Module: Batteries and Bulbs Time allocation: 10 hours

Module: Batteries and Bulbs Time allocation: 10 hours

1

IJSO Training CoursePhase II

Objectives:

Introduce a model of electrical conduction in a metal, and the concepts of resistance and internal resistance.

Define and apply the concepts of current, and the use of ammeters and voltmeters.

Draw circuit diagrams with accepted circuit symbols.

2

1. Electrical Conduction in Metals

A solid piece of metal, at room temperature, consists of metal ions arranged in a regular pattern called a crystal lattice and free electrons moving in the spaces between the ions.

The motion of the free electrons is random. We say they have random thermal motion with an average speed which increases with temperature.

3

The figure below represents a piece of metal which does not have current flowing through it. The arrows represent the random thermal motion of the electrons (their average speed at room temperature is hundreds of kms-1).

4

If a current is flowing in the piece of metal, then another motion is added to the random thermal motion (see the figure below). This motion is more regular and results in a general drift of electrons through the metal. A typical drift velocity for electrons in metals is less than 1mm/s . The magnitude of the drift velocity depends on the current, the type of metal and the dimensions of the piece of metal.

5

The resistance of a piece of metal is due to collisions between the free electrons and the metal ions.

6

During a collision, some of the kinetic energy possessed by the electron can be transferred to the ion thus increasing the amplitude of the lattice vibrations. Therefore, resistance to the flow of current causes the temperature to increase or in other words, resistance causes electrical energy to be converted into thermal energy (internal energy).

7

At higher temperature, the amplitude of the lattice vibrations increases, the collisions between the free electrons and the metal ions are more often. This suggests that the resistance of a piece of metal should increase with temperature.

8

Note: Not all materials behave in this way: the resistance of semi-conductors (e.g. silicon and germanium) decreases with temperature.

9

Conductors / Insulators

Electrical conductors readily conduct electric charges, small resistance .

Electrical insulators conduct electric charges poorly, large resistance .

Examples:

10

Good conductors

Poor conductors

Good insulators

Metals, carbon moist air, water, human body

Rubber, dry air

2. Electric Circuits

When drawing diagrams to represent electric circuits, the following symbols are used.

11

Wires crossing but not connected:

Wires crossing and connected:

Unless otherwise stated, we assume that connecting wires are made of a perfect conductor, i.e., no resistance.

12

Switch Battery A.C. supply

13

Resistor Variable resistor Push button

Filament lamp Voltmeter Ammeter

Transformer Rheostat (variable resistor)

Exercise

A:_________, B: _______, C: ________

D: _________, E: _________

push button, A.C. supply, rheostat, voltmeter, bulb

14

3. Electric Current Generally speaking, an electric current is a flow of charged particles. For examples, a current in a metal is due to the movement of electrons. In a conducting solution, the current is due to the movement of ions.

Current is measured using an ammeter. An ammeter measures the rate of flow of charge. For simplicity, an ammeter gives a reading which is proportional to the number of electrons which pass through it per second.

15

The unit of current is the Ampere, A.

16

An ammeter is always connected in series with other components. The resistance of an ammeter must be low compared with other components in the circuit being investigated.

Current in Series Circuits

A current of 2A corresponds to a certain number of electrons flowing in the circuit per second. So if I1 = 2A, I2 and I3 must also be 2A because in a series circuit, the electrons have only one path to follow.

Conclusion: The current is the same at all points in a series circuit.

17

If the three current I1, I2 and I are measured it is found that

18

I1 + I2 = I

This result is called Kirchhoff’s current law, stated as follows.

The total current flowing towards a junction in a circuit is equal to the total current flowing away from that junction.

As an analogy, consider vehicles at a road junction.

The number of vehicles passing point 1, per minute, must be equal to the number of vehicles passing point 2 per minute plus the number of vehicles passing point 3 per minute.

19

Relation between Current, Charge and Time

Another analogy is often found to be helpful. Consider a pipe through which water is flowing. If the rate of flow of water through the pipe is, for example, 25l min-1, then in 15 minutes, the total quantity of water which has moved through the pipe is 25 x 15 = 375l . The quantity of liquid is equal to the rate of flow multiplied by the time.

20

Similarly, when considering a flow of electric charges, the quantity of charge which passes is given by

the unit of charge is the Coulomb. The Coulomb can be now defined as follows:

21

quantity of charge = rate of flow of charge × time

Q = I × t

1 C is the quantity of charge which passes any point in a circuit in which a current of 1A flows for 1sec.

It should be noted that the Coulomb is a rather large quantity of charge. 1 C is the quantity of charge carried by (approximately) 6×1018 electrons!

Hence, each electron carries

–1.6022 × 10–19 C. This is the basic unit of electric charge.

22

Exercises

1. A current of 0.8A flows through a lamp. Calculate the quantity of electric charge passing through the lamp filament in 15 seconds. 12 C

2. A current of 2.5A passes through a conductor for 3 minutes. Calculate the quantity of charge passes through the conductor. 450 C

23

4. Voltage When a body is falling through a gravitational field, it is losing gravitational potential energy. Similarly, when a charge is "falling" through an electric field, it is losing electric potential energy.

Water has more gravitational potential energy at B than at A so it falls. The potential energy lost by 1 kg of water in falling from level B to level A is the gravitational potential difference (J/kg) between A and B.

24

25

The flow of water can be

maintained using a pump. A flow of electrons can be maintained using a battery. The battery maintains an electrical potential difference

between points A and B.

To measure voltage we use a voltmeter. The unit of voltage is the volt.

A voltmeter gives us a reading which indicates the amount of energy lost by each Coulomb of charge moving between the two points to which the voltmeter is connected. 1V means 1 JC-1.

26

A voltmeter is always connected in parallel with other components. The resistance of a voltmeter must be high compared with other components in the circuit being investigated.

What is an ideal voltmeter?

An ideal voltmeter can measure the potential difference across two points in a circuit without drawing any current.

27

Voltages in Series Circuits

Consider the simple series circuit above.

Energy lost by each Coulomb of charge moving from A to B is V1.

Energy lost by each Coulomb of charge moving from B to C is V2.

28

Energy lost by each Coulomb of charge moving from C to D is V3.

Obviously the total amount of energy lost by each Coulomb of charge moving from A to D must be V1 + V2 + V3 (= V).

Conclusion: The total voltage across components connected in series is the sum of the voltages across each component.

29

Voltage across Components in Parallel

30

All points inside the dotted ellipse on the right must be at the same potential as they are connected by conductors assumed to have negligible resistance.

Similarly for all points inside the dotted ellipse on the left.

So the three voltmeters are measuring the same voltage.

Conclusion: Components connected in parallel with each other all have the same voltage.

Again, this does not depend on what the components are.

31

5. Resistance

The resistance of a conductor is a measure of the opposition it offers to the flow of electric current. It causes electrical energy to be converted into heat.

The resistance of a conducting wire is given by

32

R = ρl /A

The unit of resistant is Ohms (W). It depends on the length of the piece of metal l and the cross-sectional area of the piece of metal A, and r is called the resistively, units Wm, which depends on type of metal.

In a circuit, the resistance is defined by

Where V is the voltage across the resistor and I is the current flows through it.

33

Resistant = voltage / current R=V/I

Ohm’s lawThe Ohm’s law states:

As voltage divided by current is resistance, this law tells us that the resistance of a piece of metal (at constant temperature) is constant.

Note: the resistance of a piece of metal increases as its temperature increases.

34

For a metal conductor at constant temperature, the current flowing through it is directly proportional to the voltage across it.

Exercises

1. If the resistance of a wire, of length l and uniform across sectional area A, is 10W. What is the resistance of another wire made of the same material but with dimensions of twice the length and triple cross sectional area?

unchanged

35

2. A uniform wire of resistance 4 W is stretched to twice its original length. If its volume remains unchanged after stretching, what is the resistance of the wire?

16 W

3. A current of 0.8A flows through a lamp. If the resistance of the lamp filament is 1.4W, calculate the potential difference across the lamp. 1.12 V

36

Effective Resistances

If two or more resistors are connected to a battery, the current which will flow through the battery depends on the effective resistance (or equivalent resistant), RE, of all the resistors. We can consider RE to be the single resistor which would take the same amount of current from the same battery.

37

Resistors in Series

The effective resistance of circuit A is

38

A B

RE = R1 + R2 + R3

Resistors in Parallel

The effective resistance of circuit A is

39

A B

1/RE = 1/R1 + 1/R2 + 1/R3

Exercise

1. A hair dryer consisting of two identical heating elements of resistance 70W each is connected across the 200V mains supply. The two elements can be connected in series or in parallel, depending on its setting. Calculate the current drawn from the mains in each setting.

1.4A; 5.7A

40

6. Potential Dividers

In the circuit, let v1 be the voltage across R1 and v2 the voltage across R2.

It can be shown that

Circuits of this type are often called potential dividers.

41

V1/V2 = R1/R2

Exercises

1. Two resistors are connected in series, show that and .

2. Two resistors are connected in parallel, show that and .

42

Variable ResistorsA variable potential divider can be made using all three connections of a variable resistor (also be called a rheostat) .

(i) Rotating variable resistor (internal view)

43

(ii) Linear variable resistor

44

Using Variable Resistors

In the circuit below, notice that only two of the connections to the variable resistor are used.

The maximum resistance of the variable resistor is 100W.

45

When the sliding contact, S, is moved to A the voltmeter will read 6V (it is connected directly to both sides of the supply). This is, of course the maximum reading of voltage in this circuit.

What is the reading of the voltmeter when the sliding contact is moved to B?

46

We have, in effect, the following situation.

Therefore, the voltmeter will read 3V.

47

Variable Resistor used as a Variable Potential Divider

What is the reading of the voltmeter when the sliding contact is moved to B?

The voltmeter reading can be reduced to zero by moving the sliding contact to B. The wire "x" (assumed to have zero resistance) is in parallel with the 100W resistor. This circuit is useful in experiments in which we need a variable voltage supply.

48

Exercise 1

In the circuits above, a variable resistor of resistance 100 W is connected to a 50 W resistor by means of a sliding which can be moved along the variable resistor.

(a)    Determine the maximum and minimum currents delivered by the battery, which has an e.m.f. of 10 V and negligible internal resistance, in the two circuits.

(b)   Determine also the currents delivered by the battery when the sliding contact is at the mid-point of the wire in both cases.

(a) 0.2A, 0.067A; 0.3A, 0.1A (b) 0.1A; 0.13A

49

7. Electrical Power and Energy

Any component which possesses resistance will convert electrical energy into thermal energy.

Consider the simple circuit shown below.

50

The current, I, is a measure of the number of Coulombs of charge which pass through the resistor per second.

The voltage, V, is a measure of the number of Joules of energy lost by each Coulomb of charge passing through the resistor.

So, the energy per second (power) supply by the battery is

51

P = VI

To calculate the power consumed by a resistor:

In series

In parallel

52

P = I2 R

P = V2 / R

Thus electrical energy can be expressed in the engineering unit kilowatt-hour (i.e., energy dissipated in an appliance of 1 kW rating operated for 1 hour)

Electricity is supplied to our house through the mains. The voltage supplied is alternating and correspondingly an alternating flow of charges occurs in the wire. This kind of electricity is called alternating current (a.c.) in contrast to the direct current (d.c.) as supplied by a battery.

53

Live wire: brown in colour. In Hong Kong, the voltage at the live wire changes from +220 V to -220V continuously and alternately, so that the current flows backwards and forwards round the circuit. A switch and a fuse can be installed in live wire to prevent the appliance to go 'live'.

54

Exercises

1. An electric cooker with a fuse and a switch in series with the heating element is to be connected to the pins of a socket. The correct connections should be

55

2. Three lamps A, B and C of resistance 250 W, 350 W and 600 W respectively are connected across a 200V supply as shown.

(a) Calculate the potential difference across the lamps.

(b) Calculate the current passing through the lamps.

(c) Calculate the power dissipated in the lamps.

(d)List the lamps in ascending order of brightness.

a. 200V, 117V, 83V

b. 0.33A, 0.33A, 0.33A

c. 27.8W, 38.9W, 66.7W

d. A, B, C

56

8. Battery and its Internal Resistance

The metal contacts which are used to connect a battery into a circuit are called its terminals. For this reason, when the voltage of a battery is measured, we often describe the result as the terminal potential difference of the battery.

57

A battery converts chemical energy into electrical energy.

The electrical energy given to each Coulomb of charge is called the e.m.f.1, denoted as , of the battery. So the unit of e.m.f is also Volt.

The term "e.m.f." originally came from the phrase "electro-motive force". This is now considered an inappropriate term as emf is a quantity of energy not a force. However, the abbreviation is still used.

58

In the following circuits, the voltmeter is assumed to have infinite resistance (a modern digital voltmeter has a resistance of around 107 W). The voltmeter reading is equal to the e.m.f. of the battery.

59

However, the substances of which the battery is made have some resistance to the flow of electric current. This is called the internal resistance of the battery. A more complete symbol to represent a battery is shown below.

60

The resistor, r, represents the internal resistance of the battery. The reading of the voltmeter across A and B will be

The terminal potential difference is only equal to the e.m.f. of the battery if the current flowing through the battery or the internal resistant is zero.

61

V = - Ir

Suppose there is an external resistance R in the circuit, it can be considered as in series to the internal resistance, so we have

62

= I (R + r)

Exercises:

1. A battery of e.m.f. 3 V and internal resistance 1.5 W is connected to another battery of e.m.f. 3 V and internal resistance 6 W, same polarities being wired together as shown in the figure. A student says the rate at which electrical energy is converted into internal energy is zero. Do you agree? Explain briefly.

63

2. A cell of e.m.f. 2 V is connected in series with a variable resistor of resistance R and an ammeter of resistance 0.4 W. By varying R, a series of ammeter readings, I, are taken. A graph of R against 1/I is then plotted. The value of the y-intercept is found to be -3 W. What is the internal resistance of the cell?

2.5 W

64

3. A student is given two identical batteries, each of e.m.f. 2V and negligible internal resistance and two identical resistors, each of resistance 4.5 W. Determine the current through each resistor in the circuits shown in the figure

(a) 0.444A, (b) 1.78A, (c) 0.222A, (d) 0.889A and 0.444A

65

Note: Combination of batteries 

Batteries in series

- effective e.m.f = E1 + E2+ E3

- effective internal resistance = r1 + r2 + r3

Identical batteries in parallel

- effective e.m.f. = E

- effective internal resistance = r/N, where N is the number of batteries in parallel.

- It can supply a current N times larger than that can be supplied by one battery alone.

— End —

66