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Checklist
ELECTRICITY AND ELECTRONICS
The knowledge and understanding for this unit is given below.
Electric fields and resistors in circuits
1.
State that, in an electric field, an electric charge experiences a force.
2.
State that an electric field applied to a conductor causes the free electric charges in it
to move.
3.
State that work W is done when a charge Q is moved in an electric field.
4.
State that the potential difference (V) between two points is a measure of the work
done in moving one coulomb of charge between the two points.
5.
State that if one joule of work is done moving one coulomb of charge between two
points, the potential difference between the points is one volt.
6.
State the relationship V = W/Q.
7.
Carry out calculations involving the above relationship.
8.
State that the e.m.f. of a source is the electrical potential energy supplied to each
coulomb of charge which passes through the source.
9.
State that an electrical source is equivalent to a source of e.m.f. with a resistor in
series, the internal resistance.
10.
Describe the principles of a method for measuring the e.m.f. and internal resistance of
a source
11.
Explain why the e.m.f. of a source is equal to the open circuit p.d. across the terminals
of a source
12.
Explain how the conservation of energy leads to the sum of the e.m.f.’s round a closed
circuit being equal to the sum of the p.d.’s round the circuit.
13.
Derive the expression for the total resistance of any number of resistors in series, by
consideration of the conservation of energy.
14.
Derive the expression for the total resistance of any number of resistors in parallel, by
consideration of the conservation of charge.
15.
State the relationship among the resistors in a balanced Wheatstone bridge.
16.
Carry out calculations involving the resistance in a balanced Wheatstone bridge.
17.
State that for an initially balanced Wheatstone bridge, as the value of one resistor is
changed by a small amount, the out of balance p.d. is proportional to the change in
resistance.
18.
Use the following terms correctly in context: terminal p.d., load resistor, bridge
circuit, lost volts, short circuit current.
Alternating Current and Voltage
1.
Describe how to measure frequency using an oscilloscope.
2.
State the relationship between peak and r.m.s. values for a sinusoidally varying
voltage and current.
3.
Carry out calculations involving peak and r.m.s. values of voltage and current.
4.
State the relationship between current and frequency in a resistive circuit.
Physics: Electricity and Electronics (H)
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Checklist
Capacitance
1.
State that the charge Q on two parallel conducting plates is directly proportional to the
p.d. V between the plates.
2.
Describe the principles of a method to show that the p.d. across a capacitor is directly
proportional to the charge on the plates.
3.
State that capacitance is the ratio of charge to p.d.
4.
State that the unit of capacitance is the farad and that one farad is one coulomb per
volt.
5.
Carry out calculations using C = Q/V
6.
Explain why work must be done to charge a capacitor.
7.
State that the work done to charge a capacitor is given by the area under the graph of
charge against p.d.
1
8.
State that the energy stored in a capacitor is given by2 (charge × p.d.) and equivalent
expressions.
1
9.
Carry out calculations using2 QV or equivalent expressions.
10.
Draw qualitative graphs of current against time and of voltage against time for the
charge and discharge of a capacitor in a d.c. circuit containing a resistor and capacitor
in series.
11.
Carry out calculations involving voltage and current in CR circuits (calculus methods
are not required).
12.
State the relationship between current and frequency in a capacitive circuit.
13.
Describe the principles of a method to show how the current varies with frequency in
a capacitive circuit.
14.
Describe and explain the possible functions of a capacitor: storing energy, blocking
d.c. while passing a.c.
Analogue Electronics
1.
State that an op-amp can be used to increase the voltage of a signal.
2.
State that for the ideal op-amp:
a)
input current is zero; i.e. it has infinite input resistance
b)
here is no potential difference between the inverting and non-inverting inputs;
i.e. both input pins are at the same potential.
3.
Identify circuits where the op-amp is being used in the inverting mode.
4.
State that an op-amp connected in the inverting mode will invert the input signal.
5.
State the inverting mode gain expression: V0 / V1 = -Rf / Ri.
6.
Carry out calculations using the above gain expression.
7.
State that an op-amp cannot produce an output voltage greater that the positive supply
voltage or less that the negative supply voltage.
8.
Identify circuits where the op-amp is being used in the differential mode.
9.
State that a differential amplifier amplifies the potential difference between its two
inputs.
10.
State the differential mode gain expression - Vo = (V2 - V1) Rf / Ri.
11.
Carry out calculations using the above gain expression.
12.
Describe how to use the differential amplifier with resistive sensors connected in a
Wheatstone bridge arrangement.
13.
Describe how an op-amp can be used to control external devices via a transistor.
Physics: Electricity and Electronics (H)
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Checklist
Uncertainties
1.
State that measurement of any physical quantity is liable to uncertainty.
2.
Distinguish between random uncertainties and recognised systematic effects.
3.
State that the scale-reading uncertainty is a measure of how well an instrument scale
can be read.
4.
Explain why repeated measurements of a physical quantity are desirable.
5.
Calculate the mean value of a number of measurements of the same physical quantity.
6.
State that this mean is the best estimate of a ‘true’ value of the quantity being
measured.
7.
State that where a systematic effect is present the mean value of the measurements
will be ‘offset’ from a ‘true’ value of the physical quantity being measured.
8.
Calculate the approximate random uncertainty in the mean value of a set of
measurements using the relationship:
maximum value - minimum value
approximate uncertainty in the mean =
number of measurements taken
9.
10.
11.
12.
13.
Estimate the scale reading incurred when using an analogue display and a digital
display.
Express uncertainties in absolute or percentage form.
Identify, in an experiment where more than one physical quantity has been measured,
the quantity with the largest percentage uncertainty.
State that this percentage uncertainty is often a good estimate of the percentage in the
final numerical result of the experiment.
Express the numerical result of an experiment in the form:
final value  uncertainty
Units, prefixes and scientific notation
1.
Use SI units of all physical quantities in the above checklist.
2.
Give answers to calculations to an appropriate number of significant figures.
3.
Check answers to calculations.
4.
Use prefixes (p, n, µ, m, k, M, G)
5.
Use scientific notation.
Physics: Electricity and Electronics (H)
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Summary Notes
ELECTRIC FIELDS AND RESISTORS IN CIRCUITS
Force Fields
In Physics, a field means a region where an object experiences a force without being touched.
For example, there is a gravitational field around the Earth. This attracts masses towards the
earth’s centre. Magnets cause magnetic fields and electric charges have electric fields around
them.
Electric Fields
In an electric field, a charged particle will experience a force. We use lines of force to show
the strength and direction of the force. The closer the field lines the stronger the force. Field
lines are continuous - they start on positive and finish on negative charge. The direction is
taken as the same as the force on a positive “test” charge placed in the field.
Electric Field Patterns
Positive point charge
Negative point charge
+ test charge
has a force
‘inwards’
+ test charge
has a force
‘outwards’
These are called radial fields. The lines are like the radii of a circle. The strength of the field
decreases as we move away from the charge.
Electric Field Patterns
Positive and negative point charges
Parallel charged plates
The field lines are equally spaced between the parallel plates. This means the field strength is
constant. This is called a uniform field.
Electric fields have certain similarities with gravitational fields.
Physics: Electricity and Electronics (H) – Student Material
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Summary Notes
Gravitational Fields
h
If a mass is lifted or dropped through a height then work is done i.e. energy is changed.
If the mass is dropped then the energy will change to kinetic energy.
If the mass is lifted then the energy will change to gravitational potential energy.
Change in gravitational potential energy = work done.
Electric Fields
Consider a negative charge moved through a distance in an electric field. If the charge moves
in the direction of the electric force, the energy will appear as kinetic energy. If a positive
charge is moved against the direction of the force as shown in the diagram, the energy will be
stored as electric potential energy.
Change in electric potential energy = work done
If the charge moved is one coulomb, then the work done is the potential difference or voltage.
If one joule of work is done in moving one coulomb of charge between two points in an
electric field, the potential difference, (p.d.) between the two points is one volt.
1 volt = 1 joule per coulomb
W = QV
In this section W will be used for the work done i.e. energy transferred.
Example:
A positive charge of 3 µC is moved, from A to B, between a potential difference of 10 V.
(a)
Calculate the electric potential energy gained.
(b)
If the charge is now released, state the energy change.
B
A
(c)
How much kinetic energy will be gained on reaching
+
the negative plate?
+
+Q
-6
+
(a)
W
= QV= 3 × 10 × 10
-5
+
= 3 × 10 J
(b)
Electric potential energy to kinetic energy
(c)
By conservation of energy the energy will be the same, i.e. 3 × 10-5 J.
Physics: Electricity and Electronics (H) – Student Material
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Summary Notes
Moving Charges in Electric Fields
From the previous example, when the positive charge is released at plate B then the electical
potential energy is converted to kinetic energy.
QV =
1
mv2
2
Example
An electron is accelerated (from rest) through a potential difference of 200 V.
Calculate
(a)
the kinetic energy, Ek gained.
(b)
the final speed of the electron.
(Mass of an electron = 9.1 × 10-31 kg, charge on an electron = -1.6x10-19 C)
(a)
Ek =
1 2
mv
2
= QV = 1.6x10-19 × 200
= 3.2 × 10-17 J
(b)
1 2
mv = 3.2 × 10-17
2
3.2  1017  2 3.2  1017  2
v2 

m
9.1  10 31
v = 8.4 × 106 m s-1
Applications of electric fields (for background interest)
A television involves the use of electron guns. The electrons gain kinetic energy by
accelerating through an electric field. Deflection of the electrons is usually done by
electromagnetic coils, although flat screen tubes are now dependent on electrostatic
deflection.
An oscilloscope also depends on electric fields acting on electrons.
Electrostatic Spraying makes use of electric fields. Paint or powder particles are blown from
a nozzle, where they acquire a charge. The object to be coated is earthed. The charged paint
or powder particles follow the field lines and so reach the object, some reaching the back of
the object as well as the front.
Other applications include photocopiers, ink jet and laser printers.
Physics: Electricity and Electronics (H) – Student Material
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Summary Notes
Electric Current
When S is closed, the free electrons in the conductor experience an electric field which cause
them to move in one direction.
Note: 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.
The lamp 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 the equation below.
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, for example the transistor.
Electric Power
For a given component the power P = I V where I is the current through that component and
V is the potential difference across the component.
Resistive Heating
The expression I2 R gives the energy transferred in one second due to resistive heating.
Apart from obvious uses in electric fires, cookers, toasters, etc. consideration has to given to
heating effects in resistors, transistors and integrated circuits and care taken not to exceed the
maximum ratings for such components.
V2
P = IR =
R
2
Electromotive Force (e.m.f)
The energy given to a coulomb of charge by a source of electrical energy is called the e.m.f.
of the source. This is measured in J C-1 or volts.
Note: the potential difference between two points in the external circuit is also measured in
volts, but this is concerned with electrical energy being transformed outside the source.
Physics: Electricity and Electronics (H) – Student Material
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Summary Notes
Sources of e.m.f.
E.M.F’s can be generated in a great variety of ways e.g:
chemical cells,
solar cell,
thermocouple,
electromagnetic generator.
piezo - electric generator,
Resistors in Series - Conservation of Energy
Applying the conservation of energy to resistors in series for one coulomb of charge.
Energy supplied by source = energy converted by circuit components
e.m.f. = IR1 + IR2 + IR3
Rs = equivalent
series resistance
IRS = IR1 + IR2 + IR3
RS = R1 + R2 + R3
Resistors in Parallel - Conservation of Charge
The total charge per second (current) passing through R1, R2 and R3 must equal the charge
per second (current) supplied from the cell i.e. passing through RP.
Conservation of charge gives:
Q
I = I1 + I2 + I3 (Since I =
for each resistor)
t
E
E
E
E
=
+
+
R1
R2
R3
R4
Rp = equivalent
parallel resistance
Physics: Electricity and Electronics (H) – Student Material
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Summary Notes
Internal Resistance
In choosing a suitable power supply for a circuit, you would have to ensure that it :
 gives the correct e.m.f
 is able to supply the required maximum current
When a power supply is part of a closed circuit, it must itself be a conductor. All conductors
have some resistance. A power supply has internal resistance, r.
Energy will be wasted in getting the charges through the supply (the heat from the supply will
be noticeable) and so the energy available at the output (the terminal potential difference)
will fall. There will be “lost volts”. The lost volts = I r
The greater the current, the more energy will be dissipated in the power supply until
eventually all the available energy (the e.m.f.) is wasted and none is available outside the
power supply. This maximum current is the short circuit current. This is the current which
will flow when the terminals of the supply are joined with a short piece of thick wire.
Open Circuit
When no current is taken from the power supply, no energy is wasted. The terminal
potential difference is therefore the maximum available and equals the e.m.f.
General Circuit
Any power supply can be thought of as a source of constant
e.m.f. E, in series with a small resistance, the internal
resistance.
r
With S open, the voltmeter reading gives the e.m.f. (an open
circuit).
S
E
R
V
With S closed, the voltmeter reading will fall (lost volts). The
voltmeter now gives the output voltage, the terminal potential
difference, t.p.d.
With S closed and using the conservation of energy
e.m.f. = lost volts + t.p.d.
e.m.f. = lost volts + output voltage
which is:
E
E
= Ir
= Ir
+
+
V
IR
Notice that the total resistance of the circuit is R + r, giving the equation:
E = I ( R + r)
The short circuit current is the maximum which can be supplied by a source. This occurs
when there is no external component and R = 0.
Physics: Electricity and Electronics (H) – Student Material
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Summary Notes
Example
r
E
V
R = 28 
A cell of e.m.f. 1.5 V is connected in series with a 28resistor.
A voltmeter measures the voltage across the cell as 1.4 V.
Calculate:
(a) the internal resistance of the cell
(b) the current if the cell terminals are short circuited
(c) the lost volts if the external resistance R is increased to 58 Ω.
(a)
E = Ir + IR = Ir + V
Lost volts = Ir = E - V = 1.5 - 1.4 = 0.1 V
(b)
A short circuit occurs when
r=
lost volts
0.1
=
I
1
I=
V
1.4
=
= 0.05 A
R
28
r=
0.1
=2
0.05
R = 0 (no external resistance)
I
E
1.5
=
=
= 0.75 A
R + r
r
2
(c)
Lost volts = Ir
I
E
=
= 1.5
R + r
28 + 2
= 0.05 A
Lost volts = 0.05 × 2
= 0.1 V
Physics: Electricity and Electronics (H) – Student Material
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Summary Notes
Wheatstone Bridge Circuit
Any method of measuring resistance using an ammeter or voltmeter necessarily involves
some error unless the resistances of the meters themselves are taken into account. The use of
digital voltmeters largely overcomes this as they tend to have very high resistances.
Further error may be introduced if the meter is not correctly calibrated. The only situation
where neither of these errors matter is if the meter reading is zero. The Wheatstone bridge
circuit is one such example of using a meter as a null deflection indicator.
Bridge Circuit
If VOA = VOB there is no p.d. between A and B hence no
current flows. The potentials at A and at B depend on the ratio
of the resistors that make up each of the two voltage dividers.
The voltmeter forms a ‘bridge’ between the two voltage
dividers to make up a bridge circuit.
5V
R1
A
R2
Balanced Bridge
No potential difference will exist across AB when


 V

R3
B
R
4
O
0V
R1
R3

R2
R4
The bridge is balanced when the voltmeter or galvanometer
(milliammeter) reads zero.
Alternative ‘diamond’
Representation with
galvanometer
Unbalanced Bridge
If the bridge is initially balanced, and the resistor × is altered by a small amount × then the
out of balance p.d. (reading on G) is directly proportional to the change in resistance,
provided the change is small.
Hence for small ×,
+1.5 V
Reading on G
R1
reading on G  ×
X/
R2
G
X+ X R4
Physics: Electricity and Electronics (H) – Student Material
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Summary Notes
ALTERNATING CURRENT AND VOLTAGE
Peak and r.m.s. values
The graph of a typical alternating voltage is shown below.
The maximum voltage is called the peak value.
From the graph it is obvious that the peak value would not be a very accurate measure of the
voltage available from an alternating supply.
In practice the value quoted is the root mean square (r.m.s.) voltage.
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.
Vr.m.s.= 1 Vpeak
2
and
Ir.m.s.= 1 Ipeak
2
Note: a moving coil a.c. meter is calibrated to give r.m.s. values.
Physics: Electricity and Electronics (H) – Student Material
9
Summary Notes
Graphical method to derive relationship between peak and r.m.s. values of alternating
current
The power produced by a current I in a resistor R is given by I2 R. A graph of I2 against t for
an alternating current is shown below. A similar method can be used for voltage.
2
The average value of I2 is I Peak
2
An identical heating effect (power output) for a d.c. supply = I2r.m.s R
[since I (d.c.) = Ir.m.s.]
2
Average power output for a.c. = I Peak R
2
2
I2 r.m.s R = I
Peak
2
R
hence

I2r.m.s.

=
I2 Peak
2
giving Ir.m.s. =
I Peak
2
Frequency of a.c.
To describe the domestic supply voltage fully, we would have to include the frequency i.e.
230 V 50 Hz.
An oscilloscope can be used to find the frequency of an a.c. supply as shown below.
Time base = 0.005 s cm-1
Wavelength = 4 cm
Time to produce one wave = 4 × 0.005
= 0.02 s
Frequency =
=
Physics: Electricity and Electronics (H) – Student Material
1
time to produce one wave
1
0.02
= 50 Hz
10
Summary Notes
Mains supply
The mains supply is usually quoted as 230 V a.c. This is of course 230 V r.m.s. The peak
voltage rises to approximately 325 V. Insulation must be provided to withstand this peak
voltage.
Example
A transformer is labelled with a primary of 230 Vr.m.s. and secondary of 12 Vr.m.s. What is the
peak voltage which would occur in the secondary?
V peak = 2 × V r.m.s.
V peak = 1.41 × 12
V peak = 17.0 V
Physics: Electricity and Electronics (H) – Student Material
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Summary Notes
CAPACITANCE
The ability of a component to store charge is known as capacitance.
A device designed to store charge is called a capacitor.
A typical capacitor consists of two conducting layers separated by an insulator.
Circuit symbol

Relationship between charge and p.d.
The capacitor is charged to a chosen voltage by
setting the switch to A. The charge stored can be
measured directly by discharging through the
coulomb meter with the switch set to B. In this
way pairs of readings of voltage and charge are
obtained.
Q
Charge is directly proportional to voltage.
0
Q
= constant
V
V
For any capacitor the ratio Q/V is a constant and is called the capacitance.
farad (F)
capacitance =
charge
voltage
coulombs (C)
volts (V)
The farad is too large a unit for practical purposes.
In practice the micro farad (µF) = 1 × 10-6 F and the nano farad (nF) = 1 × 10-9 F are used.
Example
A capacitor stores 4 × 10-4 C of charge when the potential difference across it is 100 V.
What is the capacitance ?
Q
4  10 C =

V
100
6
= 4 µF
Physics: Electricity and Electronics (H) – Student Material
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Summary Notes
Energy Stored in a Capacitor
A charged capacitor can be used to light a bulb for a short time, therefore the capacitor must
contain a store of energy. The charging of a parallel plate capacitor is considered below.
There is an initial surge of
electrons from the negative
terminal of the cell onto one
of the plates (and electrons
out of the other plate towards
the +ve terminal of the cell).
Once some charge is on the
plate it will repel more charge
and so the current decreases.
In order to further charge the
capacitor the electrons must
be supplied with enough
energy to overcome the
potential difference across the
plates i.e. work is done in
charging the capacitor.
Eventually the current ceases
to flow. This is when the
p.d. across the plates of the
capacitor is equal to the
supply voltage.
For a given capacitor the p.d. across the plates is directly proportional to the charge stored
Consider a capacitor being charged to a p.d. of V and holding a charge Q.
Charge
The energy stored in the capacitor is given by the area under
graph
the
Q
V p.d.
Area under graph= 1 Q x V
2
Energy stored = 1 Q x V
2
If the voltage across the
capacitor was constant work
done = Q x V, but since V is
varying, the work done =
area under graph.
Q = C × V and substituting for Q and V in our equation for energy gives:
1
1
1 Q2
2
Energy stored in a capacitor = QV = CV =
2
2
2 C
Example
A 40F capacitor is fully charged using a 50 V supply. How much energy is stored?
Energy =
1
1
CV 2 =
 40  10 6  2500
2
2
= 5 × 10-2 J
Physics: Electricity and Electronics (H) – Student Material
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Summary Notes
Capacitance in a d.c. Circuit
Charging
Consider the following circuit:-
When the switch is closed the current flowing in the circuit and the voltage across the
capacitor behave as shown in the graphs below.
p.d. across
capacitor
current
Supply voltage
0
0
time
time
Consider the circuit at three different times.
0
0
A
A
As soon as the switch is
closed there is no charge on
the capacitor the current is
limited only by the
resistance in the circuit and
can be found using Ohm’s
law.
0
A
+ +
+ +
+ +
-
--
-
As the capacitor charges a
p.d. develops across the
plates which opposes the
p.d. of the cell as a result
the supply current
decreases.
Physics: Electricity and Electronics (H) – Student Material
--
The capacitor becomes
fully charged and the p.d.
across the plates is equal
and opposite to that across
the cell and the charging
current becomes zero.
14
Summary Notes
Discharging
Consider this circuit when the capacitor is
fully charged, switch to position B
If the cell is taken out of the circuit and the
switch is set to A, the capacitor will
discharge
A
-- -++ ++
A
B
A
B
A
While the capacitor is discharging the current flowing in the circuit and the voltage across
the capacitor behaves as shown in the graphs below.
p.d. across
capacitor
Current
0
Supply voltage
0
time
time
Although the current/time graph has the same shape as that during charging the currents in
each case are flowing in opposite directions. The discharging current decreases because the
p.d. across the plates decreases as charge leaves them.
Factors affecting the rate of charge/discharge of a capacitor
 When a capacitor is charged to a given voltage the time
taken depends on the value of the capacitor. The larger
the capacitor the longer the charging time, since a larger
capacitor requires more charge to raise it to the same
C
p.d. as a smaller capacitor as V= Q
 When a capacitor is charged to a given voltage the time
taken depends on the value of the resistance in the
circuit. The larger the resistance the smaller the initial
charging current, hence the longer it takes to charge
the capacitor as Q = It
(The area under this I/t graph = charge. Both curves will
have the same area since Q is the same for both.)
Physics: Electricity and Electronics (H) – Student Material
Current
large capacitor
small capacitor
Time
Current
small resistor
large resistor
Time
15
Summary Notes
Example
The switch in the following circuit is closed at time t = 0
10 V
Vs
1 M
1 µF
Immediately after closing the switch what is:
(a)
the charge on C
(b)
the p.d. across C
(c)
the p.d. across R
(d)
the current through R.
When the capacitor is fully charged what is:
(e)
the p.d. across the capacitor
(f)
the charge stored.
(a)
(b)
(c)
(d)
(e)
(f)
Initial charge on capacitor is zero.
Initial p.d . is zero since charge is zero.
p.d. is 10 V = Vs - Vc = 10 - 0 = 10 V
I
10
= V =
= 10 3 A
R
10 6
Final p.d. across the capacitor equals the supply voltage = 10 V.
Q = CV = 2 × 10-6 × 10 = 2 × 10-5 C.
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Summary Notes
Resistors and Capacitors in a.c. Circuits
Frequency response of resistor
The following circuit is used to investigate the relationship between current and frequency in
a resistive circuit.
The results show that the current flowing through a resistor is independent of the frequency
of the supply.
Frequency response of capacitor
The following circuit is used to investigate the relationship between current and frequency in
a capacitive circuit.
The results show that the current is directly proportional to the frequency of the supply.
To understand the relationship between the current and frequency consider the two halves of
the a.c. cycle.
The electrons move back and forth around the circuit passing through the lamp and charging
the capacitor one way and then the other (the electrons do not pass through the capacitor).
The higher the frequency the less time there is for charge to build up on the plates of the
capacitor and oppose further charges from flowing in the circuit More charge is transferred
in one second so the current is larger.
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Summary Notes
Applications of Capacitors (for background interest)
Blocking capacitor
A capacitor will stop the flow of a steady d.c. current . This is made use of in the a.c./d.c.
switch in an oscilloscope. In the a.c. position a series capacitor is switched in allowing
passage of a.c. components of the signal, but blocking any steady d.c. signals.
Flashing indicators
A low value capacitor is charged through a
resistor until it acquires sufficient voltage to
fire a neon lamp. The neon lamp lights when
the p.d. reaches 100 V. The capacitor is
quickly discharged and the lamp goes out
when the p.d. falls below 80V.
1 - 2 M
120 V
Crossover networks in loudspeakers
In a typical crossover network in low cost
loudspeaker systems, the high frequencies are
routed to LS-2 by the capacitor.
LS 1
LS 2
Smoothing
The capacitor in this simple rectifier circuit is storing charge during the half cycle that the
diode conducts. This charge is given up during the half cycle that the diode does not conduct.
This helps to smooth out the waveform.
Capacitor as a transducer
A parallel plate capacitor can be used to convert mechanical movements or vibration of one
of its plates into changes in voltage. This idea forms the basis of many measuring systems,
e.g. by allowing a force to compress the plates we have a pressure transducer.
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Summary Notes
ANALOGUE ELECTRONICS
Analogue and Digital Signals
An analogue system transmits information in the form of a continuously varying signal.
A digital system has the information broken up into a series of discrete values or steps.
A simple example showing these definitions would be analogue and digital watches.


an analogue watch has a set of numbers in a circle and hands that point to them. The
hands sweep round the face in a continuous way.
a digital watch has a series of numbers that are displayed, the numbers changing at the
end of each second, minute or hour.
The Op-Amp (Operational Amplifier)
As with any amplifier, the operational amplifier (op-amp) will change the size of an input
electrical signal. A diagram of the basic set-up of an op-amp is shown below.
The op-amp has two separate inputs - an inverting input (“-” terminal) and a non-inverting
input (“+” terminal). The amplifier must have an energy supply and this is provided by the
supply voltages +Vs and -Vs across the amplifier. Often the power supply voltages are not
shown on circuit diagrams of op-amps.
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Activities
The Inverting Amplifier
If an amplifier is set up in the configuration shown below, it is said to be in the inverting
mode.
Rf
R1
V1
+
V0
0V
The input potential, V1, is applied to the inverting input (-ve input). The non-inverting
input (+ve input) is connected straight to “ground”, 0 V.
There is also a resistor, Rf, (feedback resistor) connected between the output and the inverting
input. This feedback resistor reduces the overall gain of the amplifier and allows
the gain to be stabilised controlled.
When the input voltage signal to the op-amp, d.c or a.c, is compared to the output voltage
signal it is found that the sense or ‘sign’ of the output is opposite to that of the input - the
voltage has been inverted, hence the reason why this circuit is called the “inverting mode”.
Some examples of this are given below:
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Activities
Inverting Mode Gain Equation
The ideal op-amp fulfils two conditions:
 no current flows into the op-amp. Its input resistance is infinite.
 there is no potential difference between the inverting and non-inverting inputs.
The following inverting mode gain equation can be verified by experiment.
Saturation
From the inverting mode gain equation, it would seem that by inserting any pair of resistors
R1 and Rf in the inverting amplifier circuit it is possible to provide any gain required. It could
therefore be possible to produce either very low or very high output voltages from any input
voltage.
This is not possible. The output of an op-amp circuit is limited by the size of the power
supply used (conservation of energy). In theory, the maximum output voltage possible
would be equal to the supply voltage. (However, in reality it is limited to approximately
85% of the supply voltage.)
When the maximum output voltage has been reached, the amplifier is said to be saturated.
+
+15V
V0
In theory :
-15V  V 0  +15 V
[In practice :
-13V  V 0  +13 V (13V  85% of 15 V)]
-15V
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Summary Notes
Example
An inverting mode operational amplifier is set up as shown below.
(a) If V1 is set at +0.8 V, calculate the output voltage V0.
(b) If an a.c signal of peak voltage 1.5 V is applied to V1, sketch the input voltage, V1,
and the output voltage, V0.
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Activities
The Differential Amplifier
If the amplifier is set up in the configuration shown below, it is said to be in the differential
mode.
There are two input potentials,V1 and V2, one applied to each of the input terminals of the opamp.
There is a feedback resistor, Rf, connected between the output and the inverting input. This
allows control over the gain of the amplifier as it did for the inverting mode.
When the op-amp is used in this mode, it amplifies the difference between the inputs V1 and
V2, with a gain set by the ratio
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Summary Notes
Example
A differential amplifier is set up as shown below.
For the following values shown, calculate the output voltage, V0.
(a)
V1 = +5.0 V, V2 = +4.8 V
(b)
V1 = -2.0 V, V2 = +4.5 V
(a)
V1 = +5.0 V, V2 = +4.8 V, R1 =R2 = 10 kΩ, Rf = R3 = 100 kΩ, V0 = ?
V0 =
Rf
(V - V 1)
R1 2
V0 =
V0 =
(b)
100
(4.8 - 5.0)
10
- 2.0 V
V1 = -2.0 V, V2 = +4.5 V, R1 =R2 = 10 kΩ, Rf = R3 = 100 kΩ, V0 = ?
V0 =
Rf
(V - V 1 )
R1 2
V0 =
100
(4.5 - (-2.0))
10
V 0 = + 65 V in theory.
But supply voltage = + 15 V hence output voltage, V0, will saturate at + 15 V
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Summary Notes
The differential amplifier as part of a monitoring system
+Vs
10 k
Rf
0-10 k
V1
t
10
k
V2
R1
10 k
R2
10 k

R3
100 k
+
V0
100 k
0V
0V
Wheatstone
Bridge input
Differential
Amplifier
The setting of the variable resistor can be adjusted so as to achieve an output voltage of zero
for a particular temperature setting. The bridge circuit would then be balanced, that is the
potential difference V2 - V1 = 0 V.
The potential, V2, will remain constant as long as the resistance of the variable resistor is not
changed. Any change in temperature will change the potential, V1, and will therefore
produce a potential difference between V2 and V1.
The amplifier will then amplify the difference between V2 and V1, giving an output potential,
V0. The output voltage will increase as the change in thermistor resistance, ∆Rt, increases.
The amplifier is, effectively, amplifying the out-of-balance potential difference from the
Wheatstone Bridge.
Practical Application
The output voltage could be calibrated by placing the thermistor in melting ice (0 oC), then in
boiling water (100 oC), noting the output potential, V0, for each case. The range could then be
divided into 100 equal divisions to give an electronic thermometer over the range 0 - 100 oC.
Physics Support Material: Electricity and Electronics (H) – Student Material
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Summary Notes
Control Circuits
A transistor, such as those shown below, can act as an electrical switch.
If the input voltage to these transistors, Vi, is positive, then it switches on allowing a current
to flow between the collector and emitter or source and drain, otherwise it is off.
There are two types of transistor switches:
n-channel enhancement MOSFET
bipolar n-p-n transistor
-V S
+V S
base
drain
collector
gate
emitter
Vi
0V
(i) A positive (+ve) potential (>+0.7 V)
is needed for the input, Vi, to switch
transistor on .
(ii) The collector arm is connected to a
positive (+ve) supply rail.
(iii) If the input potential is negative
(<0.7 V), the transistor is off.
source
Vi
0V
(i) A positive (+ve) potential >1.8 V is
needed for the input, Vi, to switch
transistor on .
(ii) The drain is connected to a positive
(+ve) supply rail.
(iii) if the input potential is negative or
<1.8 V the transistor is off.
These transistors can be used with a Wheatstone Bridge/differential amplifier circuit to
switch a device on or off.
Low light-level indicator
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Activities
The variable resistor in the Wheatstone Bridge is adjusted so that, at a particular light level,
the output of the op-amp is zero or less so that the transistor is off. In practice, the variable
resistor would be adjusted until the warning lamp is just off. In this situation, V1 ≈ V2.
If the light level falls, the resistance of the LDR will increase causing the voltage V1 to fall.
If V1 falls, then V1 < V2, therefore (V2 - V1) will be positive (+ve).
This will cause the output voltage, V0, to be positive also, switching on the transistor and
the Warning Lamp lights.
If the light level to the LDR increases, then the opposite will be true.
The resistance of the LDR will decrease causing the voltage V1 to increase.
If V1 increases, then V1>V2, (V2 - V1) will be negative causing the output to be negative and
the transistor will not switch on.
The gain of the circuit (about 1000) is deliberately chosen to be large so that a small variation
from the balance point of the Wheatstone Bridge produces a large enough output voltage to
switch the transistor on.
Modifications
This circuit does have its limitations however. If the output device to be used has a high
power rating requiring a current larger than the transistor can safely supply, then a relay
switch must be used with the transistor to switch on an external circuit.
An example of this type of switch is shown below. In this case, a relay can be energised by
the current through the transistor (the collector current) and its contacts can then be used to
switch on a high-power device such as a heater.
+Vs
10K
10K
V1
V2
R1
1K
R2
1K
+
R3
0V
Wheatstone
Bridge input
relay
-
100K
0-10K
t
+Vs
Rf
100K
230 V
mains
Differential
Amplifier
V0 Vi
heater
Transistor Switch
and Output
Notice that as the temperature falls the resistance of the thermistor increases, causing V1 to
fall. this means V1 < V2 giving a positive output voltage V0. The transistor will be switched
on, which will energise the relay switch. The heater will be turned on.
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Activities
Control Circuit Diagrams
The Wheatstone Bridge arrangement can be shown as two straight potential dividers or as a
diamond arrangement. They are the same circuit, just drawn a different way.
Straight potential dividers
+Vs
t
to differential
amplifier
10K
10K
V1
V2
0V
Diamond arrangement bridge circuit
V2
+Vs
10 k
t
V1
10 k
0V
to op-amp
+Vs
V2
to op-amp
10 k
t
10 k
V1
0V
All the above circuits are identical examples of the sensor bridge circuit that can be used with
a differential amplifier. It is important to recognise each circuit for what it is and how it
behaves.
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Activities
ACTIVITY 1
Title: Electric Field Patterns
Apparatus: Van de Graaff generator, petri dish with oil and seeds, set of shaped electrodes
overhead projector, some sawdust
Instructions
2-D Patterns
 Connect up the apparatus as shown and use the Van de Graaff generator to produce a high
voltage between two point electrodes in the oil.
 Draw the electric field pattern present between the two point electrodes as shown by the
seeds.
 Repeat the above but using two plane electrodes, producing a uniform electric field
between them.
3-D Patterns
 While the Van de Graaff is switched on and charging, throw a small amount of sawdust at
the dome.
 Sketch the path taken by the pieces of sawdust after they came into contact with the dome.
The sawdust is affected by the radial field around the dome.
 Explain the behaviour of the pieces of sawdust after they came into contact with the
charging dome.
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Activities
ACTIVITY 2
Title: The Cathode Ray Tube
Apparatus: Cathode Ray Deflection Tube and stand, EHT supply (for electron gun and
heater), HT supply (for deflection plates)
Instructions
Work Done by Electric Field
 Note the value of the EHT voltage used in the electron gun.
 Calculate the work done on each electron by the electric field in the electron gun.
 Hence calculate the speed of the electrons as they leave the electron gun and enter the
evacuated tube.
Deflection Plates
 Sketch the path followed by the electron beam as it passes between the deflection plates
when:
(a) bottom plate is positive, top plate is negative
(b) top plate is positive, bottom plate is negative.
 Explain the effect the following changes have on the path followed by the electron beam:
(a) decrease EHT voltage on electron gun
(b) decrease HT voltage on deflection plates
 Sketch the paths followed by the electron beam after the above changes.
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Activities
ACTIVITY 3
Title: E.m.f. and Internal Resistance with parallel circuits
Apparatus:
6 V battery, parallel circuit board, ammeter, voltmeter to measure the terminal
potential difference (t.p.d.).
V
A
L1
L2
Instruction
L3
 Connect up the circuit above, with all lamps unscrewed
so that they are off.
 Copy the table below.
 Note down the current (I) and corresponding t.p.d. (V) values when no lamps are screwed
in, and enter them into the table.
 Repeat the above for each case as one, then two, then three lamps are screwed in and light
up.
 Using Ohm’s Law, complete the final column of the table by calculating the value of the
internal resistance of the battery.
No. of
bulbs lit
Current I
(A)
t.p.d. V
(V)
“lost volts”
(V)
Internal
resistance ()
0
1
2
3
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Activities
ACTIVITY 4
Title: Internal resistance of a cell
(Outcome 3)
Apparatus: 1.5 V cell, variable resistor (1-10), voltmeter and ammeter
Instructions
 Use the voltmeter to measure the e.m.f. of the cell.
 Connect the apparatus as shown in the circuit diagram.
 Set the variable resistor to its maximum setting.
Record the readings on the ammeter and voltmeter.
 Adjust the variable resistor and take a range of readings.
 For each pair of readings determine the lost volts.
 Use an appropriate format to determine the internal resistance of the cell.
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Activities
ACTIVITY 5
Title: The Balanced Wheatstone Bridge
Apparatus: Wheatstone Bridge board, 1.5 V cell, 20-0-100 µA ammeter, 1 k/10 k
resistance boxes, set of unknown resistances A, B and C, decade resistance
board.
Schematic diagram showing position of resistances
Circuit diagram
Instructions
Part A
 Connect the 1 k resistances, decade resistance board and unknown resistance A as shown
in the diagram above.
 By finding the value of resistance of the decade resistance board which exactly balances
the Wheatstone Bridge, find the value of resistance A.
 Repeat to find the value of resistances B and C.
Part B
 Replace R2 with a 10 k resistor.
 Predict the value of the decade resistance board that will balance the Wheatstone Bridge
for each of the resistors A, B and C with this value of R2.
 Confirm your predictions by experiment.
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Activities
ACTIVITY 6
Title: The Out-of-Balance Wheatstone Bridge
Apparatus:
Wheatstone Bridge board, 1.5 V cell, 20-0-100 µA ammeter, 1 k/10 k
resistance boxes, set of resistances A, B and C, decade resistance board.
R2
R1
1 k
1 k
µA
Resistance
Board
Resistor B
Instructions
 Connect the circuit as shown in the diagram.
 Adjust the resistance of the decade resistance board so that the Wheatstone Bridge is
balanced.
 Increase the resistance of the decade resistance board in 10  steps, noting the value of the
out-of- balance current each time.
 Record your results and complete the table.
 Plot a graph of change in resistance, R (), on the ×-axis and out-of-balance current, I,
(µA) on the y-axis.
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Activities
ACTIVITY 7
Title: The Out-of-Balance Wheatstone Bridge - Applications
Apparatus: Wheatstone Bridge board, 6 V battery, 20-0-100 µA ammeter, 1 k/10 k
resistance boxes, decade resistance board, LDR, thermistor probe, strain gauge
(already set up).
Instructions
Part A - Light Meter
 Connect the circuit up as shown.
 Balance the Wheatstone Bridge circuit with
the LDR on your bench. There is no need to
“remove” the protective resistor as the
circuit is quite sensitive. Take care not to
cast shadows over the LDR when finding
balance.
 Find the out-of-balance current when the
LDR is in a light place then in a dark place.
 Explain why the measured current is greater
than zero for one condition and less than
zero for the other.
Part B -Thermometer
 Place the probe in an ice/water mixture in a
beaker. (0 °C)
 Balance the Wheatstone Bridge with the
probe in the ice/water mixture.
 Place the probe in a beaker of boiling water.
(100 °C)
 Measure and record the out-of-balance
current obtained with the probe in the
boiling water.
 Predict the current obtained when the probe
is removed and is measuring room
temperature.
 Calculate the value of room temperature
from your results.
 What assumptions are you making about the
temperature of the probe, its resistance, and
the out-of-balance current?
Continued...
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Activities
ACTIVITY 7 (continued)
Instructions (continued)
Part C - Strain Gauge
The hacksaw blade has been fitted with two strain gauges, one on each side of the blade.
The resistance of a strain gauge changes when its shape is deformed - either stretched or
compressed. When the hacksaw blade is bent, one of the strain gauges will “stretch” while
the other one will “compress”. The two strain gauges are connected to one arm of a
Wheatstone Bridge circuit.
 The strain gauge circuit should already be
set up for you.
 Balance the Wheatstone Bridge circuit
when the hacksaw blade is straight.
 Bend the hacksaw blade in one direction.
 Note the out-of-balance current.
 Bend the hacksaw blade in the other
direction.
 Note the out-of-balance current.
 Explain how the out-of-balance current is
used to show (a) the amount of bending/strain put on
the hacksaw blade
(b) the direction of the bending/strain
put on the hacksaw blade.
Physics: Electricity and Electronics (H) – Student Material
Strain
gauge (back)
Strain
gauge (front)
µA
680 
Resistance
Board
4.5 V
36
Activities
ACTIVITY 8
Title: Alternating Current – Peak and r.m.s. values
Aim: To establish a relationship between peak and equivalent direct (r.m.s.) values of
voltage.
Apparatus:
Lab pack, 6 V battery, oscilloscope, variable resistor (0-22 Ω), 2 × 2.5 V
lamps, connecting leads.
Circuit 1
Circuit 2
to
oscilloscope
to
oscilloscope
variable
a.c. supply
Instructions
 Set up Circuit 1.
 Switch the time-base on the oscilloscope OFF.
 Adjust the supply and the oscilloscope to give a measured peak alternating voltage of 1 V
on the oscilloscope
 Leave Circuit 1 switched on.
 Set up Circuit 2.
 Adjust the variable resistor until the lamp is the same brightness as the lamp in Circuit 1.
 Use the oscilloscope to measure the direct voltage across this lamp.
 Repeat the measurements for peak voltages of 2 V, 3 V, 4 V and 5 V.
 Plot a graph of direct voltage against peak voltage.
 Determine the gradient of the graph.
 State the relationship between Vd.c. and Vpeak using the value obtained from the gradient of
the graph.
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Activities
ACTIVITY 9
Title: Calibration of Signal Generator
Aim: To calibrate the frequency scale on a signal generator.
Apparatus: Oscilloscope, signal generator, connecting leads
Instructions
 Connect the output of the signal generator to the Y-inputs of the oscilloscope as shown.
 Switch the time-base ON.
 Set the signal generator to 10 Hz and switch on.
 Adjust the oscilloscope controls to obtain a recognisable waveform.
 Calculate the frequency from the trace on the screen. It is useful to record : the timebase
setting, divisions for one cycle, and time for one cycle with the frequency in your table of
readings.
 Repeat for other frequency values of 100 Hz, 10000 Hz and 10,000 Hz.
 Compare the measured and stated values of frequency.
 Include a column for percentage uncertainty in your table and complete this column
assuming the oscilloscope is 100% accurate.
 State which scale on the signal generator is most prone to uncertainty.
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Activities
ACTIVITY 10
Title: Charge and potential difference for a capacitor
Apparatus:
(Outcome 3)
Electrolytic capacitor (about 5000 µF), coulomb meter, voltmeter, 6 × 1.5 V
battery, changeover switch
A
B
V
Coulomb meter
Instructions
 Discharge the capacitor by shorting with connecting lead.
 Connect the circuit and set the switch to charge the capacitor as shown in the diagram.
Allow enough time for the capacitor to charge fully.
 Set the switch to B to fully discharge the capacitor through the coulomb meter.
 Repeat for other charging voltages.
 Use an appropriate format to show the relationship between charge and voltage.
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Activities
ACTIVITY 11
Title: Charging and Discharging Characteristics for a Capacitor
Aim: To observe the variation of the current through, and the p.d. across, a capacitor during
the charge and discharge cycles.
Apparatus
2200 µF capacitor, 10 kΩ resistor, ammeter and voltmeter, 6 V battery
stopclock
Instructions
Part 1 Charging
 Set up the circuit as shown with the switch open.
 Close the switch and start the stopclock.
 Record values of current I and p.d. V every 10 seconds until the measured p.d. becomes
constant.
 Plot graphs of current I and p.d. V against time for the charging cycle.
 Describe the change in the current and p.d. during the charging cycle.
Part 2: Discharging
 Fully charge the capacitor as in Part 1.
 Disconnect the leads from the battery and join them together, as shown above.
 Close the switch and note values for current I and p.d. V for the capacitor every 10
seconds from time t = 0.
 Plot graphs of current I and p.d. V against time for the discharge cycle.
 Explain the change in current and p.d. during the discharging cycle.
 Compare the direction of current during the charging and discharging cycles and explain
any differences.
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Activities
ACTIVITY 12
Title: Response of resistance in a variable frequency a.c. circuit
Aim: To establish a relationship between the current through a resistor and the frequency of
the a.c. supply.
Apparatus: resistor (20 - 50 ), signal generator, a.c. ammeter, a.c. voltmeter
V
A
Instructions
 Set up the circuit as shown with the supply switched off.
 Set the frequency of the supply to 300 Hz.
 Switch on and note the ammeter reading.
 Repeat in steps of 50 Hz up to 800 Hz, ensuring the p.d. of the supply is kept constant.
 Comment on the current through the resistor as the frequency is increased
 State if the resistance of a resistor is affected by the frequency of the a.c. supply.
 State why the p.d. of the supply must remain constant.
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Activities
ACTIVITY 13
Title: Current and frequency in a capacitive circuit
Apparatus:
capacitor.
(Outcome 3)
Signal generator, a.c. voltmeter or oscilloscope, a.c. ammeter, 4.7 µF
~
Instructions
 Connect the circuit as shown in the circuit diagram.
An oscilloscope may be used in place of the voltmeter.
 Set the output of the signal generator to about 3 V.
 Vary the source frequency and record readings of current and frequency using
a range of 100 Hz to 1 kHz.
 Ensure that the supply voltage remains constant.
 Use an appropriate format to show the relationship between current
and frequency.
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Activities
ACTIVITY 14
Title: Uses of Capacitors - the photographic flash
Aim: To show the principle behind the operation of a photographic flash.
In photography, where light has to be supplied by the flash unit, the light has to be supplied in
the short period of time that the shutter is open. In this time a large amount of light energy
must be emitted. This is stored as electrical energy in a capacitor until it is needed.
Apparatus: 1.5 µF capacitor, neon lamp, 100 k resistor, 120 d.c. supply, 1 SPST switch
1 push switch
120 V dc
F
100 k
S
neon lamp
Instructions
 Set up the circuit as shown.
 To switch on the “flash unit”, close switch F.
 To simulate the shutter opening for a very short time, close switch S and release quickly.
 Note what happens.
 Explain what happens to the p.d. across the capacitor when switch F is closed?
 If switch F remains closed state what will happen to the capacitor after switch S has been
released and the lamp has flashed.
 The neon lamp requires a p.d. of 100 V across it to make it light. Explain why the lamp is
able to light in this circuit.
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Activities
ACTIVITY 15
Title: Uses of Capacitors - d.c. power supply
Aim: To show the effect of capacitors in the production of a smooth d.c. supply from an a.c.
supply.
In the following circuits, the 120  resistor represents the load resistor or device being driven
by the supply e.g. a radio. The oscilloscope indicates the form of the output p.d. across the
load resistor.
Instructions
diode
12 V a.c.
120 
CIRCUIT 1
CIRCUIT 2
+
C
12 V a.c. 120 
120 
12 V a.c.
CIRCUIT 3 C = (a) 5 µF
(b) 10 µF
(c) 20 µF







120 
12 V a.c.
+ 2200 µF
CIRCUIT 4
Set up circuit 1 as shown.
Draw the circuit and sketch the waveform displayed on the oscilloscope screen.
Set up each of the other circuits, in turn.
Draw the circuits and sketch the waveforms produced.
Explain whether the waveform produced in circuit 2 is a.c. or d.c.
Describe the effect of the capacitor on the waveforms produced in circuit 3.
State what effect the size of the capacitance has on the smoothing of the supply.
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Activities
INFORMATION SHEET FOR ACTIVITIES 16, 17, 18, 19, 20, 21 AND 22
The Amplifier Circuit Board
A diagram of the Nuffield Operational Amplifier circuit board is shown below.
It is important to become familiar with the input potentiometers in order to work successfully
with the op-amp circuit board.
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Activities
ACTIVITY 16
Title: Familiarisation - Using The Input Potentiometers
Apparatus: Op-Amp Board, Dual Rail -15 - 0 - 15 V Power Supply, Multimeter and leads.
Instructions
Part A: Positive (+ve) input potential.
 Connect up the top input potentiometer and voltmeter using the connections shown.
 Adjust the top input potentiometer to confirm that you can obtain a range of voltages on
the voltmeter from 0 to +15 V.
 Repeat the above but this time connecting up the bottom input control to obtain a range of
voltages from 0 to +15 V also.
continued....
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Activities
ACTIVITY 16 (continued)
Instructions (continued)
Part B: Negative (-ve) input potential.
 Connect up the top input potentiometer and voltmeter using the connections shown below.
Negative supply rail
-V s
V
Zero volt rail
0V
 Adjust the top input potentiometer to confirm that you can obtain a range of voltages on
the voltmeter from 0 to -15 V.
 Repeat the above but this time connecting up the bottom input control to obtain a range of
voltages from 0 to -15 V also.
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Activities
ACTIVITY 17
Title: The Inverting Amplifier
Apparatus: Op-Amp Board, Dual Rail -15 - 0 - 15 V Power Supply, Multimeter and leads.
The circuit diagram for the circuit and results table are shown below.
Rf () R1 ()
V1
V0
Rf
V0
(V)
(V)
R1
V1
100K
10K
0.5
100K
10K
1.2
100K
10K
-0.3
100K
10K
-1.0
10K
10K
8.0
10K
10K
4.5
10K
10K
-6.0
10K
10K
-0.8
Instructions
 Connect up the top input potentiometer and two voltmeters using the connections shown .
The circuit board connections shown above are for positive (+ve) input potentials and a
gain of 10, since R1 = 10 k and Rf is set to 100 k.
 Connect R1 at 10 k and Rf at 100 k. Set V1 to 1.2 V.
 Measure the output voltage and record in your own table. Repeat this measurement for the
other values of V1 and Rf shown in the sample table
 Complete the last two columns of your table.
 Write a conclusion to your experiment including a comment on the polarity of V0
compared to V1.
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Activities
ACTIVITY 18
Title: Saturation
Apparatus: Op-Amp Board, Dual Rail -15 - 0 - 15 V Power Supply, Multimeter and leads.
Rf
100K
R1
10K
V1
V0
V
+
V
0V
Instructions
 Connect up the following circuit using the top input potentiometer and two voltmeters
from the circuit diagram below. This circuit is identical to the circuit used in Activity 17.
 Set the value of the input voltage, V1, to the values shown in the tables and record the
corresponding value of the output voltage, V0 in your own table.
 Graph the results of both tables from your experiment on axes similar to those below.
 State the gain setting of the inverting amplifier used?
 Describe what happens to the value of the output voltage, V0, as the input voltage, V1, is
increased.
 State the maximum output voltages available from the amplifier.
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Activities
ACTIVITY 19
Title: Square Wave Generator
Apparatus: Op-Amp Board, Dual Rail -15 - 0 - 15 V Power Supply, Multimeter and leads.
Instructions
 Connect the op-amp board as an inverting amplifier, using the resistor values shown.
 Connect the signal generator to the input of the inverting amplifier.
 Connect the oscilloscope across the inputs of the op-amp, AB.
 Set the signal generator to approximately 3 V at a frequency of 200 Hz.
 Adjust the Y-gain and time-base controls of the oscilloscope until you obtain a steady
wave pattern.
 Accurately sketch the wave pattern produced.
 Now connect the oscilloscope across the outputs of the op-amp, CD.
 Without adjusting any of the controls on the oscilloscope or signal generator, sketch the
wave pattern produced at the output.
 Accurately sketch the wave pattern produced this time.
 State how the phase of the output potential, V0, compares to that of the input potential, V1.
 Compare the frequency of the output potential, V0, to V1.
 State the gain of the amplifier in this circuit.
 Hence state the minimum value of V1 that will produce a saturated output potential, V0.
 As the input potential from the signal generator, V1, is increased, explain what happens to
the output potential, V0.
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Activities
ACTIVITY 20
Title: The Differential Amplifier
Apparatus: Op-Amp Board, Dual Rail -15 - 0 - 15 V Power Supply, Multimeter and leads.,
100 kΩ and 10 kΩ resistor panels.
The circuit diagram for the circuit is shown below and the results table is given on the
following page.
Rf
R1
In this circuit, R f and R3 will
always be taken as 100 k due
to the limitations of the board.
-
V1
R2
V2
V0
+
R3
V
0V
Instructions
 Connect up the two input potentiometers using the connections shown. The circuit board
is shown for positive (+ve) input potentials with a gain of 1, but negative (-ve) input
potentials are also used as are different settings for the gain.
(Continued...)
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Activities
ACTIVITY 20 (Continued)
Rf & R3 R1 & R2
Rf
R1
V2
(V)
V1
(V)
()
()
100K
100K
6.0
2.0
100K
100K
1.0
4.0
100K
100K
5.0
-2.5
100K
100K
-1.5
-3.0
100K
10K
3.5
3.0
100K
10K
1.5
2.5
100K
10K
-4.5
-5.0
100K
10K
-1.2
-0.8
V2 -V1
(V)
V0
(V)
Rf
R1
(V2 -V1 )
(V)
Instructions (continued)
 Set Rf and R3 at 100 kΩ.
 For each of the values of R1 and R2, V2 and V1 given on the sample table record the output
voltage V0 in your own table.
 Complete the other columns in your table.
 Write a conclusion, including a comment on the polarity of V0 compared to (V2-V1).
 Comment on the maximum gain you could obtain if V1 = 1 V and V2 = 1.5 V.
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Activities
ACTIVITY 21
Title: The Inverting Amplifier used to Control Heavier Loads
Aim: To investigate the use of a transistor and relay in a control circuit
The circuit below uses a transistor, relay and op-amp to control high current devices. Some
devices, such as motors or heaters, require a current which is too large for small transistors to
supply. The transistor can, however, be used to energise the coil of a relay which can then
switch on a separate supply to the high current device.
Apparatus: op-amp board, 2 × 5 V d.c supplies, n-p-n transistor board, relay, 6 V motor,
ammeter.
+5 V
+Vs
100 k
Rv
0-10k
1 M
V1
M
Reed
Relay
5V
supply
+
0V
0V
Instructions
 Connect up the circuit.
 Gradually increase the input voltage, V1. At some point, the relay contacts should close
and the motor will work.
 Redraw the circuit, marking in the positions of ammeters required to measure
(a) the output current from the op-amp
(b) the current in the relay coils
(c) the current in the motor.
 Ensure your circuit diagram is correct, then connect up an ammeter in the correct positions
and note the three current values above.
 State the value of the gain of this circuit.
 If the n-p-n transistor requires 0.7 V to switch on, state the value of V1 required to operate
the reed relay and also the motor.
 Explain the operation of this circuit.
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Activities
ACTIVITY 22
Title: The Differential Amplifier used to Monitor Light Level
Aim: To investigate the use of a differential amplifier with a Wheatstone Bridge to monitor
light level.
The circuit below uses a LDR as a resistive sensor in a Wheatstone Bridge circuit. With the
op-amp, it can be used to monitor light level changes. The two potential dividers, R1, R2 and
R3, R4 are connected as a Wheatstone Bridge circuit. Any out of balance potential difference
from the Wheatstone Bridge is applied across the inputs of the op-amp. The resistance of R4
can be adjusted to balance the bridge at any desired light level. The output of the op-amp is
therefore proportional to the change in resistance of the LDR caused by the change in light
level.
Apparatus: op-amp board, LDR panel, 2 × 10 kΩ resistor panels.
+Vs
100 k
R1
10 k
R3
10 k
10 k
10 k
R2
R4
0-10 k
+
100 k
0V
V
0V
Instructions
 Ideally, this experiment should be carried out in a darkened room.
 Connect up the circuit shown.
 Place a 60 W lamp facing the LDR at a distance of 50 cm, and adjust R4 until the
voltmeter reads zero.
 Move the lamp closer, 5 cm at a time, and record the output voltage readings.
 Plot a graph of output voltage against distance.
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Activities
PROBLEMS
Revision of Circuits
1.
If a current of 40 mA passes through a lamp for 16 s, how much charge has passed any
point in the circuit?
2.
A lightening flash lasted for 1 ms. If 5 C of charge was transferred during this time,
what was the current?
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.
What is the p.d. across a 2 k resistor if there is a current of 3 mA flowing through it?
5.
Find the readings on the meters in the following circuits.
10 V
(a)
A
20 
70 
10 
(b)
12 V
(c)
A
6V
15 
6.
A
3
9
5
V
V
Find the unknown values of the following resistors.
(a)
(b)
40 V
I=2A
5
40 V
A
R=?
10 
A
4
R=?
10 
20 V
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Answers
7.
Find the total resistance of the following combinations.
(a)
(b)
20 
10 
10 
10 
10 
10 
10 
(c)
(d)
5
10 
10 
4
1
20 
25 
10 
5
20 
(e)
(f)
10 
8.
5
10 
5
6
10 
12 
25 
3
If the ammeter reads 2 mA,
find the voltmeter reading.
8
A
3 k
5 k
V
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Answers
9.
Calculate the power in each of the following cases.
(a)
A 12 V accumulator delivering 5 A.
(b)
A 60 Ω heater with a 140 V supply.
(c)
A 5 A current in a 20  heater coil.
10.
An electric kettle has a resistance of 30 .
(a)
What current will flow when it is connected to a 230 V supply?
(b)
Find the power rating of the kettle.
11.
A 15 V supply produces a current of 2 A for 6 minutes. How much energy is supplied
in this time?
6V
A
Find the readings on the
ammeters and the voltmeter.
12.
6
13.
Assuming each of the four cells
cell to be identical, find:
(a) the reading on the ammeter
(b) the current through
the 20  resistor
(c) the voltage across
the 2  resistor.
6
V
6
A
5
2
20 
6V
14.
A coil has a current of 50 mA flowing through it when the applied voltage is 12 V.
Find the resistance of the coil.
15.
Write down the rules which connect the (a) potential differences and (b) the currents in
series and parallel circuits.
16.
Draw the symbol for a fuse, diode, capacitor, variable resistor, battery and a d.c. power
supply.
17.
What is the name given to the circuit
opposite.
Write down the relationship between
V1, V2, R1 and R2.
18.
10 V
R1
V1
R2
V2
Find the values of V1 and V2 of the circuit in question 17 if:
(a) R1 = 1 kΩ
R2 = 49 kΩ
(b) R1 = 5 kΩ
R2 = 15 kΩ.
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Answers
19.
Explain what would happen to the
readings on V1 and V2 if light was
shone onto the L.D.R.
R1
10 V
V2
Suppose the L.D.R. was replaced with
a thermistor which was then heated.
Explain the effect on the readings.
20.
V1
(a) What would be the polarity of A and B when connected to a 5 V supply, so
that the LED would light?
A
R
B
(b) What is the purpose of R in the circuit shown above?
(c) If the L.E.D. rating is 200 mA at 1.5 V, find the value of R.
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Answers
ELECTRICY AND ELECTRONICS PROBLEMS
Electric fields and resistors in circuits
1.
Draw the electric field pattern for the following charges:
(a)
2.
(b)
(c)
Describe the motion of the small test charges in each of the following fields.
(a)
(b)
+ test charge
+ test charge
+
(c)
+
+
+
+
-
-Q
-
(d)
+
+
+
+
+Q
-
3.
An electron volt is a unit of energy. It represents the change in potential energy of an
electron which moves through a potential difference of 1 volt. If the charge on an
electron is 1.6x10-19 C, what is the equivalent energy in joules?
4.
Mass of an electron = 9.1 10-31 kg
Charge on an electron = 1.6  10-19 C
The electron shown opposite is accelerated
across a p.d. of 500 V.
(a)
How much electrical work is done?
(b)
How much kinetic energy has it gained?
(c)
What is its final speed?
5.
+
+
+
+
+500 V
-e
-
Electrons are ‘fired’ from an electron gun at a screen. The p.d. across the gun is
2000 V. After leaving the positive plate the electrons travel at a constant speed to the
screen. Assuming the apparatus is in a vacuum, at what speed will the electrons hit
the screen?
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Answers
6.
7.
What would be the increase in speed of an electron accelerated from rest by a p.d. of
400 V?
An ×-ray tube is operated at 25 kV and draws a current of 3 mA.
(a)
Calculate
(i) the kinetic energy of each electron as it hits the target
(ii) the velocity of impact of the electron as it hits the target
(iii) the number of electrons hitting the target each second.
(mass of electron = 9.1 × 10-31 kg charge on electron = 1.6 × 10-19 C)
(b)
What happens to the kinetic energy of the electrons?
8.
Sketch the paths which
(a) an a-particle,
(b) a b-particle,
and
(c) a neutron,
would follow if each particle entered the given electric fields with the same velocity.
(Students only studying this unit should ask for information on these particles).
9.
State what is meant by
10.
Prove the expressions for the total resistance of resistors in (a) a series and (b) a
parallel circuit.
11.
In the circuit below:
(a) what is the total resistance of the circuit
(b) what is the resistance between X and Y
(c) find the readings on the ammeters
(d) calculate the p.d. between X and Y
(e) what power is supplied by the battery ?
(a) the e.m.f. of a cell
(b) the p.d. between 2 points in the circuit.
3
X
A1
12 V
A2
4
12 
Y
12.
The circuit opposite uses the 230 V
alternating mains supply. Find the
current flowing in each resistor when:
(a) switch S is open
(b) switch S is closed.
12 
230 V
Physics: Electricity and Electronics (H) – Student Material
S
8
24 
60
Answers
13.
An electric cooker has two settings, high and low. It takes 1 A at the low setting and
3 A at the high setting.
230 V
14.
Find the resistance of R1 and R2.
What is the power consumption at each setting ?
(a)
Find the value of the series resistor which
would allow the bulb to operate at its
normal rating.
Calculate the power dissipated in the resistor.
In the circuit below, r represents the internal resistance of the cell and R represents the
r
external resistance of the circuit.
When S is open, the voltmeter reads 2.0 V.
When S is closed, it reads 1.6 V and the
V
ammeter reads 0.8 A.
(a)
(b)
(c)
(d)
16.
R2
(a)
(b)
(b)
15.
R1
What is the e.m.f. of the cell ?
What is the terminal potential difference
when S is closed?
S
Calculate the values of r and R.
If R was halved in value, calculate the new readings on
the ammeter and voltmeter.
R
A
The cell in the diagram has an e.m.f. of 5 V. The current through the lamp is 0.2 A
and the voltmeter reads 3 V. Calculate the internal resistance of the cell.
r
V
A
17.
A cell of e.m.f. 4 V is connected to a load resistor of 15 W. If 0.2 A flows round the
circuit, what must be the internal resistance of the circuit?
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Answers
18.
A signal generator has an e.m.f. of 8 V and internal resistance of 4 . A load resistor
is connected to its terminals and draws a current of 0.5 A. Calculate the load
resistance.
19.
(a)
What will be the terminal p.d. across the cell in the circuit below.
(b)
(c)
Will the current increase or decrease as R is increased?
Will the terminal p.d. then increase or decrease ? Explain your answer.
20.
A cell with e.m.f. 1.5 V and internal resistance 2  is connected to a 3 Ω resistor.
What is the current?
21.
A pupil is given a voltmeter and a torch battery. When he connects the voltmeter
across the terminals of the battery it registers 4.5 V, but when he connects the battery
across a 6  resistor, the voltmeter reading decreases to 3.0 V.
(a)
Calculate the internal resistance of the battery.
(b)
What value of resistor would have to be connected across the battery to reduce
the voltage reading to 2.5 V.
22.
In the circuit shown, the cell has an e.m.f. of
6.0 V and internal resistance of 1 .
When the switch is closed, the reading on the
ammeter is 2 A. What is the corresponding
reading on the voltmeter ?
23.
In order to find the internal resistance of a cell, the following sets of results were
taken.
Voltage (V)
1.02
0.94
0.85
0.78
0.69
0.60
Current (A)
0.02
0.04
0.06
0.08
0.10
0.12
(a)
(b)
(c)
(d)
Draw the circuit diagram used.
Plot a graph of these results and from it determine
(i) the e.m.f.
(ii) the internal resistance of the cell.
Use the e.m.f. from part (b) to calculate the lost volts for each set of readings
and hence calculate 6 values for the internal resistance.
Calculate the mean value of internal resistance and the approximate random
uncertainty.
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Answers
24.
The voltage across a cell is varied and the corresponding current noted. The results are
shown in the table below.
Voltage (V)
5.5
5.6
5.7
5.8
5.9
Current (A)
5
4
3
2
1
Plot a graph of V against I.
(a)
What is the open circuit p.d?
(b)
Calculate the internal resistance.
(c)
Calculate the short circuit current.
(d)
A lamp of resistance 1.5  is connected across the terminals of this supply.
Calculate
(i)
the terminal p.d.
and
(ii)
the power delivered to the lamp.
25.
Calculate the p.d. across R2 in each case.
+5 V
(a)
+5 V
(b)
2 k
R1
4 k
8 k
R2
1 k
+5 V
(c)
500 
R2
750 
0V
t
R2
0V
0V
26. Calculate the p.d. across AB (voltmeter reading) in each case.
(a)
(b)
+5 V
+12 V
0V
3 k
A
V
3 k
3 k
9 k
B
(c)
+10 V
5 k
A
10 k
2 k
8 k
V
0V
27. (a) Calculate the reading on the voltmeter.
3 k
A
B
2 k
V
6 k
B
4 k
0V
0V
(b) What alteration could be made to
balance the bridge circuit ?
+9 V
6 k A 9 k
V
3 k B 6 k
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Answers
28.
Three pupils are asked to construct balanced Wheatstone bridges.
Their attempts are shown.
+1.5 V
+1.5 V
+1.5 V
5  A 10 
9  A 12 
7  A 14 
G
G
G
10  B 5 
Pupil A
12  B 16 
Pupil B
10  B 20 
Pupil C
One of the circuits gives a balanced Wheatstone bridge, one gives an off - balance
Wheatstone bridge and one is not a Wheatstone bridge.
(a)
(b)
(c)
29.
Identify each circuit.
How would you test that balance has been obtained ?
In the off – balance Wheatstone bridge ;
(i) calculate the potential difference across the galvanometer.
(ii) in which direction will electron current flow through the galvanometer.
Calculate the value of the unknown resistor × in each case.
2V
120 
2V
x
120 
G
G
9
25 k
15 k
4 k
G
10 k
12 k
2V
3.6 k
x
x
+1.5 V
30.
The circuit shown opposite is balanced.
(a)
What is the value of resistance ×?
(b)
Will the bridge be unbalanced if
(i) a 5  resistor is inserted next to the
10  resistor
(ii) a 3 V supply is used.
(c)
What is the function of resistor R and what is the
disadvantage of using it as shown ?
Physics: Electricity and Electronics (H) – Student Material
10 
20 
R
G
5
X
64
+1.5 V
31.
The following Wheatstone bridge circuit
is used to monitor the mechanical strain
on a girder in an oil rig.
(a)
(b)
32.
Answers
Strain gauges
Explain how the circuit can
G
be used to monitor the strain.
Sketch the graph of current
R4
through the galvanometer against the strain. R3
An automotive electrician needed to accurately measure the resistance of a resistor.
She set up a circuit using an analogue milliammeter and a digital voltmeter.
The two meter readings were:
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
(k)
(l)
(m)
What are the readings?
What is the nominal resistance calculated from these readings?
Which reading is likely to cause the greatest uncertainty?
What is the smallest division on the milliammeter?
What is the absolute uncertainty on the milliammeter?
What is the absolute uncertainty on the voltmeter?
What is the percentage uncertainty on the milliammeter?
What is the percentage uncertainty on the voltmeter?
Which is the greatest percentage uncertainty?
What is the percentage uncertainty in the resistance?
What is the absolute uncertainty in the resistance?
Express the final result as “(resistance ± uncertainty)Ω”
Round both the result and the uncertainty to the relevant number of significant
figures or decimal places.
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Answers
Alternating Current and Voltage
1.
(a)
(b)
What is the peak voltage of the 230 V mains supply?
The frequency of the mains supply is 50 Hz. How many times does the
voltage fall to zero in one second.
2.
The circuit below is used to compare the a.c. and d.c. supplies when the lamp is at the
same brightness with each supply. The variable resistor is used to adjust the
brightness of the lamp.
A B
(a)
(b)
(c)
Explain how the brightness of the lamp is changed using the variable resistor.
What additional apparatus would you use to ensure the brightness of the lamp
was the same for each supply?
In the oscilloscope traces shown below diagram 1 shows the voltage across the
lamp when the switch is in position B and diagram 2 shows the voltage when
the switch is in position A.
Y Gain set to 1 V cm-1
From the oscilloscope traces, how is the root mean square voltage numerically related
to the peak voltage.
(d)
Redraw diagrams 1 and 2 to show what would happen to the traces if the time
base was switched on.
3.
The root mean square voltage produced by a low voltage power supply is 10 V r.m.s.
(a)
Calculate the peak voltage of the supply.
(b)
If the supply was connected to an oscilloscope, Y-gain set to 5 V cm-1 with the
time base switch off, describe what you would see on the screen.
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Answers
4.
(a)
(b)
5.
The following trace appears on an oscilloscope screen when the time base is set at
2.5 ms cm-1.
(a)
(b)
6.
A transformer has a peak output voltage of 12 V. What is the r.m.s. output
voltage?
A vertical line 6 cm long appears on an oscilloscope screen when the Y gain is
set to 20 V cm-1. Calculate:
(i)
the peak voltage of the input
(ii)
the r.m.s. voltage of the input.
What is the frequency of the input including the uncertainty to the nearest Hz?
Sketch what you would see on the screen if the time base was changed to
(i)
5 ms cm-1
(ii)
1.25 ms cm-1
An a.c. input of frequency 20 Hz is connected to an oscilloscope with time base set at
0.01 s cm-1. What would be the wavelength of the waves appearing on the screen?
Capacitance
7.
A 50 µF capacitor is charged till the voltage across its plates reaches 100 V.
(a)
(b)
How much charge has been transferred from one plate to the other?
If it was discharged in 4 milliseconds, what would be the average current?
8.
A capacitor holds a charge of 3 × 10-4 C when it is charged to 600 V.
What is the value of the capacitor?
9.
A 30 µF capacitor holds a charge of 12 × 10-4 C.
(a)
What is the voltage that it is charged to?
(b)
If the tolerance of the capacitor is ± 5 µF, express this uncertainty as a
percentage.
(c)
What is the greatest voltage which could occur across the plates of the
capacitor?
10.
A 15 µF capacitor is charged from a 1.5 V battery. What charge will be stored on the
plates?
11.
(a)
(b)
A capacitor has a voltage of 12 V across its plates and stores a charge of
1.2 × 10-5 C. Calculate its capacitance.
A 0.1 µF capacitor is connected to a 8 volt direct supply. How much charge
will it store?
Physics: Electricity and Electronics (H) – Student Material
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Answers
12.
Using the circuit below a capacitor is charged at a constant charging current of
2.0x10-5 A.
The time taken to charge the capacitor is 30 s and during this time the voltage across
the capacitor rises from 0 V to 12 V.
What is the capacitance of the capacitor?
13.
A 100 µF capacitor is charged from a 20 V supply.
(a)
How much charge is stored?
(b)
How much energy is stored in the capacitor?
14.
A 30 µF capacitor stores 6 × 10-3 C of charge. How much energy is stored in the
capacitor?
15.
The circuit below is used to investigate the charging of a capacitor.
A
10 k
12 V
2000 µF
(a)
(b)
(c)
(d)
What is the response of the ammeter when switch S is closed?
How can you tell when the capacitor is fully charged?
What would be a suitable range for the ammeter?
If the 10 k resistor is replaced by a larger resistor, what will be the effect on
the maximum voltage across the capacitor?
Physics: Electricity and Electronics (H) – Student Material
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Answers
16.
In the circuit below the neon lamp flashes at regular intervals. The neon lamp
requires a potential difference of 100 V across it before it will conduct and flash. It
continues to glow until the potential difference across it drops to 80 V. While lit, its
resistance is very small compared with R.
R
120 V dc
(i)
(ii)
17.
C
Explain why the neon bulb flashes.
Suggest two methods of decreasing the flash rate.
In the circuit below the capacitor C is charged with a steady current of 1mA by
carefully adjusting the variable resistor R.
A
9 V dc
C
V
The voltmeter reading is taken every 10 seconds. The results are shown in the table.
(a)
(b)
(c)
Time
0
10
20
30
40
p.d.(V)
0
1.9
4
6.2
8.1
Plot a graph of charge against voltage for the capacitor and hence find its
capacitance (use graph paper).
Calculate the capacitance for each of the readings (ignoring readings for t = 0).
Calculate the mean capacitance and the approximate random uncertainty in the
mean to two decimal places.
Physics: Electricity and Electronics (H) – Student Material
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Answers
18.
The circuit below is used to charge and discharge the capacitor
(a)
(b)
(c)
What position should the switch be set (i) to charge and (ii) to discharge the
capacitor?
Draw graphs of VR against time for the capacitor charging and discharging,
and of VC against time for the capacitor charging and discharging.
If the capacitor has capacitance of 4.0 µF and the resistor has resistance of 2.5
M calculate:
(i)
the maximum charging current in the circuit above
(ii)
the maximum charge stored by the capacitor when fully charged in the
above circuit.
19.
3 M
3V
3 µF
(a)
(b)
20.
For the above circuit draw graphs of
(i) VC against time during charging and
(ii) VA against time during charging.
Calculate the final voltage across the capacitor and the final charged stored by
it.
For each of the circuits below state what happens to the current flowing when the
frequency is (i) increased and (ii) decreased.
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21.
In the circuit below the signal generator is set at 6.0 Vr.m.s., 1000 Hz.
The lamp operates normally.
(a)
(b)
22.
Explain why the lamp can operate normally when the plates of the capacitor
are separated by an insulator.
What happens to the brightness of the lamp when the frequency of the signal
generator is increased. Why does this happen?
For each of the following circuits sketch a graph of current against frequency.
A
A
Signal
generator
Diagram A
Diagram B
23.
A
B
Signal
generator
The supply frequency to the above circuit is increased from a very low frequency,
while the supply voltage remains constant.
What will happen to the brightness of lamp A and B?
Physics: Electricity and Electronics (H) – Student Material
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Answers
Analogue Electronics
1.
Calculate the values of V1 and V2 in the circuit shown for the following situations.
(a)
(b)
(c)
2.
VS = 12 V
VS = 6 V
VS = 10 V
R1 = 40 k R2 = 20 k
R1 = 150 k R2 = 30 k
R1 = 3 k
R2 = 5 k
V2
R1
V1
Vs
Calculate the values of R1 in the following circuits.
(a)
(b)
(c)
1 k
5 k
4V
1 k
2V
R1
12 V
V1
R1
V1
3V
3.
R1
V1
1.5 V
8.5 V
Calculate the values of V1 and V2 in the circuits shown.
(a)
(b)
(c)
+1.5 V
+1.5 V
+5 V
15 k
V2
8.2 k
V2
3.3 k
V2
3 k
V1
4.7 k
V1
2.2 k
V1
0V
0V
4.
R2
0V
Explain what happens to the value of V1 in each of the following situations (a) brightness increasing
(b) temperature increasing
+1.5 V
+1.5 V
V1
0V
(c) brightness decreasing.
+1.5 V
t
0V
Physics: Electricity and Electronics (H) – Student Material
V1
V1
0V
72
Answers
5.
The thermistor shown has a resistance of 31 k at 30 °C and 35 k at 25 °C.
(a)
If the variable resistor is set at 5 k, calculate the input voltage to the
transistor in each case.
Hence, explain how the circuit shown works.
Explain the purpose of the variable resistor.
Suggest a possible use for the circuit.
Why is the relay switch necessary?
(b)
(c)
(d)
(e)
6.
Draw a circuit diagram similar to the circuit of Question 5 that would switch on a
mains voltage lamp when the ambient light level drops below a certain level.
7.
The circuits shown in questions 5 & 6 could use a different type of transistor called an
n-channel enhancement MOSFET. Draw the symbol for this transistor and label each
terminal.
8.
The following signals are fed into an inverting amplifier with a gain of 5.
Draw the expected output trace.
(a)
(b)
(c)
(d)
(e)
(f)
Physics: Electricity and Electronics (H) – Student Material
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Answers
9.
(a)
(b)
(c)
In what mode is the op-amp being used in the circuit below.
State the relationship between V1, R1, Rf, and V0.
Find the unknown values in the table shown. All calculations should be set out
clearly.
Rf
R1
-
V1
10.
+
(a)
(b)
11.
R1(k) Rf(k) V1(V) V0(V)
(i) 10
(ii)
5
(iii)
(iv) 1000
V0
Which of the following values of Rf
will produce saturation of the output
voltage?
(i) 15 k (ii) 25 k (iii) 35 k
What is the approximate value of the
saturation voltage?
100
8
5.4
0.5
-1.4
0.4
1.5
-1.8
-0.6
Rf
+1.5V
4 k
+
0V
+12 V
-12 V
V0
Calculate the value of VO for the following situations.
+5 V
RA
RB
0V
Rf
R1
+
+15 V
-15 V
RA(k) RB(k) R1(k) Rf(k)
(a)
(b)
(c)
V0
(d)
Physics: Electricity and Electronics (H) – Student Material
1
1.2
5
5
4
4.7
2
1
10
1.5
8.2
1
20
10
27
4
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Answers
12.
Describe 3 ways to increase the
magnitude of the output voltage in
the circuit shown.
+5 V
Rf
R1
V1
+
V0
0V
13.
(a)
(b)
14.
(a)
(b)
Calculate the value of the
output voltage in the circuit
shown.
If the value of Rf is doubled,
what would be the output
voltage?
80 k
8 k
-0.8 V
V1
+
0V
An operational amplifier can be
connected in different modes.
State the operating mode of the
amplifier shown.
Calculate the gain of the circuit.
(a)
(b)
(c)
Calculate the gain of the circuit shown.
Calculate the output voltage V0
for an input voltage of 15 mV.
What happens to the gain of the circuit
if the feedback resistor is reduced.
Physics: Electricity and Electronics (H) – Student Material
+15 V
V0
-15 V
100 k
10 k
V1
1 k
V1
+
The graph shows how the voltage applied
Voltage (V)
to the input of the circuit varies with time.
0.5
(c)
Using square ruled paper, draw a graph
showing how the output voltage varies with
0
time.
10
(d)
Describe the output signal if the input voltage
is increased to 2 V.
(e)
Both resistors have an uncertainty of ± 0.01 kΩ.
Which resistor will introduce the greatest
uncertainty into the gain calculation?
(f)
What is the uncertainty in the gain?
15.
-
20
+9 V
V0
-9 V
40
30
Time (ms)
100 k
+
+15 V
-15 V
V0
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Answers
16.
The circuit below shows an operational amplifier in the differential mode.
(a)
(b)
(c)
17.
18.
Calculate the unknown values in
the table for a differential
amplifier circuit.
(a)
(b)
(c)
(d)
Rf
What is the function of this circuit ?
State the relationship that applies to this
circuit, giving the condition for this to
hold.
Find the value of VO when
V1
Rf = 10 M R1 = 10 k
R2 = 10 k R3 = 10 M
V1 = 480 mV V2 = 500 mV.
R1
+15 V
-
R2
+
V2
-15 V
V0
R3
V1 (V) V2 (V) Rf = R3 R1 = R2 V0 (V)
(a)
(b)
(c)
(d)
(e)
2.5
1.5
0.4
6.0
4.5
3.0
1.3
0.4
7.2
(k)
(k)
100
30
10
10
5
100
3
5
40
6.6
-8.0
In the circuit below, what value should be chosen for R for it to operate as a
differential amplifier?
Determine the reading on the voltmeter with the slider at position A.
If the contact is moved to a position midway between A and B calculate the
voltmeter reading.
Where should the contact be to produce an output voltage of
(i) -1.25 V
(ii) 0 V.
+1.5 V
100 k
10 k
40 k
A
5 k
16 k
+
V
R
B
0V
19.
The circuit opposite can be
shown to monitor the level
of brightness.
(a)
Explain how it operates,
mentioning why the
voltmeter reading changes.
(b)
What is the purpose of the
variable resistor?
+9 V
LDR
10 k
10 k
V1
10
k
Physics: Electricity and Electronics (H) – Student Material
100 k
0-10 k
10
k
V2
+
100 k
V
0V
76
Answers
20.
(a)
(b)
(c)
Calculate the gain of the circuit shown.
At what value of (i) V0 (ii) V1 will the transistor switch on ?
Write down a rule which will allow you to predict whether a transistor is on or
off. Remember the polarity of the base-emitter voltage is important.
-Vs
+6 V
1 M
V1
V
100
k
-
c
b
+
V0
e
V
0V
21.
State if the transistors below are switched on or off.
(a)
(b)
(c)
(d)
-5 V
+5 V
-5 V
+5 V
0V
0V
0V
0V
(e)
(f)
(g)
(h)
-5 V
+5 V
-5 V
-5 V
-3 V
+9 V
Physics: Electricity and Electronics (H) – Student Material
-2 V
-9 V
77
Answers
NUMERICAL ANSWERS
Revision of circuits
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
18.
20
Q = 0.64 C
Q = 5 × 103 A
t = 2 × 104 s.
V=6V
(a) 0.1 A
(a) 5 
(a) 15 
(f) 14.7 .
3.75 × 10-3 V
(a) 60 W
(a) 7.67 A,
9000 J
0.67 A, 4 V
(a) 0.67 A
240 
(a) V1 = 0.2 V
(c) 17.5 
(b) 0.5 A, 4.5 V
(c) 2 A, 10 V.
(b) 6 
(b) 25 
(c) 24.2 
(d) 13.3 
(b) 327 W
(b) 1763 W
(c) 500 W
(b) I (20 ) = 0.13 A, I (5 ) = 0.54 A
V2 = 9.8 V
(e) 22.9 
(c) 13.4 V
(b) V1 = 2.5 V V2 = 7.5 V
Electric fields and resistors in circuits
3.
1.6 × 10-19 J
4.
(a) 8 × 10-17 J (b) 8 × 10-17 (c) v = 1.33 × 106 m s-1
5.
v = 2.65 × 107 m s-1.
6.
v = 1.2 × 107 m s-1
7.
(a) (i) 4 × 10-15 J
(ii) 9.4 × 107 m s-1 (iii) 1.875 × 1016 electrons
11.
(a) 6  (b) 3  (c) 1.5 A (ammeter 2), 2 A (ammeter 1) (d) 6 V (e) 24 W
12.
(a) 11.5 A (b) 12.8 A (12 ), 9.6 A (8 ), 3.2 A (24 )
13.
(a) R1 = 240  R2 = 120  (b) Low - 240 W High - 720 W
14.
(a) 4  (b) 36 W
15.
(a) 2 V (b) 1.6 V (c) r = 0.5 Ω, R = 2 Ω, 1.3 A, 1.3 V
17.
10 
18.
5
19.
12 
20.
19.(a) 1.3 V
21.
0.3 A
22.
3 
23.
4V
24.
23.
(b) 1.1 V (Intercept), 4.2  (-gradient)
(c) 4.1 ± 0.02 
25.
24.
(a) 6 V
(b) 0.1 
(c) 60 A
(d)(i) 5.63 V (ii) 21.1 W
25.
(a) 4 V
(b) 1 V
(c) 3 V
26.
(a) 3 V
(b) -0.8 V
(c) 0 V
27.
(a) 0.6 V
28.
(c) (i) 0.5 V (ii) from B to A.
29.
(a) 4  (b) 45  (c) 9 
30.
(a) 10 
32.
(a,e)1.76±0.01 mA, (b,f) 1.3±0.1V,
(m) 740 ± 60 
Physics: Electricity and Electronics (H) – Student Material
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Answers
NUMERICAL ANSWERS
Alternating current and voltage
1.
(a)
Vr.m.s. = 325 V
(b)
100 times per second
Vr.m.s. = 0.71 peak
3
(a)
Vpeak = 14 V
(b)
A vertical line of height 5.7 cm is seen
4
(a)
Vr.m.s. = 8.4 cm
(b)
(i) Vpeak = 60 V
(ii) Vr.m.s. = 42 V
5.
(a)
frequency = 100 ±2 Hz
6.
wavelength = 5 cm
Capacitance
7.
(a)
Q = 0.005 C
(b)
I = 1.25 A
8.
C = 50 µF
9.
(a)
V = 40 V, (b) 0.167 %, (c) 46.7 V
10
Q = 22 µF
11.
(a)
C = 1 µF
(b)
Q = 0.8 µC
12.
C = 50 µF
13.
(a)
0.002 C
(b)
E = 0.02 J
14.
E = 0.6 J
15.
(a)
ammeter reading starts at 1.2 mA and falls steadily to zero
(b)
Ammeter reads zero
(c)
Range 0 to 2 mA
17.
(a)
C= 0.005 F
18.
(a)
(i) position 1
(ii) position 2
(c)
(i) I = 40 µA
(ii) Q = 400µC
19.
(b)
V = 3 V, Q = 9 µC
Physics: Electricity and Electronics (H) – Student Material
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Answers
NUMERICAL ANSWERS
Analogue Electronics
1.
(a) V1 = 8 V
V2 = 4 V
(b) V1 = 5 V
V2 = 1 V
(c) V1 = 3.75 V
V2 = 6.25 V
2.
(a) R1 = 15 k
(b) R1 = 3 k (c) R1 = 2.43 k
3.
(a) V1 = 0.25 V
V2 = 1.25 V
(b) V1 = 0.55 V
V2 = 0.95 V
(c) V1 = 2 V
V2 = 3 V
4.
(a) V1 decreases (b) V1 decreases (c) V1 increases
5.
(a) When T = 25 °C, V = 0.625 V. T = 30 °C, V = 0.7 V.
6.
8.
9.
(a) Inverting mode
(b) V0 = -(Rf / R1) V1
(c) (i) V0 = -5 V (ii) V0 = 2.24 V (iii) R1 = 1.2 k (iv) Rf = 0.4 M
10.
(a) Rf = 35 k (b) 10 V
11.
(a) V0 = -8 V (b) V0 = -13 V (satuaration) (c) V0 = -4.7 V (d) V0 = -3.32 V
12.
Increase Rf, decrease R1, increase V1.
13.
(a) V0 = - 8 V (b) V0 = -13 V (saturation).
14.
(a) Inverting mode
(b) Gain = 10 (f) ±0.01
15.
(a) Gain = 100
(b) V0 = -1.5 V
(c) Gain will be reduced.
16.
(a) Amplifies the difference between V1 and V2.
(b) V0 = -(Rf / R1) (V2-V1)
(c) V0 = 20 V (saturates to 13 V).
17.
(a) V0 = 5 V (b)V0 = -1.2 V (c) V0 = 0 V (d) Rf = 16.5 k (e) V2 = 3.5 V.
18.
(a) 40 k (b) V0 = 2.5 V (c) V0 = 0.625 V
(d) (i) Position B (ii) 1/3 distance from bottom.
19.
20.
(a) Gain = 0.1 (b) (i) 0.7 V approx. (ii) V1 = 7 V
21.
(a) Off
(b) On
(c) On
(d) Off
(e) Off
(f) Off
(g) On
(h) Off
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