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Topic 2.4.1 – Root Mean Square – RMS Voltage.
Learning Objectives:
At the end of this topic you will be able to;
 apply the formula Vo  Vrms 2 for a sinusoidal AC voltage;
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Module ET2
Electronic Circuits and Components.
Root Mean Square (RMS) Voltage.
In topic 2.4 we are going to be investigating some different methods of
constructing simple power supplies that would enable projects to be run from
the mains supply.
Such a simple statement actually introduces a number of issues, which need
to be addressed:
Firstly the mains electricity is an alternating current supplied at 240V. You
might expect the graph to look like the following.
Voltage
240V
20
40
60
Time (ms)
-240V
In reality if you were to display the mains voltage on an oscilloscope you would
obtain a quite different and possibly surprising result as shown below:
339V
Voltage
240V
20
-240V
-339V
2
40
60
Time (ms)
Topic 2.4.1 – Root Mean Square – RMS Voltage.
The reason for this is that the value of 240V a.c. refers to a voltage called
the root mean square (or rms) voltage. The root mean square voltage of an
a.c. signal is the value of a sinusoidal a.c. voltage which would provide the
same heating effect in a resistor as the equivalent d.c. voltage. So in this
case a 240V a.c. voltage provides the same heating effect in a resistor as a
240V d.c. voltage.
For a sinusoidal voltage it is always worth remembering that the peak value of
the voltage will always be higher than the rms voltage. The calculation of the
rms voltage or peak value for a sinusoidal wave is quite straightforward as
shown by the following equations.
Vrms 
Vo
 0.707Vo
2
or
Vo  2  Vrms  1.414Vrms
In our dealings with power supplies in the exam we will have to calculate one
of these values, and sketch the graph of the resulting a.c. waveform, labelling
the value of the peak voltage. We will see more of the graphs in the next
section of these notes when we deal with the second issue with power
supplies which is that electronic circuits only work with d.c. We therefore
have to find a way to change a.c. into d.c. but more of that later. For now we
will just concentrate on calculating rms and peak values.
3
Module ET2
Electronic Circuits and Components.
Examples:
1.
A sinusoidal voltage source has a peak value of 14V, determine the rms
voltage of this supply.
In this case we need to find the rms voltage so we use the formula:
Vrms 
Vo 14
14


 9.90V
2
2 1.414
or
Vrms  0.707Vo  0.707  14  9.90V
2.
An a.c. power supply has an rms voltage of 6V. What is the peak voltage
from the power supply.
In this case we need to find the peak voltage so we use the formula:
Vo  2  Vrms
 1.414  6
 8.49V  8.5V
Now it’s time for you to have a go:
Student Exercise 1:
1.
A sinusoidal voltage source has a peak value of 20V, determine the rms
voltage of this supply.
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Topic 2.4.1 – Root Mean Square – RMS Voltage.
2.
An a.c. power supply has an rms voltage of 9V. What is the peak voltage
from the power supply.
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3.
An a.c. power supply has an rms voltage of 12V. What is the peak voltage
from the power supply.
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4.
A sinusoidal voltage source has a peak value of 31.1V, determine the rms
voltage of this supply.
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5.
An a.c. power supply has an rms voltage of 120V. What is the peak
voltage from the power supply.
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5
Module ET2
Electronic Circuits and Components.
Solutions to Student Exercise.
1.
Vrms 
Vo
20
20


 14.14V
2
2 1.414
or
Vrms  0.707Vo  0.707  20  14.14V
2.
Vo  2  Vrms
 1.414  9
 12.73V  12.7V
3.
Vo  2  Vrms
 1.414  12
 16.97V  17V
4.
Vrms 
Vo 31.1 31.1


 21.99V  22V
2
2 1.414
or
Vrms  0.707Vo  0.707  31.1  21.99V  22V
5.
Vo  2  Vrms
 1.414  120
 169.68V  170V
No examination style questions have been set in this topic as they are
integral to longer questions on power supplies, which we are not yet in a
position to answer, so time to move on to topic 2.4.2 – Rectification.
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Topic 2.4.1 – Root Mean Square – RMS Voltage.
Self Evaluation Review
Learning Objectives
My personal review of these objectives:



apply the formula Vo  Vrms 2 for a
sinusoidal AC voltage
Targets:
1.
………………………………………………………………………………………………………………
………………………………………………………………………………………………………………
2.
………………………………………………………………………………………………………………
………………………………………………………………………………………………………………
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