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Alternating Current and Voltage
Electricity & Electronics 5:
Alternating Current and Voltage
AIM
This unit looks at several aspects of alternating current and voltage including measurement of
frequency and something called root-mean-square values.
Some of the mathematics is quite tricky, but you are not required to reproduce it in the final
exam. However, you may need to read through it several times – good luck!
OBJECTIVES
On completing this unit you should be able to:
• describe how to measure frequency using an oscilloscope.
• state the relationship between peak and rms values for a sinusoidally varying voltage
and current.
• carry out calculations involving peak and rms values of voltage and current.
• state the relationship between current and frequency in a resistive circuit.
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Electricity and Electronics
Alternating Current and Voltage
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.
A
A
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.
V
r.m
.s
.= 1V
p
e
a
k
2
a
n
d
1
Ir.m
.s
.= Ip
e
a
k
2
Note: a moving coil a.c. meter is calibrated to give r.m.s. values.
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Electricity and Electronics
Alternating Current and Voltage
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
Circuit2
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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Alternating Current and Voltage
Graphical method to derive relationship between peak and rms values of alternating
current
THIS IS NOT EXAMINABLE - IT IS HERE FOR INFORMATION ONLY!
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.
I2
The average value of I2 is Peak
2
An identical heating effect (power output) for a d.c. supply = I2r.m.s R
[since I (d.c.) = Ir.m.s.]
I2Peak
R
Average power output for a.c. = 2
I2 r.m.s R =
I2 Peak R
2
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hence
I2r.m.s. =
-4-
I2 Peak
2
giving Ir.m.s. =
I Peak
Electricity and Electronics
Alternating Current and Voltage
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 =
y in p u t
=
0 .0 0 5
ti m e b a se (s c m-1 )
1
time to produce one wave
1
0.02
= 50 Hz
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
Frequency response of resistor
The following circuit is used to investigate the relationship between current and frequency in
a resistive circuit.
A
C u rren t
(A )
S ig n a l
gen erato r
(co n s tan t
e .m .f.)
0
freq u en cy (H z )
The results show that the current flowing through a resistor is independent of the frequency of
the supply.
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Alternating Current and Voltage
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.
3 .6 c m
1 0 .2 c m
D ia g r a m 1
D ia g r a m 2
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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Alternating Current and Voltage
4.
(a)
(b)
5.
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.
The following trace appears on an oscilloscope screen when the time base is set at
2.5 ms cm-1.
6 ± 0 .1 c m
(a)
(b)
6.
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?
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Electricity and Electronics