Charge (C)I = Current (A)t = Time (s)

An external wire connected to a battery will have electrons flowing through it as shown.In completing the circuit inside the battery, the electrons need to flow from the positive to the negative terminal.They will not do this willingly and must be forced through the battery.The work done on the electrons increases their electrical energy and gives them enough energy to do another trip around the external circuit.

ElectronCurrent

Electrical Energy

11. A circuit is switched on for 6.5 minutes in that time 3.5 x 104 J of energy has been transferred to the charge carriers. If the current flowing was 11.3 amps, calculate the Potential Difference of the power supply driving that current.

9. A current of 4.2 A is being driven around a circuit by a Potential Difference of 87 V. If the circuit is allowed to operate for 36 s, how much energy has been transferred to the charge carriers ?

E = VIt = (87)(4.2)(36) = 1.32 x 104 J

10. A total of 1.2 x 103 J of electrical energy has been transferred to the charge carriers in a circuit driven by a 48V battery. If the circuit is switched on for 12 minutes, how many mA (milliamp) of current will have flowed during this time ?

E = VIt I = E/Vt = (1.2 x 103)/[(48)(12 x 60)] = 0.035 A = 35 mA

E = VIt V = E/It = (3.5 x 104)/[(11.3)(6.5 x 60)] = 7.9 V

1.7 Electrical Energy (2)The energy picked up by the charge carriers is used up in “driving” whatever device is connected to the external circuit.Here we have an incandescent light globe as part of a circuit.

Q coulombs of electricity carryingU joules of energy enter here Q coulombs of electricity leave here

Current flowing along wire

Charges have High Energy here Charges have low energy hereVoltage Drop

= V volts

U joules of electrical energy transferred to Heat and Light

1.8 Electric PowerElectric Power is DEFINED as the Time Rate Of Energy Transfer or the Time Rate Of Doing Work.

Mathematically: P = E/t

And since E = VIt Substituting we get P = VI

WhereP = Power (in Watts, W)E = Electrical Energy (J)V = Voltage (V)I = Current (A)t = Time (s)

Using Ohm’s Law (See Chapter 2.) The Power formula can be rewritten as:

P = VI = I2R = V2/R

Electrical Power

15. The kettle mentioned in Q 14 is taken on a world trip by its owner. In America (where the mains supply operates at 110V) he plugs it into a wall socket. Will the safety switch trip now ? Back up your answer with a calculation.

12. Calculate the power consumed by an electric drill operating at 240 V and 7.5 A.

13. An electric oven consumes 1.5 x 107 J of energy while cooking a roast. If the roast took 2 hours to cook, at what power is the oven operating (quote your answer in kW) ?

14. An electric kettle is rated at 3000 W. It is fitted with a 15 Amp safety switch. If it is connected to a 240 V supply will the safety “trip” (switch off) ? Back up your answer with a calculation.

P = VI = (240)(7.5) = 1800 W

P = E/t = (1.5 x 107)/ (2 x 60 x 60) = 2.1 x 103 W = 2.1 kW

No; P = VI I = P/V = 3000/240 = 12.5 A less than the 15 A safety switch rating

Yes; P = VI I = P/V = 3000/110 = 27.3 A greater than safety switch rating

1.9 Common Electrical Symbols

Diode

Single Cell Battery Earth or Ground

Crossed Wires - JoinedCrossed Wire

not JoinedFixed Resistor

Variable Resistor

Switch

Capacitor Globe

A.C. Power Supply

V

Voltmeter

A

Ammeter

G

Galvanometer LED

Electrical Components

1

G

V

2

3

4

5

A

1

2 34

56

16. Identify the numbered components in the circuits below

(a) (b)

(a) 1 = Variable Resistor, 2 = AC Supply, 3 = Ammeter, 4 = Lamp, 5 = Earth, 6 = Switch(b) 1 = Voltmeter, 2 = Battery, 3 = Capacitor, 4 = Galvanometer, 5 = Fixed Resistor

1.10 Series and Parallel

Electrical components can only be connected together in one of two ways:

Series – where components are connected end to end

Series

Parallel - where components are connected side by side.

Parallel

1.11 A Typical Electric CircuitAn electric circuit contains a number of components, typically:•A Power Supply•Connecting Wires•Resistive Elements •Meters

Circuit diagrams are usually drawn in an organized manner with connecting wires drawn as straight lines and the whole diagram generally square or rectangular in shape.

Power Supply

Resistive Element

V

This power supply is a D.C. Supply (a Battery), and it drives the current in one direction only.

Connecting wires are drawn as straight lines with right angle bends.They are always regarded as pure conductors having no resistance.

This represents the part of the circuit where electrical energy is consumed.The resistive element could be a light globe or heater or a radio or a television.

The Voltmeter measures the voltage drop across the resistive element.It is connected in parallelIt has a very high internal resistance which diverts very little current from the main circuit.

The Ammeter measures current flow in the main circuit.It is connected in seriesIt has virtually no internal resistance so as not to interfere with the current in the main circuit

Connecting Wires

A

Meters

19. Voltmeters and Ammeters differ because:A: Voltmeters have low internal resistance and are connected in series while Ammeters have high internal resistance and are connected in parallel.B: Voltmeters have high internal resistance and are connected in parallel while Ammeters have low internal resistance and are connected in series.C: Voltmeters have low internal resistance and are connected in series while Ammeters have high internal resistance and are connected in parallel.D: Voltmeters have high internal resistance and are connected in series while Ammeters also have high internal resistance and are also connected in series

17. A Galvanometer (which is a very sensitive ammeter) when included in a circuit should be connected:A: In parallelB: Across the power supplyC: In seriesD: Any way around, it doesn’t matter18. In ideal circuits the wires used to connect the circuit components together have:A: No resistanceB: A small amount of resistanceC: A large amount of resistanceD: An infinite amount of resistance.

Chapter 2

Resistance

2.0 ResistanceAll materials fall into one of three categories as far as their electrical conductivity is concerned.

They are either :1. Conductors2. Semiconductors, or3. Insulators

ALL materials exhibit some opposition to currents flowing through them.Conductors show just a small amount of opposition.Semiconductors show medium to high opposition.Insulators show very high to extreme opposition.

This opposition is called ELECTRICAL RESISTANCE.The amount of resistance depends on a number of factors:1. The length of the material.2. The cross sectional area of the material.3. The nature of the material, measured by Resistivity

Mathematically:R = ρL/A

WhereR = Resistance in Ohms (Ω)ρ = Resistivity in Ohm.Metres (Ω.m)L = Length in Metres (m)A = Cross Sectional Area in (m2)

Length = L

Wire 2

Wire 1A1

A2

Wires 1 & 2 are made from the same material (ρ is the same for each), and are the same length (L is also the same).Wire 1 has twice the cross sectional area of Wire 2. Wire 1 has ½ the resistance of Wire 2

Resistance

21. Two pieces of wire are made of the same material and are of the same cross sectional area. Wire 1 is 3 times as long as wire 2.A: Wire 1 has 3 times the resistance of Wire 2B: Wire 2 has 2/3 the resistance of Wire 1C: Wire 1 has 1/3 the resistance of Wire 2D: Wire 1 has 6 times the resistance of Wire 2

20. Nichrome wire is sometimes used to make the heating elements in electric kettles. It has a resistivity of 6.8 x 103 Ω.m. Calculate the resistance of a piece of nichrome wire of length 1.2 m and cross sectional area 2 x 10-4 m2

R = ρL/A = (6.8 x 103)(1.2)/(2.0 x 10-4) = 4.1 x 107 Ω = 41 MΩ

2.1 ResistorsResistors are conductors whose resistance to current flow has been increased. They are useful tools for demonstrating the properties of Electric Circuits.Understanding how these circuits work is an important life skill you all need to develop. Resistor is a generic term representing a whole family of conductors such as toaster elements, light bulb filaments, bar radiators and kettle elements.

They are represented on circuit diagrams as either,

1kΩ

or

1kΩ

There are only two ways to join resistors togetherIN SERIES:The resistors are connected end to end with only one path for the current to flow. The more resistors the greater the overall resistance

IN PARALLELThe resistors are connected side by side with more than one path for the current to flow. The more resistors the lower the overall resistance

2.2 Resistors in SeriesConnected end to end, this combination of resistors gives only 1 path for current flow.

The TOTAL RESISTANCE (RT) of this combination equals the sum of resistances Thus, RT = R1 + R2 + R3

In other words, the 3 resistors can be replaced in the circuit with a single resistor of size RT

Because there is only 1 path for the current to flow, the current must be the same everywhere. The current drawn from the power supply (I) is equal to the currents through the resistors. Thus I = I1 = I2 = I3

V1 V3 V2

R1 R2 R3

VS

I

I1 I2 I3

The sum of the Potential Differences across the resistors is equal to the Potential Difference of the supplyThus VS = V1 + V2 + V3

RT

Resistors in Series

(c) Determine the values of I1 and I2

24 Ω 15 Ω 11 Ω

2.9 1.8 v3

I = 0.12 A

6.0

11

12

21.(a) Calculate the equivalent resistance that could replace the resistors in the circuit.

In series resistances add. RT = R1 + R2 + R3 = 24 + 15 + 11 = 50 Ω

(b) Determine the value of V3

In a series circuit VS = V1 + V2 + V3 V3 = VS – (V1 + V2) = 6.0 – (2.9 + 1.8) = 1.3 V

In a series circuit, current is the same everywhere so I1 = I2 = I = 0.12 A

2.3 Resistors in Parallel

R1

R2

R3

VS

I

I1

I2

I3

V3

V2

V1

When connected side by side, this combination of resistors (called a parallel network) gives many paths for current flow.

The TOTAL RESISTANCE (RT) is calculated from:1/RT = 1/R1 + 1/R2 + 1/R3.In other words the three resistors can be replaced by a single resistor of value RT.The physical effect of this formula is that the value of RT is always less than the lowest value resistor in the parallel network.

The current has many paths to travel and the total current drawn from the supply (I) is the sum of the currents in each arm of the network.Thus I = I1 + I2 + I3

Each arm of the parallel network gets the full supply Potential Difference.Thus VS = V1 = V2 = V3

RT

Resistors in Parallel

(c) Determine the value of I1

I = 1.0 mA

v1

I1

v2

1 kΩ

I = 12 mA

v3

I = 17 mA

12V

3 kΩ

12 kΩ

22. (a) What single resistor could be used to replace the 3 resistors in the circuit ?

1/RE = 1/R1 + 1/R2 + 1/R3 = 1/1000 + 1/3000 + 1/12000 RE = 706 Ω

(b) Determine the values of V1, V2, and V3

In a parallel network VS = V1 = V2 = V3 = 12 V

In a parallel network I = I1 + I2 + I3

I1 = 17 – (12 + 1) = 4 mA

Resistors Combined

(b) What is the Potential Difference across and the current through the 24 Ω resistor ?

23. (a) What single value resistor could be used to replace the network shown ?

I = 0.5 A

3 V

10 Ω

15 Ω24 Ω

50 Ω

50 Ω

100 Ω

25 V

v1

10 V

Simplify parallel networks first. For 10 and 15, RE = (1/10 +1/15)-1 = 6.0 Ω. For 50,50,100 RE = (1/50 + 1/50 + 1/100)-1

= 20 Ω.

Now RE = 6 + 24 + 20 = 50 Ω

V24Ω = 25 – (10 + 3) = 12 V, I24Ω = I = 0.5 A

2.4 Ohm’s Law

The relationship between, the potential difference across, the current through, and the resistance of, a conductor was discovered by Georg Ohm and is known as Ohm’s Law

Ohm’s Law statedmathematically is:

V = IRWhere:V = Potential Difference in Volts (V)I = Current in Amps (A)R = Resistance in Ohms (Ω)

Conductors which obey Ohm’s Law are called Ohmic Conductors.

When expressed graphically, by plotting V against I, Ohm’s Law produces a straight line graph with a slope equal to resistance (R)

V

I

Rise

Run

= Slope= Resistance

RiseRun

Note the graph passes through the origin (0,0) as it must, since if both V and I are zero, resistance is a meaningless term.

Georg Ohm1789 - 1854

Ohm’s Law (1)24. A current of 2.5 mA is flowing through a resistor of 47 kΩ. What is the Potential Drop drop across the resistor ?

V = IR = (2.5 x 10-3)(4.7 x 104) = 117.5 V

25. A 12 V battery is driving a current through an 20 Ω resistor, what is the size of the current flowing ?

V = IR I = V/R = 12/20 = 0.6 amp

26. A resistor has a 48 V potential difference across it and a 2.4 A current flowing through it. What is it’s resistance ?

V = IR R = V/I = 48/2.4 = 20 Ω

Ohm’s Law (2)27. What are the readings on meters V and A ?

VP = 12 V

R = 1.2 kΩ

V

A

V = VR = VSUPPLY = 12 V I = VR/R = 12/(1.2 x 103) = 0.01 A

V1

1 kΩ

1.5 kΩ2.4 kΩ

500 Ω

500 Ω

1 kΩ

25 V

v2

v3

A1

A2

28. (a) Determine the value of the current measured by ammeter A1 (express your answer in mA)

Need to find equivalent resistance by simplifying parallel networks.

Simplify parallel networks first. For 1.0k and 1.5k, RE = (1/1000 +1/1500)-1

= 600 Ω. For 500, 500, 1k; RE = (1/500 + 1/500 + 1/1000)-1 = 200 Ω. Now RE = 600 + 2400 + 200 = 3200 Ω = 3.2 k Ω Now I = V/R = 25/3200 = 0.0078 A = 7.8 mA

Ohm’s Law (3)

V1

1 kΩ

1.5 kΩ2.4 kΩ

500 Ω

500 Ω

1 kΩ

25 V

v2

v3

A1

A2

(b) Determine the value of the potential differences measured by voltmeters V1, V2 and V3.

V1 = Voltage across 1k, 1.5k combination.RE = 600 Ω, so V1 = IRE = (7.8 x 10-3)(600) = 4.68 V; V2 = voltage across 2.4 kΩ = (7.8 x 10-3)(2.4 x 103) = 18.72 V; V3 = Voltage across the 500, 500, 1k

combination = IRE = (7.8 x 10-3)(200) = 1.56 V

(c) Determine the current measured by ammeter A2

A2 = current through 500 Ω resistor = V/R = 1.56/500 = 3.12 x 10-3 A

2.5 Short Circuits

VS

R2

Short circuits occur when the Resistive parts of a circuit are bypassed, effectively connecting the positive terminal of the power supply directly to the negative terminal providing a resistance free path for the current. The current immediately increases to its maximum.

This can be disastrous for the circuit causing rapid heating and possibly a fire.This situation is taken care of by the use of fuses, circuit breakers, “safety switches”, or residual current devices. (See chapter 5).

Chapter 3

Non Ohmic Devices

3.0 Non Ohmic ConductorsConductors which do not follow Ohm’s Law are called Non Ohmic ConductorsDevices such as diodes and transistors can be classed as non ohmics, but the best known non ohmic is the incandescent light globe. When a plot of Potential Difference against Current is drawn, it is not a straight line.

V

IAt Low Voltages, slope is shallow = Low Resistance

At High Voltages,Slope is steep= High Resistance

A Typical “Characteristic Curve” for an Incandescent Light Globe

Non Ohmics - Series29. Two non ohmic conductors with

“Characteristic Curves” as shown opposite are connected in series in a circuit as shown.

The voltage across device 1 is 6.0 V.(a) What is the current through Device 2,(b) What is the voltage across Device 2, (c) What is the voltage of the battery

powering the circuit ?

Device 2

Current (I)

Voltage (V)

Device 1

1 2

2

4

6

8

Voltage (V)

Current (I)

1 2

2

4

6

8

(a) Voltage across device 1 = 6.0 V so current through device 1 = 0.5 A (read from graph) because devices are in series current is same through both.

So I DEVICE 2 = 0.5 (a) Voltage across device 2 = 4.0 V (read up from 0.5 A on graph for device 2).(b) VSUPPLY = Sum of voltage drops around the circuit = 6.0 + 4.0 = 10.0 V

Device 1

Device 2

6.0 V

Non Ohmics Parallel30. Two devices with Characteristic Curves” shown are connected in parallel. The current through Device 1 is 1.0 A.(a) What is the voltage across Device 2 ?(b) What is the current through Device 2 ?(c) What is the voltage of the battery ?(d) the total current drawn from the battery ?

2

4

6

8

Voltage (V)

Current (I) Current (I)

Voltage (V)

2

4

6

8

1 2 1 20 0

Device 1 Device 2

(a) If current through device 1 = 1.5 A, then voltage across it = 8.0 V (read from graph). In parallel network voltage is same across each member

so VDEVICE 2 = 8.0 V (b) If voltage across device 2 = 8.0 V, current = 2.0 A (read from graph).(c) VSUPPLY = VDEVICE 1 = VDEVICE 2 = 8.0 V(d) Total Current = Sum of currents through each component = 1.5 + 2.0 = 3.5 A

Device2

Device 1

1.5 A

Chapter 4

Cells & Batteries

4.0 Cells and BatteriesElectrical Cells (as opposed to plant and animal cells) are devices which perform two functions:

1. Charge Separation.2. Charge Energisation.

Charge Separation is the process of separating positive and negative charges to produce a POTENTIAL DIFFERENCE capable of driving a current around an external circuit.

Charge Energisation is the process of providing the separated charges with the ELECTRICAL ENERGY they need to complete their journey around the circuit connected to the cell.

A Single Cell

A group of Cellsie. a Battery

Batteries have a limited ability to separate and energise charge, they eventually go “flat”. See Slide 4.5

4.1 Power SuppliesPower Supplies, (as opposed to cells and batteries) obtain their separated and energized charges from the mains supply to which they are connected, via the standard 3 pin plug.

They rely on the power generation company to separate and energize the charge carriers at the power station.The power station remains “on line” at all times, so the power supply can operate indefinitely, i.e., it does not go “flat” like a battery.

In all other senses, power supplies behave in a similar fashion to cells and batteries.

4.2 Electromotive Force (EMF)Electromotive Force (EMF) is not a true force in the Newton’s Laws sense, but it is a term used to describe the OPEN CIRCUIT VOLTAGE of a cell, battery or power supply.

“Open Circuit” means that no complete external circuit is connected to the battery or power supply and thus no current is being drawn.

R

R

With the circuit complete, a current is flowing and the P.D. across the power supply equals the P.D. across the resistor.

Consider the circuit shown

With the switch open the current stops flowing, the voltage across the resistor falls to zero and the voltage reading across the power supply rises. The Voltage reading now is the EMF of the supply

EMF is represented by the symbol (ε). The Greek letter epsilon.

Cells and Batteries

33. When a battery or power supply is switched into an external circuit the Potential Difference measured across the terminals of battery or power supply will:

A: Fall because a current is now flowingB: Rise because a current is now flowingC: Remain unchanged even through a current is now flowingD: None of these answers

31. The primary task of a battery or power supply is to:A: Supply electrons and energise themB: Provide energy for charge carriersC: Provide charge separation and energization.D: Separate electrons from protons.

32. The EMF of a battery or power supply is A: The Potential Difference of the supply when a current is being drawn.B: The Potential Difference of the supply when no current is being drawnC: The Potential Difference difference between the positive and negative terminals when they are short circuited.D: The Potential Difference difference between earth and the positive

terminal.

4.3 Internal ResistanceThe reason the Potential Difference of the power supply falls when a current is drawn from it is the “Internal Resistance” of the supply.The internal resistance is:•The “price which must be paid” for drawing a current from the supply. •A measurable quantity and, as with all resistance, is measured in Ohms (Ω).The larger the current drawn from the supply, the greater the “cost” (in terms of energy wasted inside the supply), because of the internal resistance.This means less energy is available for the charge carriers to flow around an external circuit.

A cell, battery or power supply can be represented as a pure EMF in series with a resistor, r, (representing the internal resistance).

rε

V1

Power Supply

With no external circuit connected (i.e. a so called “no load” situation), no current is drawn from the supply, and the voltmeter reading V1 will equal ε, the EMF of the supply.

4.4 Internal Resistance V2

R

rε

I

The power supply now has an external circuit connected.

This draws a current from the supply.

This current also flows through the internal resistance r.This causes a potential difference = Ir across that resistor.

The voltage measured by V2 will now be less (by an amount Ir) than the EMF (ε) of the power supply.

Mathematically: V2 = ε - IrBy replacing the fixed resistor (R) in the external circuit with a variable resistor,and changing the value of the resistance, a set of values for V2 and the corresponding current, I, can be obtained.Plotting these values gives the following.

Voltage (V2)

Current (I)

Intercept = ε

Slope = - r

This method allows you to calculate the internal resistance of the power supply, cell or battery.

Cells and Internal Resistance34. A battery or power supply can be regarded as

A: A pure P.D. source in parallel with a resistanceB: A pure P.D. aloneC: A pure P.D. source in series with a resistorD: A pure resistance in parallel with an EMF

35. A battery has an EMF of 9.0 V. When connected into a circuit drawing 25 mA the potential difference across the battery terminals in measured at 8.6 V. What is the internal resistance of the battery ?

V = ε – Irr = (ε – V)/I = (9.0 – 8.6)/(2.5 x 10-2) = 16 Ω

Internal Resistance36. A group of students set out to study the properties of a D cell battery. Using the following circuit and varying the resistance of the rheostat they collected the data shown.

I

V

2

R

I 2

Voltage (V) (volts)

Current (I)(milliamps)

0.10 120

0.25 100

0.45 70.0

0.60 50.0

0.70 35.0

0.85 15.0

Voltage (v)

Current (mA)10

20 30

40

50

60

70

80

90

100

110

120

0.1

0.20.30.4

0.50.60.70.80.9

1.0

(b) EMF = y intercept = 0.95 V

(c) Internal resistance (r) = negative of slope = -(0.85 – 0.25)/(15 x 10-3 – 100 x 10-3) = - (0.6)/(-75 x 10-3) = 8Ω

4.5 A Flat Battery

In a cell or battery, the chemical processes used to provide charge separation and energisation become less efficient as current is drawn from it.This shows up in an increase in the Internal Resistance of the battery.The internal resistance will continue to increase until the battery is no longer able to provide sufficient energy to perform its primary task (separation and energisation) and the battery is said to be flat.

In testing a battery with a multimeter, you measure the EMF, which may seem fine, because you are not drawing a current from it.To properly test a battery it needs to be placed in a resistive circuit of some kind so that a current is drawn. Measuring P.D. across the battery now will now give a more realistic picture of the battery’s condition.

Flat Battery37. Explain why is not sufficient to simply measure the EMF of a battery to check if it is “flat” ?

EMF does not measure the internal resistance of the battery and hence its ability (or otherwise) to provide a current to external circuit.

Chapter 5

Fuses and Stuff

5.0 FusesFuses are primarily Safety Devices placed in circuits to limit the current flow to a certain (predetermined) value.Limiting the current in this way reduces the chance of fire caused by overheating in a circuit carrying excessive current.A Fuse is basically a short piece of thin wire which, when too much current tries to flow through it, overheats and then melts, breaking the circuit.

In the electricity supply network fuses are present throughout the system and at the domestic or household end fuses are located in the “Fuse Box ” sometimes also called the “Meter Box”.

Modern Meter Boxes have resettable fuses called “Circuit Breakers” instead of the old style porcelain former with its separate thin wire fuse.

5.1 Residual Current InterruptIncreasingly, Meter Boxes contain Residual Current Interrupt Devices (RCI), commonly called “safety switches” and Surge Arrestors. Both are safety devices. The RCI is designed to protect people while surge arrestors protect electrical equipment.

The RCI operates using two coils to monitor the magnetic fields produced by the currents in both the active and neutral wires.Under normal conditions the Active and Neutral currents will be equal.This means the induced currents in the coils will also be equal and will cancel one another out inside the RCI.

If the two A and N currents are different, as shown with some passing down the earth wire, due to a short circuit in the toaster, the RCI reacts by opening a switch in the active wire, cutting off the current.

The RCI will respond in approx 0.03sec (less than a heartbeat)

THE RCI AND THE TOASTERG.P.O. or Power point

Earth

Neutral Wire

Active WireRCI

Coils

To understand the operation of the RCI you need to know that a current in a wire causes a magnetic field around that wire.The strength of the magnetic field depends upon the size of the current.

Safety38. Which one or more of the following act as safety devices in electric circuits ?

A: FusesB: Safety switchesC: Surge arrestorsD: Short Circuits

39. RCI’s monitor the currents inA: The Neutral and Earth LinesB: The Active and Earth LinesC: The Active Line onlyD: The Active and Neutral Lines

5.2 SwitchesSwitches break circuits by moving contacts apart.In the domestic situation switches are always placed in the Active Line.This is especially important for General Purpose Outlets (GPO’s) more often called wall sockets or power points.

There are large numbers and types of switches in use. They can be Mechanical, Electromechanical or Electronic.

Sample Mechanical Switches are shown:

SPDTSingle Pole, Double Throw

DPSTDouble Pole, Single Throw

DPDTDouble Pole, Double Throw

SPSTSingle Pole, Single Throw

Opening the switch (turning it “off”) isolates the power point from the supply.If the switch was placed in the Neutral line the power point would remain live even with the switch “off”

Switches

40. Identify each type of switch

Double Pole Double Throw

Single Pole Double Throw

Double Pole Single Throw

Single Pole Single Throw

5.3 EarthingThe Earth is a giant “sink” for electricity, it will soak up electric charge. The name given to the process of connecting a circuit to the Earth is called “earthing” and the physical connection is via the Earth Wire.

Active Line

Neutral Line

Meter Box

Main Fuse

Meter

Fuses

Power point

Neutral Bar

Earth Stake

Main Switch

To better understand earthing, an understanding of domestic wiring is needed. Below is a sample domestic wiring system showing one power point only. The Earth is also physically

connected to the neutral bar, holding it at Earth potential (voltage) of 0 volts.

The earth stake is a solid copper piece driven about 2 m into the ground

The Earth Wire provides a “resistance free” path to earth for any current that leaks from the active and/or neutral lines. Leaking current will choose to use the no (or low) resistance path to earth rather than the high resistance path through a human.

Earthing, as used in domestic wiring, is just another safety feature.

5.4 Electric ShockElectricity is dangerous! We all know this, it was drummed into us throughout our childhood.We can all remember the reaction of adults the first time they found us playing with electrical sockets at home.But exactly how dangerous is electricity and what does it do to our bodies ?

The lowest recorded voltage at which death occurred was 32 V ACDomestic electricity in Australia is supplied at 240 V AC, at 50 Hz. If you are exposed to this supply for 0.5 sec and depending on the size of the current, the following effects will be experienced.

Current (mA) Effect on Body 1 Able to be felt – slight tingling 3 Easily felt – distinct muscle contraction10 Instantly painful - Cramp type muscle reaction20 Instant muscle paralysis – can’t let go50 Severe shock – knocked from feet90 Breathing disturbed - burning noticeable

150 Breathing extremely affected200 Death likely500 Breathing stops – death inevitable

Remember that 1 mA = 1/1000th of an Amp ( 1 mA = 1 x 10-3 A)

Chapter 6

Electricity Consumption

6.0 Power ConsumptionIn general, POWER is defined as the time rate of doing WORK or the time rate of ENERGY conversion.Mathematically:

P = W/t = E/tWhereP = Power (Watts)W = Work (Joules)E = Energy (Joules)t = time (secs)

Rearranging the equation we wet:E = P.t

so 1 Joule = 1 Watt.sec

The Joule is a very small unit, too small for the Energy companies to use when it comes to sending out the bills to customers, so electricity is sold in units called kilowatt hours. (kWh). Have a look at your own electricity bill at home !

1 kW = 1000 W and 1 hour = 3600 sSo 1 kWh = 1000 x 3600 J

= 3.6 x 106 J = 3.6 MJ.

Electric Power

42. What was the power consumption of the home (in Watts) ?

41. An electricity bill indicates the household used 117.5 kWh of electricity in a week. How many megajoules were used ?

1 kWh = 3.6 MJ so 117.5 kWh = (117.5)(3.6) = 423 MJ

P = E/t, U = 423 MJ = 4.23 x 108 J t = 1 week = 7 x 24 x 60 x 60 sec = 604800 s

P = E/t = (4.23 x 108)/(6.048 x 105) = 699 W

6.1 The Kilowatt Hour

A 100 W (0.1 kW) incandescent light globe which runs for 1 hour consumes 0.1 x 1 = 0.1 kWh of electricity.

A 1500 W (1.5 kW) electric kettle which boils water in 5 minutes consumes 1.5 x 5/60 = 0.125 kWh of electricity.

A 2000 W (2 kW) oven operating for 3 hours consumes 2 x 3 = 6 kWh of electricity.

Domestic electricity costs between 12 cents and 20 cents a kWh

Cost to run @ 12c/kWh = 1.2 cents

Cost to run @ 12c/kWh = 1.5 cents

Cost to run at 12c/kWh = 72 cents

Running Costs43. If domestic electricity costs 13.5 c per kWh. Calculate the cost of running (a) a 100 W light globe run for 1 hr, (b) a 1500 W kettle run for 5 mins and (c) a 2 kW oven run for 3 hrs.

Light Globe = (0.1)(13.5) = 1.35 cents: Kettle = (0.125)(13.5) = 1.69 cents: Stove (6.0)(13.5) = 81 cents

6.2 Load CurvesThe demand on the electricity supply is not constant:•It varies from time to time during the day.•It varies from day to day during the week.•It varies from season to season during the year.This variation is best displayed on a “Load Curve”

12 Midnight 6 am 12 noon 6 pm 12 Midnight

50 %

75 %

100 %

Cold DayWinter

Mild DaySpring

Hot DaySummer

Blackouts or “Brownouts” are likely at these times

Load CurvesQuestions44. Why does the demand on a hot summer day exceed the demand on a cold winter’s day ?

45. “Blackouts”, loss of supply often occur when demand exceeds supply. At what times and on what type of day are blackouts likely to occur ?

Hot Summer days means use of air conditioners, the greatest consumers of electrical power of all household appliances. In addition commercial air conditioners must also work harder thus consuming more electricity.

Just after noon and just after 6 pm on hot summer days

Ollie Leitl 2008