Physics Factsheet
www.curriculum-press.co.uk
Number 70
AC Electricity
This Factsheet explains a.c. ("alternating current") electricity . It covers
what an alternating current is and how to give a value to alternating current
and alternating voltage by considering the electrical power delivered by this
type of electricity.
Representing the size of voltage supplied by sinusoidal a.c. electricity
A voltmeter used to measure direct current cannot be used in a circuit
supplied with alternating voltage. The graph above shows that the size of
the voltage is changing and a d.c. voltmeter would not be able to give a single
reading for this alternating voltage. Alternating voltages are best measured
with an oscilloscope. The oscilloscope is able to display a graph of voltage
against time for the alternating voltage, allowing all details of the changing
voltage to be seen.
Electrical current and the movement of charged particles
A material that will allow an electric current to flow must have charged
particles that are free to move. In metallic conductors these charged particles
are electrons. Even when there is no power supply providing a potential
difference or voltage across a metal, the electrons move freely around the
metallic structure at very high speeds (several thousand metres per second).
However, no current flows around a metal without a power supply as the
electrons move in random directions - there is no net flow of electrons in
any particular direction.
Deciding what value to give to an alternating voltage is not that straight
forward as the voltage is constantly changing.
The maximum or peak voltage is often quoted. This value is exactly as the
name suggests and represents the maximum value that the voltage reaches
during a cycle.
When a direct voltage, such as a battery, is placed across a metallic
conductor the electrons that are free to move ‘drift’ in one direction. That
is to say that the electrons continue to move randomly in all directions at
very high speeds, but superimposed on to of this motion is a ‘drift velocity’
where the electrons are seen to slowly drift in one direction. This drift
velocity is quite small, perhaps only a few millimetres per second.
This value is not that useful when trying to decide how bright a bulb will be
or what heat will be given out by a resistor as the peak voltage is only across
the circuit for an instant during each cycle.
When an alternating voltage is placed across a metallic conductor the drift
velocity changes direction at the same frequency as the alternating voltage,
in the case of electricity mains supply, this change in direction happens at
a frequency of 50 Hz.
Consider the simple circuit below where an alternating voltage is connected
across a resistor.
alternating supply voltage
A more useful value to describe the size of alternating current is arrived at
by considering the electrical power that is supplied to a component with
an alternating voltage across it.
Flow of charge in direct currents and alternating currents
In metals, the charged particles responsible for current are electrons,
which move randomly at high speeds even with no applied voltage .
• A direct current is produced when the electrons all drift in one
direction whilst continuing with their random motion.
• An alternating current is produced when the drift velocity of the
electrons constantly changes direction.
resistor
The electrical power supplied to the resistor can be determined using the
expression:
V2
P=
R
The voltage, V, in this equation does not stay constant and so we must use
the average value of V2 when making any calculations.
The effects of both direct current and alternating current are essentially the
same for many components, as both types of current can produce heat and
light when flowing through a metallic conductor such as a piece of wire or
a filament lamp.
The graphs below show how V varies and V2 varies for a sinusoidal alternating
voltage.
V0
The most common form of alternating voltage is a signal that varies
sinusoidally. This is a long word meaning that the voltage changes direction
in a smooth manner, in the same shape as sine curve, as shown in the graph
below.
time
−V0
voltage
2
V0
time
2
V0
average value of voltage2
2
time
The graph of current against time for a sinusoidal alternating power supply
would be exactly the same shape.
The average of V2 is equal to half of the maximum value - average V2 = ½(Vo)2
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Physics Factsheet
70 AC Electricity
The resistance of the resistor, R, should remain constant if it is an ohmic
conductor. This means that the average power delivered to the resistor is
given by:
½(V )2
P = Ro
Compare this with the power delivered by a direct voltage:
V2
P=
R
Sizes of alternating currents
Two methods of quoting the size of an alternating current are
generally used; the peak value, which is the maximum value of the
current, and the root mean square value, which is the peak value
divided by √2.
Exam Hint: A common error is, when asked to define the root-meansquare voltage or current, to quote its relationship with the peak value,
rather than making reference to the size of dc voltage (or current)
delivering the same power.
So if the two types of voltage supply the same amount of power then:
V2=½(Vo)2
Square rooting both sides:
V = √½ × Vo
Power supplied by an alternating power supply
The power supplied by an alternating power supply also varies and the
average power is quoted.
Which is the same as writing:
V
V= o
√2
The graph below shows how the power changes with time.
power
This represents the size of d.c. voltage that would supply the same amount
of power as the a.c. voltage with peak voltage Vo and is referred to as the root
mean square voltage, Vrms
Vrms =
Vo
√2
peak
power
V0 = Maximum value of an alternating voltage
Vrms = root mean square value of an alternating
voltage
time
Root mean square voltage is the equivalent size of d.c. voltage that would
be required to supply the same amount of power as the ac voltage.
The average power is half the peak power.
The average power can also be calculated using our root mean square values
for voltage and current with the same equations as for direct current.
Using voltmeters with alternating voltages
An oscilloscope is best used to measure an alternating voltage as it is
able to display the entire waveform as a graph of voltage against time.
P = Irms × Vrms = (Irms)2 × R =
(Vrms)2
R
P = average power delivered to the circuit (W)
Irms = root mean square current (A)
Vrms = root mean square voltage (V)
R = resistance of circuit (Ω)
Sizes of alternating voltages
Quoting the ‘size’ of an alternating voltage causes a problem as the
value of the alternating voltage changes with time. Two methods of
quoting the size of an alternating voltage are generally used; the peak
value, which is the maximum value of the voltage, and the root mean
square value, which is the peak value divided by √2.
Calculating the power delivered by alternating electricity
Always use root mean square values for voltage and current in any
equation that is used to calculate the average power delivered by an
alternating electricity supply.
Worked Example
The electricity mains supply is given at 240V rms. What is the peak voltage
supplied by the mains?
Answer
A simple substitution of values into our equation will give the answer.
V0 = √2 × Vrms = √2 × 240 = 339V
Consistency with Electrical Quantities
In any single calculation that uses values of current, voltage and power,
all of the quantities should be measured in the same way, either root mean
square values or peak values. The two types of measurement should
never be mixed and matched when carrying out circuit calculations
Representing the size of an alternating current
The same problem arises for representing the size of an alternating current
as happened for voltage. What value do you assign to current when it is
constantly changing?
Typical Exam Question
(a) Explain why it is not always useful to use the peak voltage of an
alternating supply to represent its size. (2)
Fortunately, the solution is the same as for alternating voltage. The maximum
value for current can be quoted, or peak current, as it is called. Alternatively
the root mean square value for current can also be quoted. The root mean
square current can be thought of as the equivalent direct current that could be
provided to deliver the same amount of electrical power as the alternating
current.
Irms =
average power =
½ × peak power
average
power
(b) Explain whether you would use peak or root-mean-square
current to choose the appropriate fuse for an appliance working
on an alternating supply (2)
Io
I0 = Maximum value of an alternating current (A)
√2 Irms = root mean square value of an alternating current (A)
Answer
(a) The voltage is only at its peak value instantaneously and is less than
that for the rest of the cycle
(b) Peak, because the fuse must accommodate the maximum current.
Root mean square current is the equivalent size of direct current that
would be required to supply the same amount of power as the alternating
current.
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70 AC Electricity
Physics Factsheet
A.C. Electricity and frequency
The frequency of an a.c. power supply is measured in the same way as the
frequency of any other measured quantity that repeats a set cycle. Frequency
is the number of cycles that are produced per second.
n
f=
t
Quantitative (Calculation) Test
1. Calculate the peak voltage of a sinusoidal alternating voltage with a root
mean square voltage of 4.5V.
[2]
2. Calculate the rms current of a sinusoidal a.c. with peak current of 2.5A.
[2]
f = frequency of alternating electricity (Hz)
n = number of cycles produced
t = time taken for cycles to be produced (s)
3. An sinusoidal, alternating voltage with a peak value of 12V is placed across
a 25Ω resistor. Calculate the average power delivered to the resistor.
[4]
The graph below shows a sinusoidal, alternating voltage that might be
displayed on an oscilloscope.
4. An alternating current (a.c.) power supply is connected across a 15 Ω
resistor to form a complete circuit. An oscilloscope is connected across
the resistor. The display on the oscilloscope screen is shown below.
Each one of the divisions on the time axis represents 0.02 seconds.
0V
0V
The oscilloscope settings are: Vertically, 5.0V per division.
Horizontally, 20 ms per division.
(a) What is the time period of the a.c. source?
(b) Calculate the frequency of the a.c. source.
(c) Determine the peak voltage of the a.c. source.
(d) Calculate the r.m.s. voltage.
(e) Calculate the r.m.s. current through the resistor.
(f) What is the average power delivered to the resistor?
The graph shows 4 full cycles of alternating voltage.
The four cycles are displayed over 10 horizontal divisions, giving a time of
10 × 0.02 = 0.2s
The frequency of the alternating voltage can be calculated from this display:
n 4
f= =
= 20Hz
t 0.2
Frequency can also be determined if the periodic time of the alternating
electricity is known. The periodic time is the time for a single oscillation
to be produced, and it is usually given the symbol T.
(1)
(2)
(2)
(2)
(2)
(2)
Typical Exam Question
An alternating current (a.c.) power supply is connected across a 25
Ω resistor to form a complete circuit. An oscilloscope is connected
across the resistor. The display on the oscilloscope screen is shown
below.
Consider the frequency equation once more:
n
f=
t
The oscilloscope settings are:
Vertically, 2.0V per division. Horizontally, 2.0 ms per division.
If we are observing just one cycle the n = 1 and the time taken is equal to
the periodic time, T.
1
f=
T
0V
Hence the frequency of an alternating supply and the periodic time are just
reciprocals of each other.
Frequency and periodic time
The frequency of an alternating electricity supply is the number of cycles
that are produced per second.
The periodic time of an alternating electricity supply is the time taken for
one cycle to be produced.
f = frequency (Hz)
1
f=
T = periodic time (s)
T
(a) Determine the peak voltage of the a.c. source. (2)
(b) Calculate the r.m.s. voltage. (2)
(c) Calculate the r.m.s. current through the resistor. (2)
(d) What is the average power delivered to the resistor? (2)
Answer
(a) To calculate the peak voltage the maximum value that the display
reaches must be determined and this is 3 divisions.
Vo= 3 divisions × 2 volts per division = 6.0 V
V
6
(b) Vrms= o =
= 4.2V
√2 √2
(c) Care must be taken to use the root mean square voltage in this
otherwise straightforward circuit calculation.
V
4.2
= 0.17A
Irms= rms =
R
25
(d) Again, care must be taken to use root mean square measurements in
the straightforward electrical power equation.
Average power = IrmsVrms = (0.17)(4.2) = 0.72W
Qualitative (Concept) Test
1. Describe the motion of free moving electrons in a metal conducting
(a) direct current
(b) alternating current
2. What instrument is used to display how an alternating voltage varies
with time? Why is this instrument preferred?
3. Define the root mean square voltage for an alternating supply.
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Physics Factsheet
70 AC Electricity
Exam Workshop
This is a typical poor student’s answer to an exam question. The
comments explain what is wrong with the answers and how they can
be improved. The examiner’s answer is given below.
Qualitative Test Answers
1. (a) Electrons move randomly at high speeds, with a superimposed
drift velocity (in the opposite direction to the conventional
current).
A sinusoidal, alternating voltage supply has a root mean square
voltage of 12V and a frequency of 250Hz.
(a) Calculate the time period of this supply.
(2)
1
1
T=
=
= 0.004s
f
250
(b) Electrons move randomly at high speeds, with a superimposed
drift velocity which changes direction at the same frequency as the
applied pd
2. A cathode ray oscilloscope is used rather than a dc voltmeter - the
voltmeter would not be able to give a single reading
This is a correct calculation but the student has only included one
significant figure in the answer. The answer should use the same
number of significant figures as is given in the question. In this case
that is two.
(b) Calculate the peak voltage of the supply.
V
12
Peak voltage = rms =
= 8.5V
√2
1.41
3. Root mean square voltage is the equivalent size of d.c. voltage that
would be required to supply the same amount of power as the ac
voltage.
(2)
Quantitative Test Answers
1. Vo = Vrms × √2 = 4.5 × √2 = 6.4V !
!
Io
2.5
2. Irms =
=
= 1.8A !
√2
√2 !
This equation has not been recalled correctly. A better student might
have checked that the final answer is reasonable, in this case that
means recognising that the peak voltage should be greater than the
root mean square voltage.
3. Vrms =
P=
(c) Sketch a graph of voltage on the vertical axis against time on the
horizontal axis for the alternating voltage. Include scales for
each of the axes.
(3)
8.5
0.12
0.08
0.20
t (s)
0.24
0.16
Vrms2
R
12
= 8.5V !
√2 !
8.52
= ! = 2.9W !
25
=
4. (a) T = 80 ms !
1
1!
(b) f =
=
= 1.25Hz !
T
0.8
!
!
(c) Vo = number of divisions × volts per division = 3 × 5 = 15V
V
15
(d) Vrms = o =
= 11V !
√2
√2 !
V (V)
0.04
Vo
√2
(e) Irms=
-8.5
Vrms
R
=
11
= 0.71A!
15 !
(f) P = Irms× Vrms = 0.71 × 11 = 7.5W!
!
The student has drawn a good smooth sinusoidal curve and has given
a correct scale for the peak voltage (the incorrect value from the last
part of the question would be given as an error carried forward). The
time period is incorrect. One full cycle of the wave should occur
in 0.004s and not half a cycle as the student has shown.
(d) The alternating voltage is connected across a 100Ω
Ω resistor.
Calculate the average power dissipated by the resistor.
(2)
V2 8.52
= 0.7225W
Power = =
R 100
The student has chosen to use the correct equation but has
substituted the wrong value for voltage. The average power
dissipated requires the root mean square voltage.
Examiner’s answers
1
1
(a) T = =
= 0.0040s !
f
250 !
(b) Vo = Vrms × √2 = 12 × 1.41 = 17V!
!
(c)
V (V)
Acknowledgements:
This Physics Factsheet was researched and written by Jason Slack.
The Curriculum Press, Bank House, 105 King Street, Wellington,
Shropshire, TF1 1NU
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ISSN 1351-5136
8.5
t (s)
0.04
0.08
0.12
-8.5
(d) P =
Vrms 2
R
=
122
= 1.4W !
100!
4