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Farr High School

HIGHER PHYSICS

Unit 3 Electricity

Pupil Booklet Based on booklets by Richard Orr, Armadale Academy

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Section 1

Monitoring and measuring a.c

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AC∕DC In Standard Grade, you needed to know the difference between AC and DC supplies. In Higher we will look to determine the relationship between an alternating current and a direct current that has the same effect in an electric circuit. This means that we would have different measured values of current that would give the same brightness from a lamp, say. The timebase is the time it takes the line on the oscilloscope to move across one division on the screen. In the pattern shown the width of the screen is 8 divisions so the time to produce this pattern would be 8 x 25 = 200ms [200 x 10-3s]. In this particular pattern there are 3 complete waves. This means we have 3 waves in 0.2s. Back to Standard Grade and the frequency is calculated as

time(s)

waves of number =

0.2

3 =15 Hz.

The peak voltage is calculated by multiplying the y-gain by the amplitude of the wave in divisions. This pattern has an amplitude of 2 divisions. So peak voltage = 2 div x 2V/div = 4V

The r.m.s. value of an alternating voltage or current is defined as being equal to the value of the direct voltage or current which gives rise to the same heating effect (same power output). Consider the following two circuits which contain identical lamps. The variable resistors are altered until the lamps are of equal brightness. As a result the direct current has the same value as the effective alternating current (i.e. the lamps have the same power output). Both voltages are measured using an oscilloscope giving the voltage equation below. Also, since V=IR applies to the r.m.s. valves and to the peak values a similar equation for currents can be deduced.

Timebase = 25 ms/div

Y-gain = 2 V/div

IpeakI = 1

2

VpeakV = 1

2andr.m.s. r.m.s.

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So what is r.m.s. voltage? It’s essentially a way to find an average value for an alternating signal. The actual average of any regular alternating signal will be zero. There are as many negative values as there are positive.

So finding a simple mean isn’t the way forward. However, we can get rid of the negatives by squaring all the values, all squares are positive. The pattern changes and now it is possible to work out an average for the new pattern. It turns

out that the average is

2

A2. To get back to the original we now

need to take the square root of this value which will be

2

A .

r.m.s. value =

2

peak

What we’ve done is squared our original pattern, taken the mean then taken the square root of this. In other words we have the square root of the mean of the squared value. The root mean square or r.m.s. value. Questions 1. The oscilloscope below shows the potential difference over a bulb attached to an AC power supply. The y-gain is set at 5V / div. The time base is set at 10ms / div. (a) State the peak p.d. of the trace. (b) Calculate the frequency of the supply. (c) Calculate the rms value of the p.d. 2. A peak current of 7.4 A flows through a heater of resistance 44Ω. Calculate

(a) the rms power (b) the peak voltage applied.

Text Book Page 115 Questions 1, 2 and 3 Tutorial Booklet Pages 6 and 7 Exam Questions Booklet Pages 1 to 4 Past Papers 2006 Q 26(a) 2007 Q 8, 10 2008 Q 10 2009 Q 25(a)

2010 Q 26(a) 2011 Q 11 2012 Q 19

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Section 2

Current, potential difference, power and

resistance

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Circuits Revision 1. Electric Current When S is closed, the free electrons in the conductor experience an electric field which causes them to move in one direction. The electron current will flow from the negative terminal to the positive terminal of the battery. The energy required to drive the electron current around the circuit is provided by a chemical reaction in the battery or by the mains power supply. The electrical energy which is supplied by the source is converted to other forms of energy in the components which make up the circuit. Current is a flow of electrons and is given by the flow of charge (coulombs) per second:I = Q/t The energy required to drive the electron current around this circuit is provided by a chemical reaction in the battery. The electrical energy that is supplied by the source is transformed into other forms of energy in the components that make up the circuit. 2. Voltage is defined as the energy transferred per unit charge. V = W/Q W represents the work done, i.e. energy transferred. The unit of voltage 1 volt = 1 joule per coulomb (J C–1). 3. Resistance The lamp in the circuit above has resistance R. In any circuit providing the resistance of a component remains constant, if the potential difference V across the component increases the current I through the component will increase in direct proportion. This is Ohm’s Law which is summarised by: V = IR A component which has a constant resistance when the current through it is increased is said to be ohmic. Some components do not have a constant resistance, their resistance changes as the p.d. across the component is altered, e.g. a transistor. 4. Resistors in series and parallel You should remember the equations for resistors in series and in parallel and be able to carry out calculations using these: Resistors in series: RT = R1 + R2 + R3 Resistors in parallel: 1/RT= 1/R1 + 1/R2 + 1/R3.

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5. Electric Power For an electrical component, power P is given by P = I V where I is current through the component and V is the potential difference across the component. The power of a component is the energy transformed each second. P=E/t A power of 1 watt (W) means that 1 joule (J) of electrical energy is converted into other forms of energy every second, e.g. electrical energy light/heat energy. The equivalence between IV, I2 R and V2 / R: Since P=IV and V=IR, two more equations for power can be generated by replacing V with IR and replacing I with V/R.

1. P = IV = I x (IR) = I2R 2. P = IV = (V/R) x V = V2/R

Depending on the quantities given in a question, the value of P,V, I and R can then be found directly.

P = IV = I2R = V2/R = E/t 6. Rules for current and voltage in series and parallel circuits Series circuits Current: In a series circuit the current is the same at all positions. Voltage: In a series circuit the potential differences across each component add up to equal the supply voltage. Parallel Circuits Current: In a parallel circuit the sum of the current the branches equals the supply current. Voltage: In a parallel circuit the voltage across each branch is the same for each component. It is important that you have a good grasp of these rules. They are summarised below. Summary of circuit rules for current/voltage

Series Parallel

Current same at all points sum to supply current

Voltage sum to supply voltage same across each branch

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Next reminder, potential dividers. Voltage dividers were used to set an input voltage to a control circuit; do you remember the transistor circuits? Never mind, at some point you won’t wake up screaming!! Potential dividers are simple series circuits with two or more resistors in. The bigger resistor will have the bigger p.d. across it. Calculating the size of the p.d. across R2 in a voltage divider depends on your own preference.

VIR twice method:

Calculate the current flowing in the circuit using I =

T

SR

V Then calculate the voltage across your

desired resistor [R2] using V = IR2. Ratios method: Ratio of resistance is R1:R2, voltage splits same ratio

VR2 = R2 x

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SRR

V

, which is essentially the same as the first method only in one equation.

Questions 1. Calculate

a) The current flowing b) The power transferred c) The total charge flowing in 5 mins

2. Calculate

a) the current flowing b) The potential difference ( voltage ) across each resistor

3. Calculate

a) The current through each resistor b) The effective resistance

Tutorial Booklet Pages 2 – 5 Exam Question Booklet Pages 5 to 7 Past Papers 2006 Q 8, 25(a) 2007 Q 9, 11 2008 Q 9, 11 2009 Q 9 2010 Q 9 2011 Q 9 2012 Q 18

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Section 3

Electrical Sources and Internal Resistance

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Internal resistance Up until now we have made the assumption that sources of electrical energy have no resistance component. When we have been calculating the total resistance of a circuit we have combined all the relevant resistances in the external circuit. If we think about it carefully it makes sense that a cell/battery/power supply should have some resistance, they are after all made of ‘stuff’ and all ‘stuff’ has resistance.

As always we can make the situation a bit easier for ourselves if we make an assumption that the supply has two parts, a

source of electrical energy [E] and a separate internal

resistance[r].

We can think of any circuit as a source, an internal resistance and an external resistance[R]. This external resistance could be made up of a number of separate resistances. This then becomes a straightforward series circuit with a supply and two resistors. The e.m.f. is the energy supplied to each coulomb of charge which passes through the supply. This is what you have previously called the voltage of the supply. If a voltmeter is connected across a supply then it will read the e.m.f. only if no current flows in the circuit. This is known as the open circuit voltage. When a current, I, is flowing there will be a p.d. across the internal resistance [Ir]. Because this p.d. is generated inside the supply and is not available to the external circuit it is known as the ‘lost volts’. This means that the voltage generated in the external circuit is less than the e.m.f. The external voltage is known as the terminal potential difference [t.p.d.] which strangely enough is the p.d. measured across the terminals of the source!! This is the same as the p.d. across the external resistance [IR]. Applying the simple circuit rule for voltages in a series circuit we obtain:

E = IR + Ir

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A circuit is set up as shown below. The external resistance is altered to give a series of different current readings. For each current reading the reading on the voltmeter [t.p.d.] also is recorded. The results are then plotted onto a graph as shown below.

The graph shows that as the current flowing in the circuit increases the t.p.d. decreases. This is because the lost volts [current x internal resistance] will increase as the current increases, leaving less voltage available for the external circuit. The e.m.f. can be established from the graph. The e.m.f. is the measured voltage when no current flows. This corresponds to the intercept of the line with the p.d. axis. In this case the e.m.f. is 5.0 volts. The internal resistance can now be calculated from the graph by choosing any current value, in this case 1.2A, and determining what the lost volts for that current is.

Internal resistance =

current

volts lost =

1.2

1.5-5 =

1.2

3.5 = 2.9Ω

If the supply is short circuited, then the only resistance in the circuit will be the internal resistance. The short circuit current can be determined by

ISC =

r

Since internal resistances tend to be small, this generally gives rise to a large current. This large current can deplete a battery fairly quickly.

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Power matching [application of internal resistance] Note that: Electrical impedance is the measure of the opposition that a circuit presents to the passage of a current when a voltage is applied, specifically when the current is alternating. 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, but fortunately you do not need to be able to reproduce this explanation.

Power = I2R but I =

r) (R

so P =

2r) (R

2

x R = 22 r Rr2 R

R2

Differentiating with respect to R gives

dR

dP = 2

3r)(R

2R - r)(R

Now, you know when

dR

dP = 0 you can find the maximum. This turns out to be when R + r = 2R

so R = r for maximum power. This is used when matching speakers to amplifiers in an audio system. The output impedance of the amplifier (internal resistance of the source) should be equal to the impedance of the speakers (external load). Activity Consider a cell of emf 12·0 V and internal resistance 10·0 Ω connected to load resistor R. Let R = 2·00 Ω

Total resistance RT = 10·0 + 2·00 = 12·0 Ω

Current: A 00.112

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TR

EI

Power in load: W 00.200.200.1 22 RIPL

Total power produced: W 00.1200.1200.1 22 TT

RIP

The table shows how the value of the load resistor affects the current in the circuit, the power in the load resistor and the total power in the circuit.

Load resistor (Ω) 0.00 2.00 4.00 6.00 8.00 10.00 12.00 14.00 16.00 18.00 20.00

Current (A) 1.20 1.00 0.86 0.75 0.67 0.60 0.55 0.50 0.46 0.43 0.40

Power in load resistor (W) 0.00 2.00 2.94 3.38 3.56 3.60 3.57 3.50 3.41 3.31 3.20

Total Power (W) 14.40 12.00 10.29 9.00 8.00 7.20 6.55 6.00 5.54 5.14 4.80

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The graphs show how total power varies with the value of the load resistance and how the power delivered to the load varies with the value of the load resistance.

0.00

2.00

4.00

6.00

8.00

10.00

12.00

14.00

16.00

0.00 5.00 10.00 15.00 20.00 25.00

To

tal

po

wer

PT

(W

)

Load resistance (ohms)

Load resistance vs Total power

0.00

0.50

1.00

1.50

2.00

2.50

3.00

3.50

4.00

0.00 5.00 10.00 15.00 20.00 25.00

Lo

ad

Po

we

r P

L (

W)

Load resistance (ohms)

Load resistance vs Power delivered to the Load

This shows us that:

1. As the load resistance increases the total power of the cell decreases. 2. The power delivered to the load resistor peaks when the load resistance is equal to the

internal resistance.

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Example In the circuit below, the reading on the voltmeter is 5 V when switch S is open and 3 V when it is closed. a) What is the EMF of the cell? 5V b) Calculate the current flowing in the circuit when the switch is closed. V = IR so I = V/R = 3/15 = 0.2A

c) What is the internal resistance of the cell? E = IR + Ir so r = (E – IR)/I = (5 – 3)/0.2 = 10Ω

Questions

1. A car battery has an e.m.f. of 12.0V. When it is used to operate the starter motor it supplies

a current of 50.0A and its t.p.d is 9.0V

a) Find the internal resistance of the battery?

b) What is the resistance of the starter motor?

2. The graph shown to the right was drawn after measuring

the t.p.d. of a battery when different currents were

taken from it.

a) What is the e.m.f. of the battery?

b) Find the battery’s internal resistance.

c) What is the maximum current that could be taken

from the battery?

Text Book Page 121 Questions 1 to 6 Tutorial Booklet Pages 8 to 13 Exam Questions Booklet Pages 8 to 17 Past Papers 2006 Q 9 2007 Q 25 2008 Q 24 2009 Q 24 2011 Q 24 2012 Q 17, 31

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Section 5

Capacitors

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Capacitors A and B are two metal plates. If electrons are moved from A to B then A will become positively charged and B negatively charged by the same amount. The transference of charge will produce a potential difference V between the plates and also an electric field E. To transfer more electrons from A to B, work must be done against the electric field (plate B, being negative, will repel electrons). This work or energy becomes the electrical potential energy stored in the plates. The charge on the plates will stay there unless there is an external circuit along which electrons can return from B to A. We say that the plates have a CAPACITY for storing charge. The pair of plates is called a CAPACITOR. A capacitor is a device which consists of two conducting plates seperated by an insulator. Charge can be moved onto the plates, building up an electric field and storing energy in the field. There is a direct relationship between the charge stored on the capacitor plates and the potential difference across the plates. This relationship can be verified by using the circuit opposite. The capacitor is charged to a known potential by setting the switch to position A. The switch is the moved to position B where the capacitor discharges through the coulombmeter which records the charge stored on the capacitor. This process is repeated for a range of potential values. A graph is then plotted of the potential across the capacitor against the charge stored on the capacitor. The resulting graph is shown opposite. The ratio of charge to potential difference is known as the capacitance. This value is the same as

the gradient of the line. C =

V

Q [eqn. 1]

In order to move charge from one plate to the other work must be done by the supply. The work done can be calculated by working out the area under the charge voltage graph. [This is the same principle as velocity time graph area is the displacement and force time graph area being impulse.] This is a triangle of base V and altitude Q, giving an expression for the energy stored as

E = ½QV [eqn. 2]

A bit of mathematical jiggery pokery allows two other expressions to be derived.

Eqn. 1 can be written as Q = CV [eqn. 3] or V =

C

Q [eqn. 4]

Sub [3] into [2] E = ½(CV)V = ½CV2

Sub [4] into [2] E = ½Q

C

Q = ½

C

Q2

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Charging and Discharging Capacitors When a capacitor is charging up electrons are removed from one plate, moved around the circuit and deposited on the opposite plate. As was stated above, this process requires energy. When the switch is closed the capacitor begins to charge. As the capacitor charges the negative charge on the right hand plate will increase. This makes it more difficult to add further electrons as they have a greater charge to overcome. The rate at which electrons are added to the plate reduces until the supply is unable to supply enough energy to overcome the repulsive force between the electrons and the negative plate. When this point is reached we say that the capacitor is fully charged. The voltage across the capacitor will be VS at this point. Charge/Discharge Graphs The graphs of charging voltage against time and charging current against time describe the description given on the previous page.

The voltage across the capacitor increases rapidly at the start because the repulsive force of the charges on the plates is smaller. As charge builds up on the plates the rate decreases. The current flowing in the circuit depends on the voltage across the resistor. This is largest at the instant the switch is closed.

Imax =

T

SR

V

The dashed line represents the graph obtained if the resistance is increased. Bigger resistance means the current flowing will be reduced, so increasing the time required to supply charges to the plates of the capacitor. It could also indicate a larger capacitance.

The discharge graphs follow a similar pattern to the charging graphs. The voltage across the capacitor will decrease as it discharges, the greatest rate of change being at the start when the voltage, and consequently the current flowing in the circuit, is greatest.

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Questions Text Book Page 151 Questions 1 to 5 Page 159 Questions 1 to 4 Page 161 Questions 1 to 4 Tutorial Booklet Pages 14 to 19 Exam Questions Booklet Pages 18 to 28 Past Papers 2006 Q 11, 12, 25(b) 2007 Q 26 2008 Q 25 2009 Q 11, 12, 26 2010 Q 11, 12, 24(b), 26(c)(d) 2011 Q 13, 14, 25 2012 Q 32

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Section 5

Semiconductors

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Conductors, semiconductors and insulators We already know that conductors allow a current to flow through them and insulators do not allow a current to flow through them. So what about semiconductors? These are materials that can conduct a bit, with their ability to conduct increasing with temperature. Why does this happen? Let’s think about atoms. A nucleus [positive] is surrounded by electrons [negative]. Each electron can have an energy determined by the energy levels of the particular atom. When atoms combine to form a solid the electrons align themselves into energy bands. The two highest energy bands are known as the valence band and the conduction band. The conduction band is the higher of the energy bands. The electrons are held tightly in the valence band, but are able to move freely in the conduction band. So if there are electrons in the conduction band the solid will conduct. If there are no electrons in the conduction band, the solid will not conduct. We can represent the three types of solid as diagrams showing these two bands:

In a conductor the bands overlap so the conduction band contains electrons free to move. These electrons can move to produce the current when an e.m.f. is applied to the solid.

In an insulator the valence band is full but no electrons are in the conduction band. There is a large energy gap between the bands [band gap]. It is so large that electrons almost never cross the gap and the solid never conducts. If we supply enough energy the solid will conduct but often the large amount of energy ends up destroying the solid.

The semiconductors are like the insulators in that the valence band is full. However the gap between the two bands is small and at room temperature some electrons have enough energy to jump the gap. The solid will conduct to an extent. Where the temperature increases, the number of electrons in the conduction band increases and so the semiconductor conducts better.

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Semiconductors Semiconductors can be made by adding impurity atoms to insulators. This changes the number of electrons around some of the atoms, reducing the band gap from the large gap in insulators to the small gap in semiconductors. The elements that are typically used for semiconductors are silicon and germanium. In their pure form, germanium and silicon are poor conductors of electricity. Their atoms have four outer shell electrons. This means that they can form four chemical bonds with other identical atoms. In a crystal of pure germanium each germanium atom is surrounded by four other germanium atoms. In this state the germanium will not conduct. Pure silicon can be doped by tiny quantities of arsenic which has five outer shell electrons. The impurity atoms will fit into the crystal structure but there will be one electron left over. The extra electron causes the material to conduct and the silicon is an n-type semiconductor. If indium (with three outer shell electrons) is added to germanium or silicon, the atoms will fit into the crystal structure but there will be one electron missing. These areas in the crystal where there is an electron missing are called (positive) holes. This impurity causes the material to conduct and is called a p-type semiconductor.

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Section 6

p-n junctions

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p-n junctions When semiconductors are joined together to form a p-type:n-type sandwich then some very useful things start to happen. A layer forms between the two types of semiconductor, known as the depletion layer. It is in this layer that the interesting stuff happens. Depending on the materials used to make the p-n junction different effects can be observed. Solar cell: The materials are chosen such that when light is shone on the cell a voltage is produced across the junction. As you will find out later in the course light is made up photons, it is these photons which supply the energy which ultimately ends up with the production of a voltage. LED: This device is really the opposite of the solar cell. This time if a voltage is applied across the junction, photons are produced and emitted at the junction. The colour of light emitted from the LED depends on the value of the band gap in the semiconductor materials. Photodiodes: Diodes can also be made so that the junction will absorb photons of light. A photon of light will cause an electron hole pair to be created in the junction allowing the diode to conduct. This is a photodiode. If a photodiode is connected in a circuit in reverse bias it will only conduct when light shines on the junction. This is known as the photoconductive mode. If a photodiode is connected as a power source this is known as the photovoltaic mode. Reverse Bias If the diode is connected as shown below, the electrons move towards the positive terminal and the holes move towards the negative terminal. This causes the depletion layer to get bigger and so the diode does not conduct. This diode is said to be reverse biased.

Forward bias If the diode is connected as shown below, the electrons move towards the positive terminal and the holes move towards the negative terminal. In this case electrons and holes both head for the depletion layer and fill it with charge carriers. This causes the depletion layer to disappear and the diode conducts. This diode is said to be forward biased.