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BASIC ELECTRICITY AND ELECTRONICS
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2.17 RMS Value of an ac Waveform Assignment
2.17.1 Objectives

To investigate the power carried by ac waveforms.

To determine dc equivalent power for a sinusoidal waveform.

To calculate the rms value of a sinusoidal waveform.
2.17.2 Prerequisite Assignments

Power

Time Constant

Alternating Waveforms
2.17.3 Knowledge Level

See Prerequisite Assignments.
2.17.4 Equipment Required
Qty
Apparatus
1
Basic Electricity and Electronics Module 12-200-A
1
Power Supply Unit, 0 – 20 V variable dc regulated and 12 V ac
(eg, Feedback Teknikit Console 92-300).
2
Multimeters
1
2-Channel Oscilloscope
OR
Feedback Virtual Instrumentation may be used in place of one
of the multimeters and the oscilloscope.
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2.17.5 Background
You will now investigate the power producing properties of some waveforms.
First of all, consider the direct current waveform of fig 1, i = I (constant).
Fig 1
Suppose the current is passed through a resistor of value R ohms.
At all times the current value is I amps and the resistor value is R ohms, thus the power
dissipated in the resistor will be:
Power = I2R watts
Suppose now that a voltage with the sinusoidal waveform shown in fig 2 is applied across
the resistor R.
This waveform alternates in value between positive and negative value voltages and is
thus termed an alternating waveform.
Fig 2
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Because, for a resistor, the current is directly proportional to the voltage (V = IR), the
current in the resistor will have the same shape, as is shown in fig 3.
Fig 3
For this waveform, consider the instantaneous powers at times t 1, t2, t3……..etc.

At time t1 the power will be i12 R watts

At time t2 the power will be i22 R watts

At time t3 the power will be i32 R watts

At time tn the power will be in2 R watts
The average power produced will be:
2R
i2R + i2R + i2R +...in
1
2
3
=
watts
n
This is the average power produced in the positive half-cycle shown in fig 3. The power
produced in the negative half-cycle will be the same because the currents, although
negative, will have positive squares,
ie:
(-i1)2R = i12R
Thus there will be an effective power value of current for the ac waveform.
The effective value I is that value of direct current which produces the same power as the
average power produced by the ac.
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I12R  I22R  I23R  ...In2R
i.e where I R 
n
2
I12  I22  I23  ...In2
I 
n
2
I 
I12  I22  I23  ...In2
n
The effective value of current is found from the root of the mean (average) of the squares
of the currents in the ac waveform, and is thus termed the ROOT MEAN SQUARE value,
or the rms value of the ac waveform.
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2.17.6 Practical 1
The circuit that you will be using is shown in fig 4.
Fig 4
In this Practical you will apply a dc voltage to the circuit (with the switch to the left) and
note the brightness of the lamp. You will then switch so that an ac voltage is applied to
the lamp and you will vary the position of the potentiometer wiper to get same level of
brightness of the lamp.
This will indicate that the same power is being dissipated by the lamp for both positions of
the switch.
You will measure the dc and ac conditions and calculate the equivalent ac power and
relate this to the dc power.
Connect up the circuit as shown in the Patching Diagram for this Practical.
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Practical 1 Patching diagram
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2.17.6.1 Perform Practical
Ensure that you have connected up the circuit as shown in the Patching Diagram for this
Practical and that it corresponds to the circuit diagram of fig 7.
Ensure that the variable dc voltage control is fully counterclockwise, and switch off the
psu.
Fig 7

Set the potentiometer to its mid position and switch on the psu.

Set the slide switch initially to the left.

Set the variable dc voltage to about 5 V as shown on the meter. The lamp should
be on, but dim. Notice the intensity of the lamp.

Switch the slide switch to the right and adjust the potentiometer until the intensity of
the lamp is roughly the same as before.

Switch the slide switch left and right, adjusting the potentiometer at the same time,
until the intensity of the lamp is identical for both switch positions.

Take readings of the dc voltage and current, and the ac peak-to-peak voltages
seen on the scope.
Go to the Results Tables section of this Assignment and copy fig 8 to tabulate your
results.
Calculate the peak-to-peak ac current.

As the lamp is at the same intensity for the ac and the dc waveforms, what
can you say about the power in the ac and dc waveforms?
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
What is the effective value of the alternating current?

What is the rms value of the alternating voltage?

What is the peak value of the alternating voltage?

What is the peak value of the alternating current?
Calculate the relationships between the rms values and the peak values for current and
voltage.
The ratio:
peak value
rms value
is called the Peak Factor of the waveform.

What is the peak factor of the current waveform?

What is the peak factor of the voltage waveform?
Construct a sine wave using the method you used in the Alternating Waveforms
Assignment, Practical 3.
Fig 9

Draw such a sine wave, and mark off the horizontal scale in 20° intervals, i.e. 20°,
40°, 60°, 80°, ............... 160°, 180° to meet your waveform.

Draw vertical lines at the mid-ordinates of these intervals, i.e. at 10°, 30°, 50°,
.......... 150°, 170°.
Calling the peak value of the sine wave unity (i.e. the radius of the circle r = 1), graduate
the vertical axis between -1 and +1.
Go to the Results Tables section of this Assignment and copy fig 10 to tabulate your
results.
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Calculate:
y210 + y230 + y250 + y270 +.........y2170
In the above, the number of mid-ordinates is 9.

Calculate
y2 + y2 + y2 + y2 +... y2
10
30
50
70
170
9

Square root this to find:
2  y 2  y 2  y 2 ... y 2
y10
30
50
70
170
9
This will give you the rms value of a sine wave whose peak value is unity.

How does this compare with the peak factor that you calculated earlier?

Multiply the rms value above by the peak value of the alternating voltage used
before.

How does this value compare with the rms value found from the lamp
experiment?

Do the same for the alternating current.
You should have found that the peak factor for a pure sine wave is given by:
Peak factor =
2
= 1.414
Also the rms value of a sine wave with unit peak value should be:
Rms =
1
2
Thus you can say:
For a sine wave Vrms = 0.707 V pk
If a source of triangle and/or square waves is available, repeat the experiment to find the
rms values of those waveforms. Construct graphically triangle and square waves and
verify your experimental results.
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2.17.6.2 Questions
1. What is the rms value of a sinusoidal waveform whose peak value is 5 V?
2. What is the peak voltage of a sinusoidal waveform whose rms value is 240 V?
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2.17.7 Results Required
When you have performed this Assignment you should have:

observed that you can have an equivalent power for an ac waveform as for a dc,
 determined the rms and peak values of current and voltage for an ac waveform,
 determined the peak factor for the waveform,
 derived the relationship between rms value and peak value for a sinusoidal
waveform.
Your report should contain:

the circuit that you investigated,

the results that you achieved,

the plot of the sinusoidal waveform,

the calculations for determining rms and peak values,

conclusions on your findings in the Assignment.
To produce your report you should use a word processing package.
To achieve the calculated values you could use a spread sheet package.
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2.17.8 Practical Considerations and Applications
When an ac waveform is rectified and measured by a moving-coil meter the deflection of
the needle is proportional to the average value of the waveform. The average value is the
normal average taken over a positive or negative half cycle.
For a sinusoidal waveform the average value is 0.636 Vmax.
The calibration on the face of rectifying moving coil meters, and indeed on most ac
meters, is given in terms of the rms value, and its accuracy is dependent on the fixed ratio
between the average and rms values of a sinusoidal waveform. Thus if any wave with a
shape other than sinusoidal is measured with an ac meter the indicated value will be
incorrect, as this ratio will be different for different waveforms.
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2.17.9 Results Tables
angle
yn
yn2
10
30
50
70
90
1.0
110
130
150
170
Fig 10
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2.17.10 Further Work
If a source of triangle and/or square waves is available, repeat the experiment to find the
rms values of those waveforms.
Construct graphically triangle and square waves and verify your experimental results.
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2.18 Resistive Circuit at ac Assignment
2.18.1 Objectives

To investigate a resistive circuit at ac.

To show that the current and voltage in a resistor are in phase at ac.

To plot the frequency response of a resistor.
2.18.2 Prerequisite Assignments

RMS value of an ac waveform
2.18.3 Knowledge Level

See Prerequisite Assignments.
2.18.4 Equipment Required
Qty
Apparatus
1
Basic Electricity and Electronics Module 12-200-A
1
Power Supply Unit, 0 – 20 V variable dc regulated and 12 V ac
(eg, Feedback Teknikit Console 92-300).
1
Function Generator, 50 – 1 kHz Sine 20 V pk-pk
2
Multimeters
OR
Feedback Virtual Instrumentation may be used in place of one
of the multimeters
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2.18.5 Background
Resistors are used at ac in similar ways to those dc applications of earlier assignments.
There is no phase shift across a true resistance, and the ratio of voltage to current shown
by a resistance is constant with frequency, so the behaviour at ac is no different from its
behaviour at dc
When calculating power dissipation at ac, rms values of voltage and current must apply.
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2.18.6 Practical 1
The circuit that you will be using is shown in fig 1.
Fig 1
In this Practical you will apply an ac voltage to the circuit and you will measure the
resulting voltage and current.
You will vary the input voltage and observe the variation in current and determine the
relationship between them.
Connect up the circuit as shown in the Patching Diagram for this Practical.
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Practical 1 Patching Diagram
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2.18.6.1 Perform Practical
You will now see how a resistor behaves at ac.
Ensure that you have connected up the circuit as shown in the Patching Diagram for this
Practical and that it corresponds to the circuit diagram of fig 2.
Fig 2

Ensure the variable dc voltage control knob is turned fully counterclockwise, then
switch on the psu.

Vary the dc control slowly, and observe the two meters.

Do the readings on the two meters keep in step with each other?

Vary the control somewhat faster and observe the meters.

Do the readings on the two meters still keep in step with each other?
You should find that when the current is zero the voltage is zero; and, by Ohm's Law, the
current is proportional to the voltage, so that if ac is applied to a resistor the current
waveform and the voltage waveform will be of the same shape. See fig 3
Fig 3
The two waveforms will have their zeros at the same time, and their maximum values at
the same time.
In a resistor, the current and voltage are said to be in phase.
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2.18.7 Practical 2
You will be using the same area on your 12-200-A workboard.
The circuit that you will be using is essentially the same as for the first Practical, except
that you will be using the function generator to provide the input voltage.
This is shown in fig 4.
Fig 4
You will apply sinusoidal ac voltages to the circuit at a number of frequencies and you will
measure the resulting voltage and current.
You will then determine how the resistor performs at different frequencies.
Connect up the circuit as shown in the Patching Diagram for this Practical.
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Practical 2 Patching Diagram
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2.18.7.1 Perform Practical
Ensure that you have connected up the circuit as shown in the Patching Diagram for this
Practical and that it corresponds to the circuit diagram of fig 5.
Fig 5

Set the generator frequency to 50 Hz, with an output amplitude of 4 V rms, as read
on the 0-10 V ac meter.
Go to the Results Tables section of this Assignment and copy fig 6 to tabulate your
results.

Calculate the resistance of the resistor at 50 Hz and tabulate this also.
Repeat the readings for a frequency of 100 Hz, and for 100 Hz increments thereafter, up
to 1 kHz.

Plot a graph of resistance against frequency using your results.

How do the current, voltage and resistance change with frequency?
It should be seen from this experiment that the current and voltage in a resistive circuit are
in phase, and that the resistance of a resistive circuit does not change with frequency (see
fig 7).
Fig 7
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2.18.8 Results Required
When you have performed this Assignment you should have:
 observed that the current and voltage in a resistor remain in phase when an ac is
applied,
 taken readings of current and voltage for a resistive circuit at various frequencies,
 calculated the resistance value at those frequencies,
 plotted resistance against frequency to obtain a frequency response for the resistor.
Your report should contain:

the circuit that you investigated,

the results that you achieved,

the plot of frequency response,

conclusions on your findings in the Assignment.
To produce your report you should use a word processing package.
To achieve the calculated values you could use a spread sheet package.
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2.18.9 Results Table
frequency
(Hz)
voltage
(V) rms
current
(mA) rms
resistance
(ohms)
50
100
200
300
400
500
600
700
800
900
1000
Fig 6
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2.19 Capacitive Circuit at ac Assignment
2.19.1 Objectives

To investigate a capacitive circuit at ac.

To display current and voltage waveforms for a capacitor.

To determine the phase relationship between current and voltage for a capacitor.
2.19.2 Prerequisite Assignments

Capacitors
2.19.3 Knowledge Level
Before working this assignment you should:

Know how to describe varying ac quantities by vectors (phasors).

See Prerequisite Assignments.
2.19.4 Equipment Required
Qty
12-200S
Apparatus
1
Basic Electricity and Electronics Module 12-200-A
1
2-channel Oscilloscope
1
Function Generator, 250 Hz sine 10 V pk-pk
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2.19.5 Background
You will know, from the Capacitors Assignment, that the relationship between charge,
voltage and capacitance is:
Q = CV
Also you know:
Q=It
where I is current and t is time
From these you can say that if a capacitor of C farad is charged from 0 V to V volts, in t
seconds then:
charging current,
I=
Q coulombs
t seconds
charging current = capacitance x rate of increase of voltage
As the current waveform reaches it positive peak value 90° before the voltage waveform
reaches that value, you say that:
The current in a capacitive circuit leads the voltage by 90°.
Often the 90° is referred to as


. As there are 2π radians in 360°, then 90° is
radians.
2
2
Mathematically, if the voltage waveform is denoted by the formula:
v = Vmax sin
t
then as i = C x rate of change of voltage
i=C
= CVmax
i = CVmax
cos
dv
(sin
dt
t)
t
Thus if v is sinusoidal, i will be cosinusoidal; ie, the same shape, but leading by 90°.
This is because cos
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In the RMS value of an ac waveform Assignment you saw how a sine wave can be plotted
by taking projections from a rotating vector.
Fig 1
If you were to plot the voltage and current waveforms in the capacitor by that method you
would require two vectors. Both vectors would rotate while keeping a constant 90° angle
between them, as shown in fig 1.
As you go on you will find it useful to think in terms of these vectors, but rather confusing if
they are always rotating. Usually it is the relationships between them that are important, as
for instance the 90° angle between those in fig 1.
These relationships can be studied conveniently in a diagram where the vectors are
shown at rest. The vectors are then said to be represented by 'phasors' and the diagram is
a 'phasor diagram'. Fig 2 is a phasor diagram corresponding to fig 1.
Fig 2
The voltage phasor is taken as the reference and is drawn horizontally pointing to the right
(3 o'clock). The current phasor leads the voltage phasor by 90° and is thus drawn 90°
counterclockwise from the reference.
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When an alternating voltage is applied across a capacitor an alternating current flows. Yet
when a dc voltage is applied, after an initial flow of charging current, no dc current flows.
This behaviour is different from that of a resistance. Nevertheless if an ac voltage and an
ac current can exist, the ratio between them is likely to be of interest, and the ratio is
therefore given a different name.
In an ac circuit the ratio of voltage to current is called 'impedance' and is denoted by Z.
Thus in an ac circuit
Z=
V
I
You will examine this idea further in other assignments. For the moment it may be noted
that impedance may be taken as the ratio of two phasors, and therefore has both
magnitude and phase.
Magnitude = |Z| =
Vmax
Imax
Phase of the impedance is the angle between the phasors. The impedance of a capacitor

has a phase of - 90° or - radians
2
Impedances of ±90° phase angle have special properties and are given the special name
reactance.
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2.19.6 Practical 1
The circuit that you will be using is shown in fig 3.
Fig 3
In this Practical you will apply an ac voltage to the circuit and you will measure the
resulting voltage and current waveforms using an oscilloscope.
You will plot the voltage and current and determine the relationship between them.
Connect up the circuit as shown in the Patching Diagram for this Practical.
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Practical 1 Patching Diagram
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2.19.6.1 Perform Practical
Now you will see what happens when a sinusoidal alternating voltage is applied to a
capacitor.
Ensure that you have connected up the circuit as shown in the Patching Diagram for this
Practical and that it corresponds to the circuit diagram of fig 4.
Fig 4

Set the function generator to give a 10 V peak-to-peak sine waveform at 250 Hz.

Set the oscilloscope as follows:


Y1 channel 1 V/cm.

Y2 channel 500 mV/cm.

Timebase to 1 ms/cm
Zero both the traces and then observe the two waveforms on the oscilloscope.
Carefully draw the two waveforms, showing their relative positions with respect to each
other.
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2.19.6.2 Questions
1.
Where on the voltage waveform is the rate of change of voltage a positive maximum?
2. Where on the voltage waveform is the rate of change of voltage zero?
3.
Where on the voltage waveform is the rate of change of voltage a negative maximum?
4. We have said that the current is proportional to the rate of change of voltage. Does
your current curve substantiate this?
5.
Fig 5
Using the same notation as in fig 5 (ie, 360° in one cycle), how many degrees are the
two waveforms apart?
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What proportion of a cycle is that?
7. Remembering that the time increases towards the right, which of the two waveforms
reaches its positive maximum first?
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2.19.7 Results Required
When you have performed this Assignment you should have:

observed the relationship between the current and voltage in a capacitor when an
ac is applied,

plotted the current and voltage waveforms.
Your report should contain:

the circuit that you investigated,

the results that you achieved,

the plot of current and voltage,

conclusions on your findings in the Assignment.
To produce your report you should use a word processing package.
To achieve the calculated values you could use a spread sheet package.
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2.19.8 Practical Considerations and Applications
The applications section of the Capacitors Assignment gives details of capacitors in
general, but there are a few more points which appear at high frequencies (hf).
Due to the form of construction, wound capacitors of any type possess appreciable
inductance, thus do not act as pure capacitors. The effects of this inductance are greater
at high frequencies and thus the wound form of capacitor is inferior at hf.
Mica capacitors, ceramic tubular and disc capacitors, and air dielectric capacitors are ideal
for high frequency work.
Capacitors normally have a stated maximum voltage rating above which they cannot be
used safely. When used at ac the peak or crest value of the voltage must be used to
determine whether the capacitor is within rating.
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Notes
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2.20 Inductive Circuit at ac Assignment
2.20.1 Objectives

To investigate an inductive circuit at ac.

To display current and voltage waveforms for an inductor.

To determine the phase relationship between current and voltage for an inductor.
2.20.2 Prerequisite Assignments

Inductors
2.20.3 Knowledge Level
Before working this assignment you should:

Know how to describe varying ac quantities by phasors.

Know how a capacitive circuit behaves at ac.

Know how to apply Kirchhoff’s Law

See Prerequisite Assignments.
2.20.4 Equipment Required
Qty
12-200S
Apparatus
1
Basic Electricity and Electronics Module 12-200-A
1
2-channel Oscilloscope
1
Function Generator, 250 Hz Sine 10 V pk-pk
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BASIC ELECTRICITY AND ELECTRONICS
STUDENT’S MANUAL
2.20.5 Background
From the Inductors Assignment you have found out that the relationship between induced
emf, current and inductance in a system is:
e=-L
di
dt
or: (Induced emf) = - (inductance)(rate of change of current)
If the current waveform is denoted by
I = Imax sin
then as e = -L
t
di
dt
e = - LImax
cos
t
Here you must be careful. You have found the induced emf, and included the minus sign
as a reminder that it opposes the change of current. But when you look at ac applied to a
resistor or dc applied to it, you look at the applied voltage.
In the same way, with the inductance, you must look at the applied voltage if you are to be
consistent. Since V around the circuit is zero (Kirchhoff’s Law) , this applied emf is equal
and opposite to the induced emf,
ie, applied emf = +L
= LImax
cos
di
dt
t
So you can say that a positive applied emf produces a positive current which induces an
opposing (negative) emf in the inductor.
This should be evident by a look at fig 1, in which the polarity of the emfs is indicated by
arrows in the direction of action.
Fig 1
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The results show that the voltage waveform is the same as that of the current, but
leads it by 90°, since cos t = sin ( t + 90°).
Thus the voltage in an inductive circuit leads the current by 90°.
To remember which waveform leads which in either a capacitive or an inductive circuit, the
following mnemonic is useful.
CIVIL
ie, in a capacitive circuit (C) the current (I) leads the voltage (V)
CIV..
in an inductive circuit (L) the voltage (V) leads the current (I)
..VIL
The phasor representation of voltage and current is given in fig 2.
Here the current is taken as the reference and the voltage leads it by 90°.
Fig 2
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2.20.6 Practical 1
The circuit that you will be using is shown in fig 3.
Fig 3
In this Practical you will apply an ac voltage to the circuit and you will measure the
resulting voltage and current waveforms using an oscilloscope.
You will plot the voltage and current and determine the relationship between them.
Connect up the circuit as shown in the Patching Diagram for this Practical.
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Practical 1 Patching Diagram
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2.20.6.1 Perform Practical
Now you will see what happens when a sinusoidal alternating voltage is applied to an
inductor.
Ensure that you have connected up the circuit as shown in the Patching Diagram for this
Practical and that it corresponds to the circuit diagram of fig 4.
Fig 4

Set the function generator to give a 10 V peak-to-peak waveform at 250 Hz.

Set the oscilloscope as follows:


Y1 channel (voltage) to 1 V/cm

Y2 channel (current) to 500 mV/cm

Timebase to 1 ms/cm.
Zero both traces, then observe the two waveforms on the oscilloscope.
Draw the two waveforms you see, showing their relative positions with respect to each
other.
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2.20.6.2 Questions
1. Where on the waveform is the rate of change of current a positive maximum?
2. Where is it zero?
3. Where is it a negative maximum?
4.
Does the voltage waveform correspond to the rate of change of current?
5.
How far apart, in degrees, are the voltage and current waveforms?
6.
How far apart are they in radians?
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Which waveform is leading?
8. Is this the same as for a capacitive circuit?
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2.20.7 Results Required
When you have performed this Assignment you should have:
 observed the relationship between the current and voltage in an inductor when an
ac is applied,
 plotted the current and voltage waveforms.
Your report should contain:

the circuit that you investigated,

the results that you achieved,

the plot of current and voltage,

conclusions on your findings in the Assignment.
To produce your report you should use a word processing package.
To achieve the calculated values you could use a spread sheet package.
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2.20.8 Practical Considerations and Applications
By its nature of operation an inductance will create a magnetic field around it. With an
alternating current flowing through the inductor the magnetic field around it will be
alternating. If the inductor is in the presence of other components or conductors this
alternating magnetic field will link with these conductors and induce emfs in them. These
emfs will generally be unwanted, and give rise to noise, hum, or interfering signals. For
this reason, inductors are often magnetically screened in cans, or housings made from a
non-magnetic material such as mumetal or aluminium.
When designing inductors care must be taken in the selection of type and gauge of wire
used. Obviously the lowest possible resistance is desired, and at low frequencies this
means the thickest possible wire gauge consistent with a reasonable sized winding,
however at high frequencies multi-stranded litz wire is often better to minimise the skin
effect.
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2.21 Capacitive Reactance Assignment
2.21.1 Objectives

To measure current and voltage in a capacitor when a sinusoidal waveform is
applied.

To investigate the impedance of a capacitor to an ac sinusoidal waveform.

To plot the variation of impedance with frequency for a capacitor.
2.21.2 Prerequisite Assignments

Capacitive Circuit at ac.
2.21.3 Knowledge Level
Before working this assignment you should:

Know the meaning of peak or crest factor and effective or rms value of an ac
quantity.

Know the relationship between frequency and time period, measured in degrees
and radians.

See Prerequisite Assignments.
2.21.4 Equipment Required
Qty
Apparatus
1
Basic Electricity and Electronics Module 12-200-A
2
Multimeters
OR
Feedback Virtual Instrumentation may be used in place of one
of the multimeters
1
12-200S
Function Generator, 50 Hz - 1.6 kHz sine 20 V pk-pk
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2.21.5 Background
In a capacitor a sinusoidal current and voltage are always 90° out of phase with one
another, so that the phase of the impedance is constant. The magnitude varies however.
Consider the circuit diagram of fig 1.
Fig 1
The input voltage is sinusoidal, so you can represent it mathematically as:
v = Vmax sin  t
and you know from the Capacitive Circuits at ac Assignment that:
i = CVmax  sin ( t +

)
2
Thus, when i = Imax
ie:
when sin ( t +

)=1
2
Imax = CVmax
1
 Vmax 
Imax C
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Now, for a sine wave you know:
Vmax
=
1.414 Vmax = 1.414Vrms
=
1.414 Imax = 1.414Irms
Vrmsx
and
Imax
Irms
Thus
V max
=
Imax
Therefore
V rms
Vrms
Irms
=
Irms
1
ωC
Zc =
1
ωC
Therefore, as = 2πf
Zc =
1
2fC
Thus the impedance of a capacitor is inversely proportional to both the frequency and the
capacitance.
The impedance of a capacitor is normally termed capacitive reactance, and is given the
symbol Xc.
In comparing your experimental impedance values with this calculated value, bear in mind
that the capacitor has a ±20% tolerance, and that any errors in measuring the voltage, the
current and the frequency may increase further the discrepancy between experimental
and calculated values.
Your results should however show the correct form of the impedance-frequency
relationship quite well.
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2.21.6 Practical 1
The circuit that you will be using is shown in fig 2.
Fig 2
In this Practical you will apply an ac voltage to the circuit and you will measure the
resulting voltage and current.
You will determine the relationship between current and voltage and will see if this
relationship holds true for a number of applied voltages.
Connect up the circuit as shown in the Patching Diagram for this Practical.
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Practical 1 Patching Diagram
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2.21.6.1 Perform Practical
Ensure that you have connected up the circuit as shown in the Patching Diagram for this
Practical and that it corresponds to the circuit diagram of fig 3.
Fig 3

Set the frequency of the function generator to 800 Hz.

Adjust the output of the generator to give 1 V rms as read on the meter.

Take the current reading for this voltage.
Go to the Results Tables section of this Assignment and copy fig 4 to tabulate your
results.

Reset the voltage output of the generator to 2 V rms.

Record the resultant current.
Repeat this procedure for voltage of 3 V, 4 V, 5 V and 6 V rms.

Record your results and calculate the ratio of rms voltage to rms current.
The magnitude of the impedance Z can be defined as the ratio of the rms voltage to
the rms current.
(For the remainder of the assignment 'the magnitude of the impedance' will be
abbreviated to 'impedance').
Minimise your spreadsheet window. Don’t close it; you will be using it later.
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2.21.6.2 Questions
1. From your results, is there any relationship between rms voltage and rms current?
2.
What is the average value of the impedance?
3. Thus we can say that the impedance of the capacitor is ......ohms at a frequency of
800 Hz?
4. We know that the impedance of a pure capacitor at dc is infinite, and we have now
found the impedance of the capacitor at 800 Hz. Are they the same?
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2.21.7 Practical 2
You will be using the same area on your 12-200-A workboard.
The circuit that you will be using is the same as for Practical 1.
In this Practical you will apply a fixed ac voltage to the circuit at a number of different
frequencies and you will measure the resulting voltage and current.
You will then determine how the capacitor performs at different frequencies.
2.21.7.1 Perform Practical
It would seem reasonable to assume that the impedance of a capacitor is proportional in
some way to the frequency of the voltage across it.
Now you will investigate the impedance at other frequencies.

Set the frequency of the generator to 50 Hz.

Take readings of current for voltage settings of 1 V, 2 V, 3 V, 4 V, 5 V and 6 V rms.
Go to the Results Tables section of this Assignment and copy fig 5 to tabulate your
results.

Set the generator to 100 Hz.
Repeat the readings.

Do the same for frequencies of 200 Hz, 400 Hz and 1600 Hz.
Go to the Results Tables section of this Assignment and copy fig 6 to tabulate your
results.

Tabulate your results and work out the impedance for each step.

Calculate the average impedance for each frequency and enter the results in the
table.

Plot a graph of impedance against frequency for the capacitor using axes as in
fig 7.
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Fig 7
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2.21.7.2 Question
1. Does the impedance vary with frequency?
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2.21.8 Results Required
When you have performed this Assignment you should have:

taken readings of current and voltage for a capacitive circuit at various frequencies,

calculated the average impedance values at those frequencies,

plotted impedance against frequency to obtain a frequency response for the
capacitor.
Your report should contain:

the circuit that you investigated,

the results that you achieved,

the plot of frequency response,

conclusions on your findings in the Assignment.
To produce your report you should use a word processing package.
To achieve the calculated values you could use a spread sheet package.
12-200S
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2.21.9 Practical Considerations and Applications
Capacitors show an infinite impedance to dc but their impedance is finite to ac and
decreases as the frequency of the ac increases. Therefore one of their uses is in coupling
between circuits or parts of circuits, where it is wanted to block any dc transmission but to
allow ac signals to pass.
Large value electrolytic capacitors are often used in 'smoothing' circuits in which a dc
voltage which has an ac ripple voltage superimposed upon it is applied to the capacitor.
The capacitor has no effect on the dc component of the waveform, as its impedance to dc
is infinite, but bypasses to earth the ac. This is shown in fig 8.
Fig 8
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Results Tables
f = 800 Hz
rms voltage
(V)
rms current
(mA)
rms voltage
rms current
1
2
3
4
5
6
Fig 4
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Frequency 50 Hz
voltage
(V)
current
(mA rms)
impedance
(k)
1
2
3
4
5
6
Frequency 100 Hz
1
2
3
4
5
6
Frequency 200 Hz
1
2
3
4
5
6
Frequency 400 Hz
1
2
3
4
5
6
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Frequency 800 Hz
voltage
(V)
current
(mA rms)
impedance
(k)
1
2
3
4
5
6
Frequency 1600 Hz
1
2
3
4
5
6
Fig 5
frequency
(Hz)
average impedance
(k)
50
38.4
100
16.6
200
8.1
400
4.0
800
2.0
1600
1.0
Fig 6
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2.21.11 Further work
Use the formula for capacitive reactance derived in the Background section of this
Assignment and calculate the theoretical values of impedance (reactance) for each
frequency.
Compare these theoretical values with the practical values that you found in Practical 2.
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2.22 Inductive Reactance Assignment
2.22.1 Objectives

To measure current and voltage in an inductor when a sinusoidal waveform is
applied.

To investigate the impedance of an inductor to an ac sinusoidal waveform.

To plot the variation of impedance with frequency for an inductor.
2.22.2 Prerequisite Assignments

Inductive Circuits at ac.
2.22.3 Knowledge Level
Before working this assignment you should:

Know the meaning of the term impedance, Z, and that:
|Z| = Vrms
Irms

Know the meaning of the term capacitive reactance, Xc.

See Prerequisite Assignments.
2.22.4 Equipment Required
Qty
Apparatus
1
Basic Electricity and Electronics Module 12-200-A
2
Multimeters
OR
Feedback Virtual Instrumentation may be used in place of one
of the multimeters
1
12-200S
Function Generator, 4-16 kHz 20 V pk-pk sine
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2.22.5 Background
You know that impedance, |Z|, is given by:
|Z| = Vrms
Irms
and that, for a capacitor, its impedance is termed capacitive reactance, X c.
Similarly the inductive reactance of an inductor can be determined.
Inductive reactance is given the symbol XL.
You know that (from the Inductive Circuits at ac Assignment) for an inductor, if
i = Imax sin
t
then the applied voltage, v, will be given by:
v = LImax
and when cos
cos
t
t = 1, v = Vmax
thus
Vmax = LI max
 Vmax  ωL
Imax

Z=
L

X = 2πfL
Thus the impedance (inductive reactance) of an inductor is directly proportional to
the frequency.
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2.22.6 Practical 1
The circuit that you will be using is shown in fig 1.
Fig 1
In this Practical you will apply an ac voltage to the circuit and you will measure the
resulting voltage and current.
You will determine the relationship between current and voltage and will see if this
relationship holds true for a number of applied voltages.
You will then apply a fixed ac voltage to the circuit at a number of different frequencies
and you will measure the resulting voltage and current.
From this, you will determine how the inductor performs at different frequencies.
Connect up the circuit as shown in the Patching Diagram for this Practical.
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Practical 1 Patching Diagram
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2.22.6.1 Perform Practical
Ensure that you have connected up the circuit as shown in the Patching Diagram for this
Practical and that it corresponds to the circuit diagram of fig 2.
Fig 2

Adjust the generator to give a frequency of 4 kHz and an output of 1 V rms sine
wave, as shown on the meter.

Take the current reading for this voltage.
Go to the Results Tables section of this Assignment and copy fig 3 to tabulate your
results.

Readjust the output amplitude to 2 V rms

Record the resultant current.
Repeat this for voltage steps of 3 V, 4 V, 5 V and 6 V rms.

Record your results and calculate the impedance for each step.
Repeat the above procedure for frequencies of 8 kHz, 12 kHz, 16 kHz, 20 kHz and
24 kHz.
Go to the Results Tables section of this Assignment and copy fig 4 to tabulate your
results.

Calculate the average impedance for each frequency.
Go to the Results Tables section of this Assignment and copy fig 5 to tabulate your
results.
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Plot a graph of impedance against frequency, with axes as in fig 6.
Your graph should have the correct form, although you may find the value of L obtainable
from it differs from the marked value. This is simply because the inductor is a widetolerance component.
Fig 6
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2.22.6.2 Questions
1.
What is the average value of impedance of the coil at 4 kHz?
2.
How does the impedance vary with frequency?
3. The impedance (inductive reactance) of an inductor is directly proportional to
frequency. Does the shape of your graph substantiate this?
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2.22.7 Results Required
When you have performed this Assignment you should have:

taken readings of current and voltage for an inductive circuit at various frequencies,

calculated the average impedance values at those frequencies,

plotted impedance against frequency to obtain a frequency response for the
inductor.
Your report should contain:

the circuit that you investigated,

the results that you achieved,

the plot of frequency response,

conclusions on your findings in the Assignment.
To produce your report you should use a word processing package.
To achieve the calculated values you could use a spread sheet package.
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2.22.8 Practical Considerations and Applications
The inductor is the converse of the capacitor in that it shows an impedance which is
practically zero to dc, but increases with frequency. Like a capacitor, an inductor can also
be used in smoothing circuits, but it is used in series with the dc line instead of across it.
See fig 7. The dc is passed without effect while the ac is greatly impeded by the inductor.
In this application the inductor is often called a 'Choke'.
Fig 7
When an inductor contains magnetic material such as iron, it can only maintain its
inductance over a limited range of current. Excessive current causes the 'iron' to saturate
(ie, fail to permit more than a limited amount of flux). The inductance therefore decreases.
This is why you were advised to omit any readings for which the current exceeded 60 mA.
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2.22.9 Results Tables
Frequency 4 kHz
rms voltage
(V)
rms current
(mA)
impedance
(k)
1
2
3
4
5
6
Fig 3
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Frequency 8 kHz
voltage
(V)
current
(mA rms)
impedance
(k)
1
2
3
4
5
6
Frequency 12 kHz
1
2
3
4
5
6
Frequency 16 kHz
1
2
3
4
5
6
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Frequency 20 kHz
voltage
(V)
current
(mA rms)
impedance
(k)
1
2
3
4
5
6
Frequency 24 kHz
1
2
3
4
5
6
Fig 4
frequency
(kHz)
ave. impedance
(k)
4
8
12
16
20
24
Fig 5
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2.22.10 Further Work
Use the formula for inductive reactance derived in the Background section of this
Assignment and calculate the theoretical values of impedance (reactance) for each
frequency.
Compare these theoretical values with the practical values that you found in the Practical.
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Notes
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2.23 The Series CR Circuit Assignment
2.23.1 Objectives

To investigate the series CR circuit and determine the relationships governing
amplitude and phase shift.

To plot the phasor diagrams for a series CR circuit for a number of resistor values.

To plot the locus diagram of a series CR circuit for various values of resistance.
2.23.2 Prerequisite Assignments

Resistive Circuit at ac

Capacitive Reactance Assignments
2.23.3 Knowledge Level

See Prerequisite Assignments.
2.23.4 Equipment Required
Qty
Apparatus
1
Basic Electricity and Electronics Module 12-200-A
1
Function Generator, 50 Hz 20 V pk-pk sine
(eg, Feedback FG601)
1
2-Channel Oscilloscope
OR
Feedback Virtual Instrumentation may be used in place of the
oscilloscope.
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2.23.5 Background
This is represented in fig 1.
Fig 1
VR and Vc are both 'phasor quantities', ie, they both have a magnitude and phase angle.
Both of these must be taken into account when combining VR and Vc to obtain the
resultant total voltage.
The phasor sum of VR and VC is Vin,
but Vin  VC+ VR.
Phasor summation may be done graphically by drawing VC and VR to scale, with the
relevant phase angle between them, and then completing the parallelogram (in this case a
rectangle). The phasor sum of VR and Vc is then given by the diagonal of the
parallelogram (see fig 2a).
Fig 2
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An alternative way of getting the same result is to regard the addition of phasors as joining
the tail of each phasor to be added on to the head of the previous one. The magnitude
and direction of each phasor remains as before, but the position (which has no special
significance) is adjusted to enable addition to be done.
Whichever way is chosen, the phasor diagram serves as a graphical calculator for adding
together sinusoidal quantities which are out of phase with one another.
The angle θ in fig 2 corresponds to the phase shift between the input and the capacitor
voltages. It might well be useful to know the phase shift between the input and output
voltages of the R-C circuit, in which case the phase of the current is not directly of interest.
Although fig 2 takes the current as a reference for angles, Vin could just as well be taken
as the reference, as in fig 3. Do this for your results in the table, by plotting them on a
piece of graph paper, showing the phase angle, and with the length of the phasor
representing the amplitude of the output voltage V c (see fig 3).
Fig 3
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2.23.6 Practical 1
The circuit that you will be using is shown in fig 4.
Fig 4
As you can see, the circuit comprises a capacitor and a number of different values of
resistor connected in series with it.
In this Practical you will apply an ac voltage to the circuit with a 1 kΩ resistor connected
and you will measure the resulting voltage and current.
You will determine the phase relationship between current and voltage.
Connect up the circuit as shown in the Patching Diagram for this Practical.
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2.23.6.1 Perform Practical
You know that voltage and current are in phase in a resistive circuit, and that, for a
capacitive circuit, the current leads the voltage by 90°.
You will now investigate what happens in a circuit consisting of both capacitance and
resistance, connected in series.
Ensure that you have connected up the circuit as shown in the Patching Diagram for this
Practical and that it corresponds to the circuit diagram of fig 5.
Fig 5

Set the generator frequency to 50 Hz, and its output to 20 V peak-to-peak, sine
wave as shown on the Y1 channel of the oscilloscope.

Have the oscilloscope set to:

Y1 channel (Vin) to 5 V/cm

Y2 channel (Vc) to 5 V/cm

Timebase to 5 ms/cm

Zero both the oscilloscope traces.
With link 1 connected as shown in fig 5, ie, bypassing the resistor chain so that the
resistor is zero, observe the waveforms for Vin and Vc. They should be superimposed.
Measure their amplitude.
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
Now connect link 1 between points A and D to leave the 1kΩ resistor in series with
the capacitor.

Measure the amplitude of Vc and the phase shift between Vc and Vin.
The phase shift may be measured as follows:
Set the timebase switch until one cycle of the waveform occupies approximately the
whole width of the screen of the oscilloscope, then adjust the timebase variable
control until one cycles occupies six main horizontal divisions on the X scale (see
fig 6).
Fig 6
Now 6 cm = 1 cycle = 360°

n cm = 60n degrees
Thus the phase shift may be found easily.

Reconnect link 1 in turn between points A and C, A and B and finally B and E, and
for each circuit measure the amplitude of V c and phase shift.
Go to the Results Tables section of this Assignment and copy fig 7 to tabulate your
results.
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From previous assignments you know that the voltage across the resistor will always be in
phase with the current through it, and that the voltage across the capacitor will be lagging
the current by 90°.
This is represented in fig 8.
Fig 8
You have, however, measured a phase shift between the applied voltage, V in, and the
capacitor voltage Vc.
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2.23.6.2 Questions
1. Does the capacitor voltage, Vc, lead or lag the applied voltage, Vin?.
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2.23.7 Practical 2
You will be using the same area on your 12-200-A workboard and the circuit that you will
also be the same as for the first Practical.
In this Practical you will apply an ac voltage to the circuit with a number of different value
resistors connected and you will measure the resulting voltage across the resistor.
You will then plot a locus graph showing how the resistor value affects the voltages and
phase angle.
Connect up the circuit as shown in the Patching Diagram for this Practical.
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2.23.7.1 Perform Practical

Use the oscilloscope to measure the voltage across the resistor by reconnection of
the circuit as in the Patching Diagram for this Practical.
Leave the Y2 oscilloscope channel disconnected.

Measure the amplitude of VR for the five values of resistance as before.
Follow the link from the first practical and fill in the table for this Practical, or go to the
Results Tables section of this Assignment and copy fig 9 to tabulate your results.
From your results in figs 7 and 9 draw, to scale, phasor diagrams of the type in fig 10
showing the relative amplitudes of Vin, Vc and VR, and the correct phase shift θ, for the
four values of resistance greater than zero.
Fig 10
The angle θ in fig 10 corresponds to the phase shift which you have recorded in your table
(fig 7).
It might well be useful to know the phase shift between the input and output voltages of
the R-C circuit, in which case the phase of the current is not directly of interest. Although
fig 9 takes the current as a reference for angles, Vin could just as well be taken as the
reference, as in fig 11.
Do this for your results in the table, by plotting them on a piece of graph paper, showing
the phase angle, and with the length of the phasor representing the amplitude of the
output voltage Vc (see fig 11).
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Fig 11

Mark on your graph R = 0 against the point plotted for zero resistance.

Mark R = 10kΩ for the point plotted for the 10 kΩ resistor.

Put the point of a .pair of compasses half way along the R = 0 phasor, and draw a
semi-circle with a radius equal to half the R = 0 value.
These results show that as the value of R is increased the phase shift between V in and Vc
increases, but the amplitude of Vc, the output voltage, decreases. This is shown in fig 12.
Fig 12
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2.23.7.2 Questions
1. Is there any visible connection between the horizontal and vertical components of Vin,
and the lengths of Vc and VR ?
2. By Pythagoras' theorem, what is the relationship between V in and VR and Vc?
3. What amplitude would you expect Vc to approach when R tends to infinity (ie, R gets
very large)?
4. Name a use to which an RC circuit might be put.
5. Do the points you have plotted map out any recognisable shape?
6. What value does the phase shift tend to as R is increased towards infinity?
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7. What would be the amplitude value approached as R tends to infinity?
8. Does the semi-circle drawn join up the points plotted?
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2.23.8 Results Required
When you have performed this Assignment you should have:

displayed waveforms of applied voltage and capacitor voltage for a series CR
circuit,

determined the phase shift for a number of resistor values,

plotted phasor diagrams showing VR, VC and Vin for each of these resistor values,

plotted the locus diagram to show how VC and phase angle vary with resistor value.
Your report should contain:

the circuit that you investigated,

the results that you achieved,

the phasor diagram plots and the locus plot,

conclusions on your findings in the Assignment.
To produce your report you should use a word processing package.
To achieve the calculated values you could use a spread sheet package.
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2.23.9 Results Tables
resistor
value
phase shift
 (deg)
Vc amplitude
(V pk–pk)
0
1k0
2k0
4k2
10k0
Fig 7
resistor
value
VR
(V pk–pk)
0
1k0
2k0
4k2
10k0
Fig 9
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2.24 The Series LR Circuit Assignment
2.24.1 Objectives

To investigate the series LR circuit and determine the relationships governing
amplitude and phase shift.

To plot the phasor diagrams for a series LR circuit for a number of resistor values.

To plot the locus diagram of a series LR circuit for various values of resistance.
2.24.2 Prerequisite Assignments

Inductive Reactance

The Series CR Circuit
2.24.3 Knowledge Level

See Prerequisite Assignments.
2.24.4 Equipment Required
Qty
Apparatus
1
Basic Electricity and Electronics Module 12-200-A
1
Function Generator, 5 kHz 20 V pk-pk sine
(eg, Feedback FG601)
1
2-Channel Oscilloscope.
OR
Feedback Virtual Instrumentation may be used in place of the
oscilloscope.
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2.24.5 Practical 1
The circuit that you will be using is shown in fig 1.
Fig 1
As you can see, the circuit comprises an inductor and a number of different values of
resistor connected in series with it.
In this Practical you will apply an ac voltage to the circuit with a 1kΩ resistor connected
and you will measure the resulting voltage and current.
You will determine the phase relationship between current and voltage.
Connect up the circuit as shown in the Patching Diagram for this Practical.
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2.24.5.1 Perform Practical
You have investigated what happens in a series RC circuit when R is varied. You will now
see what happens when R is varied in an LR circuit.
Ensure that you have connected up the circuit as shown in the Patching Diagram for this
Practical and that it corresponds to the circuit diagram of fig 2.
Fig 2

Set the generator frequency to 5 kHz, and its output to 20 V peak-to-peak sine
wave as shown on the Vin channel of the oscilloscope.

Set the oscilloscope to:


Y1 channel (Vin) to 5 V/cm

Y2 channel (VL) to 5 V/cm

Timebase to 50 µs/cm
Zero both oscilloscope traces.
With link 1 connected as shown in fig 2; ie, bypassing the resistor chain so that the
resistance is zero; observe the waveforms for Vin and VL. They should be superimposed.

Measure the amplitude of VL and the phase shift between Vin and VL, as given by
the method used in The Series CR Circuit Assignment.
Now connect link 1 between points A and D to leave the 1kΩ resistor in series with the
inductor.

2-386
Measure the amplitude of VL and the phase shift.
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Reconnect link 1 in turn between points A and C, A and B and finally B and E, and for
each circuit measure the amplitude of VL and the phase shift.
Go to the Results Tables section of this Assignment and copy fig 3 to tabulate your
results.
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2.24.6 Practical 2
You will be using the same area on your 12-200-A workboard and the circuit that you will
also be the same as for the first Practical.
In this Practical you will apply an ac voltage to the circuit with a number of different value
resistors connected and you will measure the resulting voltage across the resistor.
You will then plot a locus graph showing how the resistor value affects the voltages and
phase angle.
Connect up the circuit as shown in the Patching Diagram for this Practical.
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2.24.6.1 Perform Practical

Use the oscilloscope to measure VR by reconnection of the circuit as in the
Patching Diagram for this Practical.
Leave the Y2 oscilloscope channel disconnected.

Measure the amplitude of VR for the five values of resistance as before.
Follow the link and fill in the table for this Practical, or go to the Results Tables section of
this Assignment and copy fig 3 to tabulate your results.

Draw phasor diagrams to scale, showing the relative amplitudes of V in, VL and VR
and the correct phase shift
(as in fig 4).
Fig 4

Also plot VL and
on graph paper (similar method as for V in the Series CR Circuit
Assignment). This time
will be positive.

Draw in the shape that the plotted points map.
Label your plot in a similar fashion as done for the Series CR Circuit Assignment.
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2.24.6.2 Questions
1. What is the relationship between Vin and VR and VL?
2. What shape do the plotted points map?
3. What values do the amplitude of VL and phase
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approach as R tends to infinity?
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2.24.7 Results Required
When you have performed this Assignment you should have:

displayed waveforms of applied voltage and inductor voltage for a series LR circuit,

determined the phase shift for a number of resistor values,

plotted phasor diagrams showing VR, VL and Vin for each of these resistor values,

plotted the locus diagram to show how VL and phase angle vary with resistor value.
Your report should contain:

the circuit that you investigated,

the results that you achieved,

the phasor diagram plots and the locus plot,

conclusions on your findings in the Assignment.
To produce your report you should use a word processing package.
To achieve the calculated values you could use a spread sheet package.
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2.24.8 Results Table
resistance value
(k)
phase shift 
(deg)
VL
(Vpk–pk)
VR
(Vpk–pk)
0
1k0
2k0
4k2
10k0
Fig 3
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Notes
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2.25 Impedance of a Series CR Circuit Assignment
2.25.1 Objectives

To investigate the impedance of a series CR - circuit to an ac sinusoidal waveform.

To determine how the impedance varies with frequency.
2.25.2 Prerequisite Assignments

Capacitive Reactance

The Series CR Circuit
2.25.3 Knowledge Level
Before working this assignment you should:

Know how a potential divider works.

See Prerequisite Assignments.
2.25.4 Equipment Required
Qty
Apparatus
1
Basic Electricity and Electronics Module 12-200-A
2
Multimeters
OR
Feedback Virtual Instrumentation may be used in place of one
of the multimeters
1
12-200S
Function Generator, 50 Hz - 1.6 kHz 20 V pk-pk sine
(eg, Feedback FG601)
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2.25.5 Background
As the circuit is a series circuit, the same current must flow through all parts of the circuit.
Fig 1
You know from previous work that:
VR = IR
Vc = I Xc or I.
1
C
and
Vin = I Z
where Z is the impedance of the circuit.
Thus the phasor diagram for the RC circuit found in the Capacitive Reactance Assignment
may be relabelled as in fig 2.
Fig 2
By Pythagoras:
|Vin| =
I |Z| =
2-396
V C  VR
2
2
I2
+ I2R 2
2
2
 C
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Impedance is a phasor quantity. This can be seen simply by redrawing fig 2 as in fig 3 on
a scale such that the factor I is removed. Each phasor then represents an impedance and
is scaled to represent so many ohms. Since the direction as well as the magnitude of each
phasor is important, an impedance is not completely specified unless both the magnitude
and the phase angle are stated.
The relevant phase angle is the angle shown as theta, between the input voltage and the
current. This is the same as the angle in an impedance diagram between the impedance
in question and a pure resistance.
Fig 3
The value of Ф can be seen to be given by:
Ф = tan-1
XC
R
1
 Ф = tan-1 C
R
 Ф = tan-1
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CR
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2.25.6 Practical 1
The circuit that you will be using is shown in fig 4.
Fig 4
As you can see, the circuit comprises a capacitor and a resistor connected in series with it.
In this Practical you will apply an ac voltage to the circuit and you will measure the
resulting voltage and current.
You will determine the phase relationship between current and voltage.
Connect up the circuit as shown in the Patching Diagram for this Practical.
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2.25.6.1 Perform Practical
You have investigated the impedance (or reactance) of a purely capacitive and an
inductive circuit, and arrived at formulae relating impedance with frequency and
capacitance or inductance.
You will now see if you can arrive at some formulae for the series CR circuit of fig 5.
Ensure that you have connected up the circuit as shown in the Patching Diagram for this
Practical and that it corresponds to the circuit diagram of fig 5.
Fig 5

Set the frequency of the function generator to 100 Hz.

Adjust the output of the generator to give 1 V rms as read on the meter.

Take the current reading for this voltage.

Reset the voltage output of the generator to 2 V rms.

Record the resultant current.
Repeat this Procedure for voltages of 3 V, 4 V and 5 V rms.
Go to the Results Tables section of this Assignment and copy fig 6 to tabulate your
results.
Work out the magnitude of impedance of the circuit at 100 Hz, by calculating the value at
each voltage step and taking an average of these.
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2.25.6.2 Questions
1. The impedance of the circuit was ....... ohms at a frequency of 100 Hz?
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2.25.7 Practical 2
The circuit that you will be using the same circuit as for the first Practical, reproduced
again in fig 7.
Fig 7
In this Practical you will vary both the applied ac voltage to the circuit and its frequency
and you will measure the resulting currents that flow.
You will determine the impedance of the circuit for the different frequencies.
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2.25.7.1 Perform Practical
You have found that the impedance of a capacitor varies with frequency and that the
impedance of a resistor (resistance) does not vary with frequency. You will now investigate
the variation of a combined CR circuit.

Set the frequency of the generator to 150 Hz.

Take readings of current for voltage settings of 1 V, 2 V, 3 V, 4 V and 5 V rms.

Set the generator to 200 Hz.

Repeat the readings.
Do the same for frequencies of 400 Hz, 800 Hz and 1200 Hz.
Go to the Results Tables section of this Assignment and copy fig 8 to tabulate your
results.

Work out the impedance for each step.
Calculate the average magnitude impedance for each frequency, and tabulate the results
(Practical 2b tab on the worksheet or fig 9 in the Results Tables section).
Plot a graph of impedance against frequency, for the circuit using axes as in fig 10.
Fig 10
1
the theoretical reactance of the capacitor for each frequency
2CR
at which measurements were taken.
Calculate, using Xc =
Fill in the table and compare the measured values with the theoretical ones (fig 11 in the
Results Tables section).
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2.25.7.2 Questions
1. From your plotted graph, does the impedance vary with frequency?
2. From the equation below,, what is the relationship between |Z|, R and
I |Z| =
1
?
C
I2
+ I2R 2
 2 C2
3. Do your average measured impedances agree with those calculated from
|Z| =
2-404
R 2 + X C2
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2.25.8 Results Required
When you have performed this Assignment you should have:

taken readings of ac current and voltage for a series capacitor-resistor circuit at
various frequencies,

determined the phase relationship between current and voltage for the series CR
circuit,

calculated the impedance value at those frequencies,

plotted impedance against frequency to obtain a frequency response for the series
CR circuit.
Your report should contain:

the circuit that you investigated,

the results that you achieved,

the plot of frequency response,

conclusions on your findings in the Assignment.
To produce your report you should use a word processing package.
To achieve the calculated values you could use a spread sheet package.
12-200S
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2.25.9 Practical Considerations and Applications
As well as a phase shifting network, the RC circuit can be used as a frequency dependent
circuit as part of a 'filter'. A filter is a circuit which will pass some frequencies but attenuate
(reject) others.
Consider the circuit in fig 12.
Fig 12
By the potential divider relationship
X
IV2I = IV1I ( C )
Z
=
=
V1
C
1 2
R2 + (
)
C
V1
(CR)2 + 1
Now when CR «1,
IV2I  IV1I
and when CR »1,
IV2I 
V1
CR
At low frequencies the capacitor takes little current, so the volt drop in R is small. At high
frequencies the capacitor short-circuits the output, and most of the input voltage is
dropped across R.
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When
CR = 1,
f=
1
2CR
and at this frequency
|V2| =
V1
2
ie,
0.707 x |V1|
Around the frequency a transition occurs between these conditions, as shown in fig 13.
Fig 13
This type of circuit is known as a low-pass filter, as it passes all frequencies below the
frequency known as the cut-off frequency, fc, and attenuates all those above f c.
Now consider the circuit of fig 14.
Fig 14
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In this case :
IV2I = IV1I
R
Z
V1 R
=
 1 
R2 + 

 C 
=
 V1CR
 RC 2 +
2
1
Now when CR «1,
IV2I  CR IV1I
and when CR »1,
IV2I  IV1I
At low frequencies the high reactance of the capacitor restricts the current flowing in R,
and therefore also the output voltage. At high frequencies the capacitor virtually connects
the output and input directly together.
Similarly around the frequency
1
there is a transition between these two states, as
2CR
shown in fig 15.
Fig 15
2-408
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This type of circuit is known as a high-pass filter, as it passes all frequencies above the
cut-off frequency, fc, and attenuates all those below f c.
Filters are used extensively in electronics. High-pass and low-pass types are both
common, as is a combination of the two types, called a band-pass filter, which passes all
frequencies within a certain band, called the passband, and attenuates all others.
Drawings of a band pass filter and its response are given in figs 16 and 17.
Fig 16
Fig 17
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2.25.10
Results Tables
Frequency 100 Hz
rms voltage
(V)
rms current
(mA)
1
2
3
4
5
Fig 6
2-410
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frequency
(Hz)
voltage
(V rms)
150
1
current
(mA rms)
impedance
(k)
2
3
4
5
200
1
2
3
4
15
400
1
2
3
4
5
800
1
2
3
4
5
1200
1
2
3
4
5
Fig 8
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frequency
(Hz)
ave. impedance
(k)
100
150
200
400
800
1200
Fig 9
frequency
(Hz)
ave. Zmeas
(k)
Xc2
(M)
Xc
(k)
R2
(M)
Ztheory
(k)
100
150
200
400
800
1200
Fig 11
2-412
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2.26 Impedance of a Series LR Circuit Assignment
2.26.1 Objectives

To investigate the impedance of a series LR- circuit to an ac sinusoidal waveform.

To determine how the impedance varies with frequency.
2.26.2 Prerequisite Assignments

The series LR Circuit

Impedance of a Series CR Circuit
2.26.3 Knowledge Level

See Prerequisite Assignments.
2.26.4 Equipment Required
Qty
Apparatus
1
Basic Electricity and Electronics Module 12-200-A
2
Multimeters
OR
Feedback Virtual Instrumentation may be used in place of one
of the multimeters
1
12-200S
Function Generator, 800 Hz - 6.4 kHz 20 V pk-pk sine
(eg, Feedback FG601)
2-413
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2.26.5 Background
Like CR circuits, LR circuits may be used as filter networks.
Consider the circuit of fig 1.
Fig 1
With this circuit:
IV2I =
=
R
R2 + XL2
R
R2 + (2fL)2
IV2I =
. IV1I
. IV1I
| V1|
2
 L 
 R  +1
 
This gives a low-pass response similar to that given in fig 13 in the Impedance of a series
CR circuit assignment.
The inductor has a low impedance to low frequencies (l.f.), thus the circuit behaves as a
shunt resistance at l.f., and at high frequencies (h.f.) the inductor has a high impedance
and attenuates h.f. signals.
The converse of this is shown in the circuit of fig 2.
Fig 2
2-414
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Here :
IV2I =
=
IV2I =
XL
R2 + XL2
L
R2 + (L)2
L
R
 L 
1  
R
2
. IV1I
. IV1I
.IV1I
This is the equation for a high-pass filter. At low frequencies the inductor shunts the output
to earth, whilst at high frequencies it has a high impedance and the circuit appears just as
a series resistor.
The bandpass form of the LR circuit is shown below in fig 3.
Fig 3
12-200S
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2.26.6 Practical 1
The circuit that you will be using is shown in fig 4.
Fig 4
As you can see, the circuit comprises an inductor and a resistor connected in series with
it.
In this Practical you will apply an ac voltage to the circuit and you will measure the
resulting voltage and current.
You will determine the phase relationship between current and voltage.
Connect up the circuit as shown in the Patching Diagram for this Practical.
2-416
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Practical 1 Patching Diagram
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2.26.6.1 Perform Practical
Using a similar technique to that employed in the Impedance of a series CR circuit
assignment, you will now investigate the impedance of a series LR circuit.
Ensure that you have connected up the circuit as shown in the Patching Diagram for this
Practical and that it corresponds to the circuit diagram of fig 5.
Fig 5

Set the frequency of the function generator to 800 Hz, and adjust the sinusoidal
output of the generator to give 1 V rms as given on the meter.

Take the current reading for this voltage.

Reset the voltage output to 2 V rms and record the resultant current.
Repeat this procedure for outputs of 3 V, 4 V and 5 V.
Go to the Results Tables section of this Assignment and copy fig 6 to tabulate your
results.

2-418
Work out the average impedance for the LR circuit at this frequency.
12-200S
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Questions
1. The impedance of the LR circuit was found to be ....... ohms at 800 Hz?
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2.26.7 Practical 2
The circuit that you will be using the same circuit as for the first Practical, reproduced
again in fig 7.
Fig 7
In this Practical you will vary both the applied ac voltage to the circuit and its frequency
and you will measure the resulting currents that flow.
You will determine the impedance of the circuit for the different frequencies.
2-420
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2.26.7.1 Perform Practical

Repeat Practical 1 for frequencies of 1600 Hz, 3200 Hz and 6400 Hz.
Go to the Results Tables section of this Assignment and copy fig 8 to tabulate your
results.
Calculate the average magnitude impedance for each frequency, and tabulate the results
(fig 9 in the Results Tables section).
Plot a graph of impedance against frequency for the circuit using axes as in fig 10.
Fig 10
As :
VR=I R
IVLI = I XL =
L
and,
IVinI = I IZI
Draw a representative phasor diagram for the LR circuit labelled correctly, and derive an
expression relating |Z| to R and XL for the LR circuit.
12-200S
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2.26.7.2 Questions
1.
2.
Derive an expression relating |Z| to R and XL for the LR circuit. The expression is
Z = ...?
What is the corresponding expression for the phase angle,
?
3. Do your average measured impedances agree with those calculated from
IZI =
2-422
R  XL
2
2
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2.26.8 Results Required
When you have performed this Assignment you should have:

taken readings of ac current and voltage for a series inductor-resistor circuit at
various frequencies,

determined the phase relationship between current and voltage for the series LR
circuit,

calculated the impedance value at those frequencies,

plotted impedance against frequency to obtain a frequency response for the series
LR circuit.
Your report should contain:

the circuit that you investigated,

the results that you achieved,

the plot of frequency response,

conclusions on your findings in the Assignment.
To produce your report you should use a word processing package.
To achieve the calculated values you could use a spread sheet package.
12-200S
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2.26.9 Results Tables
Frequency 800 Hz
rms voltage
(V)
rms current
(mA)
1
2
3
4
5
Fig 6
frequency
(Hz)
voltage
(V rms)
1600
1
current
(mA rms)
impedance
(k)
2
3
4
5
3200
1
2
3
4
5
6400
1
2
3
4
5
Fig 8
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frequency
(Hz)
av. impedance
(k)
800
1600
3200
6400
Fig 9
frequency
(Hz)
ave. Zmeas
(k)
XL
()
XL2
(k)
R2
(k)
Ztheory
(k)
800
1600
3200
6400
Fig 11
12-200S
2-425
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Notes
2-426
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2.27 Impedance of a Series LCR Circuit Assignment
2.27.1 Objectives

To investigate the impedance of a series LCR circuit to an ac sinusoidal waveform.

To determine how the impedance varies with frequency.

To compare the impedance of the circuit with those of its constituent components.
2.27.2 Prerequisite Assignments

Impedance of a Series CR Circuit

Impedance of a Series LR Circuit
2.27.3 Knowledge Level

See Prerequisite Assignments.
2.27.4 Equipment Required
Qty
Apparatus
1
Basic Electricity and Electronics Module 12-200-A
2
Multimeters
OR
Feedback Virtual Instrumentation may be used in place of one
of the multimeters
1
12-200S
Function Generator, 20 Hz - 1 kHz 20 V pk-pk sine
(eg, Feedback FG601)
2-427
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2.27.5 Background
From the work you have done on series CR and LR circuits, you know the relationships
between the current through the components and the voltages across them. In a series
LCR circuit the current flowing through each component is the same and can be taken as
a reference in the phasor diagram.
Also, from Kirchhoff's Law, the phasor sum of V R, VL and VC will equal Vin.
Fig 1
The voltage phasor diagram is of the form of fig 1.
You know that:
VR = IR
From fig 1 you can see that
|Vin|2 = VR2 + (|VL| - |VC| )2
From this it might seem reasonable to write
(IZ)2 = (IR)2 + (IXL - IXC)2
This would be quite right if you continued to treat XC as a positive quantity (as you have in
effect done so far by ignoring the question of its sign). However, not every circuit will tell us
so easily which is a reactance of the XL type, in which voltage leads current by 90°, and
which is of the XC type, in which voltage lags current by 90°. To avoid this problem a
convention is established, by which a reactance of the XC type is defined to be negative.
Thus from now on you will write the reactance of a capacitance C as:
XC = -
2-428
1
C
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With this convention reactances can be simply added to find the total effective reactance
in a series circuit, in just the same way as resistances are added for a series circuit. (This
will not upset the results in earlier assignments since the appropriate phase angles have
already been noted, and the amplitude calculations have always involved taking the
square of the reactance).
From the equation for Vin2 you can therefore derive
(I|Z|)2 = (IR)2 + (IX)2
where
X = XL + XC
=
L-
1
C
It follows that
|Z|2 = R2 + X2
ie:
|Z| =
R2 + X2
This formula is of very general application in series circuits. You know that the total
resistance of several resistances in series is their sum: in just the same way the total
effect of several reactances in series is also their sum (provided that the sign convention is
properly observed).
This is easy to verify by starting with resistances R1 R2 etc and reactances X1, X2 etc all in
series and deriving the formula for Z as above.
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2.27.6 Practical 1
The circuit that you will be using is shown in fig 2.
Fig 2
As you can see, the circuit comprises an inductor, a capacitor and a resistor connected in
series.
In this Practical you will apply an ac voltage to the circuit and you will measure the
resulting voltage and current.
You will determine the phase relationship between current and voltage.
Connect up the circuit as shown in the Patching Diagram for this Practical.
2-430
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Practical 1 Patching Diagram
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2.27.6.1 Perform Practical
Now you will investigate a series circuit containing inductance, capacitance and resistance
as in fig 3.
Fig 3
You know, from Kirchhoff's Voltage Law, that the phasor sum of V R, VL and VC will equal
Vin.
Ensure that you have connected up the circuit as shown in the Patching Diagram for this
Practical and that it corresponds to the circuit diagram of fig 4.
Fig 4

Connect the voltmeter across the input to the circuit (points marked P and S on
fig 4) and adjust the generator output to give 4 V rms output at 500 Hz.

Record the resultant current.

Transfer your voltmeter to read the voltage across the resistor (measure across
points P and Q).

Record this voltage.

Measure and record the inductor voltage (points Q and R) and the capacitor voltage
(points R and S).
Go to the Results Tables section of this Assignment and copy fig 5 to tabulate your
results.
Draw to scale, a phasor diagram showing Vin, VR, VL and VC.
2-432
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Questions
1. What is the phase relationship between VL and VC?
2. How do antiphase voltages combine together?
3. What is the resultant voltage of the combination of V L and VC?
4. What is the phase relationship between the resultant of V L and VC and the resistor
voltage VR?
5.
How do you find the phasor sum of the voltages V L, VC and VR?. Draw it..
6.
What is this sum?
12-200S
2-433
BASIC ELECTRICITY AND ELECTRONICS
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7.
What is the phase of Vin with respect of I?
8.
What are the values of XL and XC at 500 Hz?
Chapter 2
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9. Calculate VL from IXL and VC from IXC. .Do these calculations agree with the
measured values?
10. Using |Z| = R2 + X2 ,does this value agree with one calculated from Ohm's Law:
ie, |Z| =
2-434
Vin
?
I
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2.27.7 Practical 2
The circuit that you will be using the same circuit as for the first Practical, reproduced
again in fig 6.
Fig 6
In this Practical you will vary both the applied ac voltage to the circuit and its frequency
and you will measure the resulting currents that flow.
You will determine the impedance of the circuit for the different frequencies.
12-200S
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2.27.7.1 Perform Practical

Reconnect the voltmeter across the input of the circuit (points P and S).

Set the generator output to 4 V rms at 20 Hz and record Vin and I.

Reset the generator to 4 V at 40 Hz and repeat the measurements.
Go to the Results Tables section of this Assignment and copy fig 7 to tabulate your
results.

Do the same for the frequencies listed in fig 7
Calculate |Z| for each frequency, and plot a graph of |Z| against f.

What is the frequency at which |Z| is a minimum?

What is the magnitude of |Z| at this frequency?

Measure VR and VC at the frequency at which the magnitude of the impedance is a
minimum and draw the circuit phasor diagram showing VL, VC, VR and Vin.
A graph showing the curves of XL, XC, R and Z versus frequency is given in fig 8.
Fig 8 Impedance of series RLC circuit
Notice that: at the frequency of minimum impedance XC is equal in magnitude to XL.
This frequency is called the Resonant Frequency (or frequency of resonance) and is
usually given the symbol f0, as shown in fig 8.
2-436
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2.27.7.2 Questions
1. What relationship is there between VL and VC at the frequency at which |Z| is a
minimum?
2. How are Vin and VR related?
3. How do Z and R compare at this frequency?
12-200S
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2.27.8 Results Required
When you have performed this Assignment you should have:

taken readings of ac current and voltage for a series inductor-capacitor-resistor
circuit at various frequencies,

determined the phase relationship between current and voltage for the series LCR
circuit,

calculated the impedance value at those frequencies,

plotted impedance against frequency to obtain a frequency response for the series
LCR circuit,

compared the impedance of the total circuit to that of the individual components.
Your report should contain:

the circuit that you investigated,

the results that you achieved,

the plot of frequency response,

conclusions on your findings in the Assignment.
To produce your report you should use a word processing package.
To achieve the calculated values you could use a spread sheet package.
2-438
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2.27.9 Practical Considerations and Applications
The LCR network can be used as a low-pass, a high-pass, or a band-pass network
depending on which component the output voltage is taken across.
Consider the circuit of fig 9.
Fig 9
Here :
|V2| =
=
XC
2
R2   XL  XC 
1
C
1 

R2   L 


C 
2
This gives a low-pass response, but above the cut-off frequency both L and C influence
the rate of cut-off. At high frequencies the inductor has a high impedance and the
capacitor has a low impedance, thus the attenuation above f c is at a greater rate than for
the CR or LR filter.
2.27.9.1 High-Pass Circuit Form
For a high-pass filter the connection is as in fig 10.
Fig 10
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This gives a high-pass filter response, as at high frequencies the capacitor is virtually
a short circuit, and the inductive reactance is very high.
The circuit acts like a series resistance at h.f.
2.27.9.2 Bandpass Circuit Form
The bandpass circuit form is shown in fig 11.
Fig 11
The centre frequency of the bandpass response for such a circuit will be the resonant
frequency of the circuit.
2-440
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2.27.10
Results Tables
frequency
(Hz)
Vin
V(rms)
VR
V(rms)
VL
V(rms)
VC
V(rms)
I
(mA rms)
500
Fig 5
frequency
(Hz)
I
(mA)
Vin
(V)
Z
()
20
40
60
80
100
160
200
240
300
400
500
600
700
800
900
1000
Fig 7
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2-442
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2.28 Parallel Impedances Assignment
2.28.1 Objectives

To investigate the impedance of parallel connected components to a sinusoidal
alternating current.

To determine the formula for the impedance of parallel connection of components.

To introduce the concepts of admittance, conductance and susceptance.
2.28.2 Prerequisite Assignments

Impedance of a Series LCR Circuit
2.28.3 Knowledge Level

See Prerequisite Assignments.
2.28.4 Equipment Required
Qty
Apparatus
1
Basic Electricity and Electronics Module 12-200-A
2
Multimeters
OR
Feedback Virtual Instrumentation may be used in place of one
of the multimeters
1
12-200S
Function Generator, 20 Hz - 1 kHz 20 V pk-pk sine
(eg, Feedback FG601)
2-443
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2.28.5 Background
So far we have investigated series circuits of RC, RL and RLC, and found formulae for
their impedances, and how these impedances vary with frequency. Let us now look at
parallel connected impedances.
For series connected impedances the reference that is common to all components is that
of current, and the phasor sum of the voltage drops around the loop is zero.
For parallel connected impedances the common reference is voltage, and the phasor sum
of the currents at any node is zero. This is illustrated in fig 1.
Fig 1
The voltage v is applied to all of the impedances Z 1, Z2 and Z3 in common, and
considering instantaneous currents,
i1 + i2 + i3 - i = 0
Fig 2
2-444
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In the circuit of fig 2 the total resistive current is simply IR as marked in the diagram, and is
given by:
V
R
IR =
where R is the resistance, V the voltage.
Fig 3 shows a phasor diagram for fig 2. The total reactive current IX is found by the
phasor addition of IC and IL (taking account of the signs).
Fig 3
The total current is then:
|I|=
2
IR  IX
2
where IR is the total resistive current and IX is the total reactive current.
The reactive current IX is given by:
V
V
V


XL XC X
where XL = L is the reactance of the inductor,
and XC = –
1
is the reactance of the capacitor.
C
X is the combined reactance of the parallel combination.
Notice that this combined reactance of two parallel reactances is found by a similar
formula to that used for parallel resistances,
ie,
12-200S
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

X X1 X2
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This does not mean that resistances and reactances can be mixed in the same formula. It
does make it easy to calculate reactances however, and this is one of the reasons why
capacitive reactance is defined to have a negative value. If it were not so the formula
would be different, and one would have to know whether each of X 1 and X2 were inductive
or capacitive.
If Z is the impedance of the whole parallel combination then:
|I| =
V
Z
Also:
IR =
V
V
and |IX| = ,
R
X
Substitute these in the expression for I gives:
1
1
1


Z
R2 X 2
Now, we call the reciprocal of impedance admittance, given by the symbol Y.
The reciprocal of resistance is called conductance, with the symbol G.
The reciprocal of reactance is called susceptance, the symbol being B.
Thus the above expression becomes:
Like impedance, admittance is a phasor quantity.
The unit of admittance, conductance and susceptance is the siemens (S).
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2.28.6 Practical 1
In this Practical you will investigate the currents that flow in parallel connected impedances
such as shown in fig 4.
Fig 4
The circuit that you will be using is shown in fig 5.
Fig 5
As you can see, the circuit comprises an inductor, a capacitor and a resistor connected in
parallel.
In this Practical you will apply an ac voltage to the circuit and you will measure the
currents in the various branches of the circuit. To do this you will remove the links, in turn,
and replace each with a meter to read the rms ac current in each branch.
Connect up the circuit as shown in the Patching Diagram for this Practical.
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2.28.6.1 Perform Practical
Ensure that you have connected up the circuit as shown in the Patching Diagram for this
Practical and that it corresponds to the circuit diagram of fig 6.
Fig 6

Set the function generator to give an output voltage of 4 V rms at 1600 Hz, as
shown on the meter.

Disconnect link 1 and insert the 0-5 mA milliammeter.

Record the total current, I. Disconnect the 0-5 mA meter and replace link1.

Remove link 2 and connect the meter instead. Record the resistor current, I R.
In similar fashion, measure and record the inductor current, IL, and the capacitor current,
IC.
Go to the Results Tables section of this Assignment and copy fig 7 to tabulate your
results.
Fig 8
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Draw a phasor diagram of the currents IR, IL and IC. Use V as your reference
direction. You can use either the completion of a parallelogram or drawing each
phasor from the tip of the previous one to find the resultant.
Fig 8 combines both techniques in order to show how the reactive currents must be
phasor-added to find the total reactive current; if several resistors were connected in
parallel their currents would be numerically added to find the total resistive current. The
total current is then
|I|=
2
IR  IX
2
where IR is the total resistive current and Ix is the total reactive current.
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2.28.6.2 Questions
1. What is the magnitude and direction of the resultant current, as taken from your phasor
diagram?
2.
How does this compare with the measured value of I?
3. Draw a copy of your phasor diagram and, dividing the current by the voltage, calculate
the admittance corresponding to each phasor. Label the phasors to make a phasor
diagram of admittances (ie, of susceptances, conductance and resultant admittance).
What would be the admittance of a parallel combination of conductances G 1, G2 and
susceptances B1, B2?
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2.28.7 Results Required
When you have performed this Assignment you should have:

taken readings of ac current in the branches of a parallel inductor-capacitor-resistor
circuit at a fixed frequency,

determined the phase relationship between current and voltage for the parallel LCR
circuit,

become acquainted with the concept of the admittance of a component,

plotted a phasor diagram for the parallel LCR circuit.
Your report should contain:

the circuit that you investigated,

the results that you achieved,

the phasor diagram plot,

conclusions on your findings in the Assignment.
To produce your report you should use a word processing package.
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2.28.8 Practical Considerations and Applications
Circuits that include a number of parallel branches are often more easily analysed using
the admittance formulae rather than the impedance formulae.
Calculations should be carried out in terms of the admittance of the branches and then, if
a final impedance value is required, the reciprocal of the final admittance should be taken.
When taking the reciprocal of a phasor quantity it is necessary to find both the magnitude
and the angle of the reciprocal. The magnitude of Z is simply found as the reciprocal of the
magnitude of Y. The phase angle of Z is minus the phase angle of Y.
Fig 9 illustrates this. V and I are the total circuit voltage and current respectively in the two
diagrams shown, but in one diagram V is the reference phasor, and in the other I is the
reference phasor. Since the phase angle is reckoned from the reference to the other
phasor the signs must be different in the two cases.
Fig 9
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2.28.9 Results Table
total current
I (mA)
IR
IL
(mA) (mA)
IC
(mA)
Fig 7
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2.29 Power at ac Assignment
2.29.1 Objectives

To investigate the power dissipated at ac by various components.

To appreciate the meaning of power factor.

To determine if power and power factor are frequency dependent.
2.29.2 Prerequisite Assignments

Power

Series LR Circuit
2.29.3 Knowledge Level

See Prerequisite Assignments.
2.29.4 Equipment Required
Qty
Apparatus
1
Basic Electricity and Electronics Module 12-200-A
1
Multimeter
1
2-Channel Oscilloscope
OR
Feedback Virtual Instrumentation may be used.
1
12-200S
Function Generator, 150 Hz – 1 kHz 20 V pk-pk sine
(eg, Feedback FG601)
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2.29.5 Background
The power dissipated by a component at any instant is given by the product of the voltage
across that component and the current through it, at that instant
ie, P = V I
In an ac circuit both V and I are continuously varying, and P therefore also varies
continuously. What is usually of interest is the average power.
When a resistor and an inductor are connected together in a circuit there will be a phase
shift between the voltage and the current (see The Series LR Circuit assignment). This is
shown in fig 1.
Fig 1
The relationship between the power V, I , and
is given by:
Average Power = V I cos
where V and I are rms values.
The factor, cos
resistance.
is called the power factor of the circuit, and is unity for a pure
For a purely inductive circuit the power factor is 0 and similarly for a purely capacitive
circuit.
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2.29.6 Practical 1
The circuit that you will be using in this Practical is shown in fig 2.
Fig 2
As you can see, the circuit comprises an inductor and a resistor connected in series.
In this Practical you will apply an ac voltage to the circuit and you will measure the input
voltage and the voltage across the resistor. You will use an oscilloscope to do this.
You will measure any phase difference between the two voltages and then draw the
voltage and power waveforms.
You will use the multimeter to measure the rms voltage across the resistor and you will
calculate the power dissipated by the resistor.
Connect up the circuit as shown in the Patching Diagram for this Practical.
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2.29.6.1 Perform Practical
Ensure that you have connected up the circuit as shown in the Patching Diagram for this
Practical and that it corresponds to the circuit diagram of fig 3.

Set the signal generator to sine-wave output. Set the frequency to 150 Hz and the
output amplitude to exactly 100 mV rms, as measured on the digital multi-meter
(ideally, on its 200 mV range).
Fig 3

Display Vin and VR on the two traces (Y1 and Y2) of the oscilloscope and measure
the phase difference between them as accurately as you can.

Measure the voltage across the resistor, VR, with the digital multimeter.
Draw two sine waves on the same axes, with the correct phase relationship, one to
represent the current round the circuit and one the resistor voltage. They do not have to
be to scale.
As P = VI, construct the power curve on the same axes (remember that -V.-I = +P).
You should have drawn a power curve that always is positive and has two cycles for every
cycle of the current or voltage curves. The average power dissipated by a resistor supplied
with an alternating current will thus always be a positive, finite value.
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Fig 4 Purely resistive circuit

Calculate I, from VR and the value of the resistor.

Using PR = VR.I, calculate the power dissipated by the resistor.
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2.29.6.2 Questions
1. What is the phase relationship between the current through and the voltage across a
resistor?
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2.29.7 Practical 2
The circuit that you will be using in this Practical is the same as for the last.
In this Practical you will apply an ac voltage to the circuit and you will examine the voltage
and current in the inductor. You will use an oscilloscope to do this.
You will measure any phase difference between the two voltages.
You will draw the voltage, current and power waveforms for the inductor.
You will also draw the waveforms for the complete circuit and relate the total power
dissipated to the currents and voltages in the circuit.
2.29.7.1 Perform Practical
Now, you will examine the voltage and current in the inductor.

Measure the resistance of the inductor, RL, with the digital multimeter.

Using PL = I2.RL, calculate the power dissipated by the inductor.
As before, draw voltage and current curves for an inductor, showing the relative phase.
Construct the power curve. You should have a power curve which has two cycles for every
cycle of current but which is symmetrical about the voltage axis.
Thus for each cycle of current the average power dissipated is zero.
When a resistor and an inductor are connected together in a circuit there will be a phase
shift between the voltage and the current (see the Series LR Circuit assignment). This is
shown in fig 5.
Fig 5
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Construct the power wave for such a case. You will see that the power wave cuts the
voltage axis, but is not symmetrical about it. The amount by which the power wave cuts
the axis is proportional to Ф.
In fact the relationship between the power V, I , and Ф is given by:
Average Power = V I cos Ф
where V and I are rms values.
The factor, cos Ф is called the power factor of the circuit, and is unity for a pure
resistance. For a purely inductive circuit the power factor is 0 and similarly for a purely
capacitive circuit.

Calculate the total power dissipated in the circuit from: P T = PL + PR

From your earlier measured value of Φ, calculate the total average power given by
V I cos Φ

Compare this with your result from PL + PR.
Ф is also given by the formula
Ф = tan-1 XL/R
where XL = 2fL.

Calculate Ф from this expression

Using that value, calculate V I cos Ф

Compare this with the Experimental values.
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2.29.7.2 Questions
1. What is the phase relationship between the current through and the voltage across an
inductor?
2. What is the average power dissipated by the inductor in each cycle of current?
3. Theory says that the average power dissipated by an inductor is zero. Does this agree
with your experimental findings? If not, why?
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2.29.8 Practical 3
The circuit that you will be using in this Practical is the same as for the last.
In this Practical you will repeat the procedure of the first two Practicals, but at a different
frequency.
2.29.8.1 Perform Practical
Repeat the Procedures of Practicals 1 and 2 for a 1 kHz, 100 mV input to the circuit.
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2.29.8.2 Questions
1. Do the voltages measured at 1 kHz differ significantly from those found for 150 Hz?
2. Do the powers measured at 1 kHz differ significantly from those found for 150 Hz?
3. Do the powers measured at 1 kHz comply with the same rules as those found for
150 Hz?
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2.29.9 Results Required
When you have performed this Assignment you should have:

taken readings of ac voltage across the components of a series LR circuit,

determined the phase relationship between voltages measured,

drawn the voltage, current and power waveforms for the resistor in the circuit,

drawn the voltage, current and power waveforms for the inductor in the circuit,

determined if there are differences in results for two frequencies of operation.
Your report should contain:

the circuit that you investigated,

the results that you achieved,

the waveform plots,

the comparison for different frequencies,

conclusions on your findings in the Assignment.
To produce your report you should use a word processing package.
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2.29.10 Practical Considerations and Applications
Power at ac is given by the expression
P = V I cos Φ watts
the units of power are the watt (W) and the kilowatt (kW), 1000 W.
The product V I is called the apparent power, and may be designated by the symbol S.
The units of S are volt-ampere (VA) and kilovolt-ampere (kVA).
The product V I sin Ф is called the reactive power (or wattless power) and is indicated by
the symbol Q.
The units of Q are also volt-ampere (VA) and kilovolt-ampere (kVA), although it is common
practice to add the letter R to distinguish reactive power. Thus while S is always
expressed in VA and its multiples, Q may be expressed in VAR and its multiples.
A power phasor triangle may be constructed as in fig 6.
Fig 6
Only the real power, P, is taken as energy from the supply.
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2.30 Series Resonance Assignment
2.30.1 Objectives

To investigate further the series LCR circuit, especially around the point of
minimum impedance.

To determine the resonant frequency for a series LCR circuit.

To see the effect of the resistance on the response of the circuit.

To determine the Q of the circuit.
2.30.2 Prerequisite Assignments

Impedance of a Series LCR Circuit
2.30.3 Knowledge Level

See Prerequisite Assignments.
2.30.4 Equipment Required
Qty
Apparatus
1
Basic Electricity and Electronics Module 12-200-A
2
Multimeters
OR
Feedback Virtual Instrumentation may be used in place of one
of the multimeters
1
12-200S
Function Generator, 100 Hz – 1 kHz 20 V pk-pk sine
(eg, Feedback FG601)
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2.30.5 Background
You have seen from the Impedance of a Series LCR Circuit assignment that the
impedance of a series LCR circuit varies with frequency, and that the form of variation is
given in fig 1.
Fig 1
The frequency at which the circuit impedance in a series LCR circuit is a minimum is
termed the resonant frequency of the circuit.
At the resonant frequency of the circuit the capacitive reactance is equal to the inductive
reactance, and they cancel each other out. The circuit impedance at the resonant
frequency is thus just the resistance of the circuit.
The formula for XC is:
XC =
1
2fC
and for XL:
XL = 2fL
at resonance |XC| = |XL|.
Thus, at resonance,
1
= 2fL
2fC
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If the resonant frequency is denoted by the symbol f o.
Then,
1
= 2foL
2foC
fo =
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2 LC
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2.30.6 Practical 1
The circuit that you will be using in this Practical is shown in fig 2.
Fig 2
As you can see, the circuit comprises an inductor, a capacitor and a resistor connected in
series.
In this Practical you will apply an ac voltage to the circuit and you will measure the input
voltage and the current through the circuit.
You will vary the frequency of the input signal between about 100 Hz to 300 Hz and note
how the current varies.
Connect up the circuit as shown in the Patching Diagram for this Practical.
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2.30.6.1 Perform Practical
Ensure that you have connected up the circuit as shown in the Patching Diagram for this
Practical and that it corresponds to the circuit diagram of fig 3.
Fig 3

Connect the voltmeter across the input to the circuit (points marked P and S on
fig 3) and adjust the generator output to give 4 V rms at 200 Hz.

Vary the frequency of the generator from about 100 Hz to 300 Hz and note the
variation in current and voltage.

Find the frequency at which the circuit impedance is a minimum. This is where the
current is a maximum and the voltage is a minimum.

What is this frequency?
This frequency is termed the resonant frequency of the circuit

Measure the current and voltage at resonance and calculate the impedance at
resonance.

Does the calculated value agree with the value expected?
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2.30.6.2 Questions
1. If
1
= 2foL From this derive an expression for fo in terms of L and C. fo = ............?
2foC
2. Substitute L = 100x10-3H (100 mH) C = 2.2 x 10-6F (2.2 µF) in your expression and
work out fo. Does your calculated value agree with the frequency found previously for
minimum impedance?
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2.30.7 Practical 2
In this Practical you will be using the same circuit as for the first Practical.
You will vary the signal frequency over a wider range, in a series of steps, and measure
the resulting circuit current, inductor voltage and capacitor voltage at each frequency.
You will plot your results on in graphical form with a logarithmic frequency scale.
You will then repeat the procedure for the circuit with the resistor removed.
2.30.7.1 Perform Practical

Set the generator frequency to 20 Hz and the output voltage to 4 V rms. The
generator frequency may be taken from the dial if the calibration is sufficiently
accurate, or a digital frequency meter used for more accuracy.

Take readings of the circuit current |I|, the inductor voltage |VL|, and the capacitor
voltage |VC|.
Go to the Results Tables section of this Assignment and copy fig 4 to tabulate your
results.

Increase the frequency to 40 Hz, and readjust the generator output to give 4 V rms
again.

Take readings of V, I, VL and VC as before.
Repeat the procedure for the frequencies given in the table.
Ensure that the input voltage is 4 V rms for each frequency setting.

Find the resonant frequency again (the frequency at which the current is a
maximum) and take a set of readings at this frequency, f o.
On a sheet of 2 cycle logarithmic graph paper, draw curves of I, VL and VC against
frequency, using the axes shown in fig 5.
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Fig 5
Calculate:
ω0 L
R
for your circuit and compare it with the value found for Q from the graphs.
Calculate:
1
ω0 CR
and compare it with the graphical Q.
Now examine your graphical results for the circuit with R removed.

Remove the resistor from the circuit and connect points P and Q together.

Set the generator frequency back to 20 Hz, but now set the output amplitude to
1.0V rms as shown on the meter.
Note: as you have removed the resistor from the circuit a lower voltage will be required to
obtain a reasonable flow of current in the circuit. This is why the generator amplitude has
been set to 1 V.

Take measurements as before, and copy another table as in fig 4 (but with Vin
1 V), using the same frequencies, but always ensuring that the generator output
is 1.0 V rms.
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Go to the Results Tables section of this Assignment and copy fig 4 to tabulate your
results.
Plot the second set of I, VL and VC curves, on axes as used before for fig 5.
Notice the differing shapes of the two sets of curves.

From the curves plotted with R = 1 kΩ in circuit, what is the resonant
frequency of the circuit (where |I| is a maximum)?

What is the value of |I| at this frequency?

What is the value of |VL| at fo?

What is the value of |VC| at fo?

Do they have any relationship to each other, if so why?

What is the ratio of |VL| or |VC| at fo to the input voltage?
The ratio of V L
V in

2-478
or V C at fo is termed the Quality Factor, or Q, of the circuit.
V in
What is the Q of your circuit?
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2.30.7.2 Questions
1. Examine your graphical results for the circuit with R removed What is the resonant
frequency?
2.
Does it differ from fo when R = 1 k
?
3. What are the value of I, VC, and VL, at fo?
4. What is the Q of this circuit, calculated from the graphs?
5. Using Q =
ω oL
1
or
, what is the calculated Q of the circuit?
ω o CR
R
6. Is the Q of the circuit when R = 0 higher or lower than that when R = 1 k
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7. With R = 0 the calculated value of Q is infinite. Obviously this is unreasonable, as there
would then be an infinite voltage across the inductor and the capacitor. Where do you
think resistance is present to limit the Q to the value obtained experimentally?
8. What would the value of this resistance have to be?
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2.30.8 Results Required
When you have performed this Assignment you should have:
 determined the frequency of minimum impedance for the circuit,
 determined the resonant frequency of the circuit,
 measured the currents and voltages in the circuit for a number of frequencies for
resistance values of 1 kΩ and zero,
 plotted the voltage and current curves with frequency for these circuits,
 determined the Q of the circuits.
Your report should contain:

the circuits that you investigated,

the results that you achieved,

the plots of current and voltage against frequency,

the calculations of Q,

conclusions on your findings in the Assignment.
To produce your report you should use a word processing package.
To achieve the calculated values you could use a spread sheet package.
12-200S
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2.30.9 Practical Considerations and Applications
The phenomenon of resonance is found in many branches of physics when the physical
properties of a system allow oscillations to occur much more severely at one particular
frequency, so that when the system is excited by some outside source at this frequency
the vibrations build up to a large amplitude.
You may have seen pictures of the bridge at Tacoma Rapids, USA, as it shattered itself.
This was due to the resonant effect excited by the wind. The bridge started to vibrate due
to the wind, which was of such a velocity to reinforce the vibration, and the whole bridge
became resonant, and literally whipped itself into fragments.
In electrical circuits resonance occurs when there are both inductance and capacitance in
circuit at the frequency at which the inductive reactance is equal to the capacitive
reactance. At resonance the response of the circuit is only limited by the losses of the
circuit due to resistance, etc.
There are several sources of loss in a resonant circuit. Power losses in an inductor are:
copper loss, due to the resistance of wire; losses due to hysteresis and eddy current
losses in the core; losses due to the induction of currents in screening cans and objects in
close proximity, causing eddy currents to be set up in these, and dissipating power.
At high frequencies the coil former may show appreciable dielectric loss also. Losses due
to the capacitor are: dielectric losses, and resistance due to the capacitor plates.
The full equivalent circuit of a series tuned circuit, representing the loss mechanisms by an
equivalent resistor, is given in fig 6.
Fig 6
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These losses are frequently represented in terms of a single equivalent series or parallel
resistance for each component.
The tuned circuit (resonant circuit) is used extensively in electronics for its frequency
selective properties. It can be seen from your graphs that the impedance of the series
tuned circuit varies widely with frequency, and that the impedance is a minimum at the
resonant frequency.
The series resonant circuit will thus form an acceptor circuit which will pass frequencies
near to the resonant frequency and attenuate other frequencies.
If the series resonant circuit is connected across a signal line, as in fig 7, it will shunt to
earth signals of frequency near its resonant frequency. This facility is useful if it is wished
to cut out a particular interfering signal, and a series tuned circuit could be used.
Fig 7
The degree to which a resonant circuit responds to one frequency rather than another is
termed its selectivity. From your results it should be seen that circuits with a high Q have
also high selectivity and vice-versa. Normally, high selectivity is desired, and thus resonant
circuits should be designed to have the highest Q (ie, lowest losses) possible.
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2.30.10
Results Tables
frequency
(Hz)
input voltage
Vin (V)
current I
(mA)
VC
(V)
VL
(V)
20
40
60
80
100
160
200
210
220
230
240
250
300
400
500
600
700
800
900
1000
f0 =240
Fig 4
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2.31 Parallel Resonance Assignment
2.31.1 Objectives

To investigate further the parallel LCR circuit, especially around the point of
maximum impedance.

To determine the resonant frequency for a parallel LCR circuit.

To see the effect of the resistance on the response of the circuit.

To determine the Q of the circuit.
2.31.2 Prerequisite Assignments

Parallel Impedances

Series Resonance
2.31.3 Knowledge Level

See Prerequisite Assignments.
2.31.4 Equipment Required
Qty
Apparatus
1
Basic Electricity and Electronics Module 12-200-A
1
2-channel oscilloscope
OR
Feedback Virtual Instrumentation may be used in place of the
oscilloscope
1
12-200S
Function Generator, 100 Hz – 1 kHz 20 V pk-pk sine
(eg, Feedback FG601)
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2.31.5 Background
Fig 1
Referring to fig 1, at resonance the inductive reactance is numerically equal to the
capacitive reactance, thus IC and IL will be equal and opposite. They will cancel each
other out, as shown in the phasor diagram of fig 2.
Fig 2
The resultant current at resonance is therefore only that required to supply the resistance,
R.
Referring to fig 1 at resonance
|IC| = |IL| thus IR = I at fo.
For a parallel circuit you define the Magnification Factor Q as the ratio of reactive current
to resistive current, at resonance:
ie:
Q = IL  IC
I
I
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Now
IL 
V
XL
and I 
V
R
V R R
 IL  . 
I
XL V XL
Q=
R
ω0 L
Similarly:
Q = 0CR
Compare these expressions with those found for Q in the series resonant circuit.
Remember, in the series case the R is the series resistance in the circuit, in the parallel
case the R is the parallel resistance.
For clarity you may denote these by Rs and Rp for series and parallel resistance
respectively.
The equations then become:
Series circuit:
Q=
 oL
1
=
 o CRs
Rs
Q=
Rp
= o CRp
 oL
Parallel circuit:
The formula for the impedance of a parallel RLC circuit is given by:
1
=
Z
1
1
(from Parallel Impedances)

Rp2 X 2
But at resonance the reactive terms cancel each other out, and the total resultant
reactance is zero.
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at fo
1
=
Z
1
1
=
Rp2 Rp
Z = Rp
In the case when there is no parallel resistance connected across the capacitor and
inductor then Rp =
(infinity).
There must therefore be some resistance present to limit Q and Z.
An equivalent circuit may be drawn showing all the parallel resistances. This is shown in
fig 3.
Fig 3
All these parallel paths cause a reduction in Q and in the impedance at resonance.
The inductor also has resistance. This may be represented as in fig 4.
Fig 4
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r represents the internal resistance of the inductor.
An appreciable internal resistance in the inductor also drastically reduces the Q of a
parallel tuned circuit.
The resonant frequency of a parallel tuned circuit can be shown to be:
2
1
1 r
f0 =
 2
2π LC
L
which is approximately:
f0 
1
2 LC
for small values of r2.
Note that this is the same formula as for resonant frequency of a series tuned circuit.
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2.31.6 Practical 1
The circuit that you will be using in this Practical is shown in fig 5.
Fig 5
As you can see, the circuit comprises an inductor and a capacitor connected in parallel.
The 10 kΩ resistor is present to increase the output resistance of the function generator.
Without this resistor in circuit the low resistance of the generator would lower the Q so
much (see the Background section for an explanation) that the results achieved would be
meaningless.
In this Practical you will apply an ac voltage to the circuit and you will measure the input
voltage and the voltage across the parallel LC circuit.
You will vary the frequency of the input signal between about 100 Hz to 1 kHz and note
how the voltages vary.
Connect up the circuit as shown in the Patching Diagram for this Practical.
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Practical 1 Patching Diagram
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2.31.6.1 Perform Practical
The full circuit to be investigated in this Assignment is shown in fig 6. However, you will
start with the simpler case where R is omitted, as shown in fig 7.
Fig 6
Ensure that you have connected up the circuit as shown in the Patching Diagram for this
Practical and that it corresponds to the circuit diagram of fig 7.
Fig 7

Set the function generator to give an output, V1, of 6 V pk-pk at a frequency of
100 Hz.
Move the oscilloscope set to :

Y1 channel (V1) to 1 V/cm

Y2 channel (V2) to 200 mV/cm

Timebase to 1 ms/cm
Vary the frequency of the generator slowly from 100 Hz to 1 kHz, and notice the variation
of the two voltages shown on the oscilloscope.
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
Does the voltage output of the generator vary appreciably?

Does the voltage across the capacitor and inductor vary?

What is the relationship between V1, V2 and I (fig 7)?

Does I vary with frequency?
Set the generator to the frequency that gives a minimum in I (maximum V2) and measure
the frequency. This is the frequency of resonance (alternatively called the resonant
frequency).

What is the resonant frequency?

Is V2 a maximum or minimum at the resonant frequency of the circuit?

Is I a maximum or a minimum at fo?
Calculate the current I at resonance from:

I = V1 V 2
10k 
Since the oscilloscope is being used for measurements, it will be found convenient to work
in peak-to-peak values of voltage and current throughout. The values of impedance
calculated will be the same as if rms values were used, provided that the same kind of
measure is used for both voltage and current.
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2.31.6.2 Questions
1. What is the impedance of the parallel LC circuit at resonance? (Find this from V2/I)
2. Is the impedance high or low at resonance?
3. How does this compare with the series resonant circuit?
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2.31.7 Practical 2
The circuit that you will be using in this Practical is the same as for the first one.
You will vary the frequency of the input signal in steps between 150 Hz and 1 kHz and
measure the voltage across the LC circuit at each frequency.
You will calculate the current and the impedance at each frequency and then plot a curve
of impedance against frequency for the circuit.
2.31.7.1 Perform Practical
Ensure that you have connected up the circuit as shown in the Patching Diagram for the
first Practical and that it corresponds to the circuit diagram of fig 8 (this is the same as for
Practical 1).
Fig 8

Set the generator frequency to 150 Hz and the output amplitude to give 8 V pk-pk.
If the dial accuracy of the generator used is not thought sufficient, a digital frequency
meter may be used for greater accuracy.

Measure the voltage across the parallel LC circuit, V2.
Go to the Results Tables section of this Assignment and copy fig 9 to tabulate your
results.

Increase the generator frequency to 200 Hz, and reset the output amplitude to 8V
pk-pk.

Measure and record the resulting V2.
Repeat this Procedure for frequencies of, 250, 300, 350, 400, 450, 500, 550, 600, 700,
800, 900 and 1000 Hz.
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Ensure that the amplitude of V1 remains constant for each frequency setting.

Find the resonant frequency again and take a set of readings at f o.

Calculate I and Z for each step, and enter your results in the appropriate spaces.
On a sheet of single-cycle logarithmic graph paper, draw a curve of Z against frequency,
using the axes shown in fig 10.
Fig 10
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2.31.8 Practical 3
The circuit that you will be using in this Practical is virtually the same as for the first one,
except that you will connect a resistor in parallel with the LC circuit. This is to simulate the
effect of having losses in the circuit and thus a lowering of Q.
As before, you will vary the frequency of the input signal in steps between 150 Hz and
1kHz and measure the voltage across the LC circuit at each frequency.
You will calculate the current and the impedance at each frequency and then plot a curve
of impedance against frequency for the circuit.
You will then compare this curve with the one plotted in Practical 2, without any resistor in
circuit.
2.31.8.1 Perform Practical
Now connect a 1 kΩ resistor between points A and B, in parallel with the resonant circuit,
as shown in fig 11.
Fig 11

Set the generator frequency back to 150 Hz and the output to give 8 V pk-pk as
before, and measure and record the resultant V 2.
Repeat the Procedure for the same frequency steps as in Practical 1, and draw the
impedance curve on the same piece of graph paper as used for the previous part of the
assignment.

Notice the different shapes of the two impedance curves.

Determine, from your curves, the bandwidths of the two circuits.
Using the expression relating Q with bandwidths and f o, determine the Q values of the two
circuits.
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2.31.8.2 Questions
1. Which circuit (Practical 2 or Practical 3) has the higher Q?
2. What is the resonant frequency of the circuit with R = 1 k  inserted?
3. Does this differ from when R was excluded?
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2.31.9 Practical 4
In this Practical you will investigate the effect of the internal resistance of the inductor on
the Q and impedance by exaggerating the value of r by adding resistance in series with
the inductor.
The circuit that you will be using in this Practical is shown in fig 12.
Fig 12
As you can see, it is virtually the same circuit as investigated in Practical 2, except that
there is a resistor in series with the inductor in the circuit.
As before, you will vary the frequency of the input signal in steps between 150 Hz and
1kHz and measure the voltage across the LC circuit at each frequency.
You will calculate the current and the impedance at each frequency and then plot a curve
of impedance against frequency for the circuit.
You will then compare this curve with the ones plotted in the other Practicals.
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2.31.9.1 Perform Practical

Remove the 1 kΩ resistor and connect a 100Ω resistor between points C and D as
shown in fig 13.
Fig 13
As previously, take readings of voltage for the same frequencies between 150 Hz and
1 kHz, and record them in another copy of the results table as in fig 6.

Draw the impedance curve on the same sheet of graph paper as before.

Find the resonant frequency of the circuit.
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2.31.9.2 Questions
1. How does the Q of the circuit tested in this Practical compare with those of the other
circuits?
2. What would you say must be done if a high Q circuit is to be achieved?
3. Is the resonant frequency the same as for the other two circuits?
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2.31.10 Results Required
When you have performed this Assignment you should have:

determined the frequency of maximum impedance for the circuit,

determined the resonant frequency of the circuit,

measured the currents and voltages in the circuit for a number of frequencies for
resistance values of 1kΩ and zero,

plotted the voltage and current curves with frequency for these circuits,

determined the Q of the circuits.
Your report should contain:

the circuits that you investigated,

the results that you achieved,

the plots of current and voltage against frequency,

the calculations of Q,

conclusions on your findings in the Assignment.
To produce your report you should use a word processing package.
To achieve the calculated values you could use a spread sheet package.
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2.31.11 Practical Considerations and Applications
The points concerning losses in a resonant circuit which were enumerated in the Practical
Considerations and Applications notes in the Theory section of the Series Resonance
assignment apply equally for parallel tuned circuits as for series circuits. However the
equivalent circuit becomes as in fig 14.
Fig 14
Again, these are normally represented by a single parallel loss resistance, Rp.
The difference in resonant frequency caused by the taking into account the series
resistance of the coil is normally very small. This is because:
But:
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Normally Q is in the region 10 to 200. With a typical value of Q = 50 the term
1
2
makes
Q
the equation
inaccurate by only one part in 2500,
so
with an accuracy in this case of one part in 5000.
This is the same equation as used in the series resonance case, and in practice provides
a very close approximation to the true f o.
The impedance of a parallel tuned circuit is at its highest at resonance. Thus the circuit is
sometimes referred to as a 'rejector circuit'.
The parallel resonant circuit is perhaps more often used than the series circuit, and it is
commonly used to provide a frequency dependent load across which a high voltage will be
present at resonance, but a low voltage at frequencies away from resonance. This is
shown in fig 15.
Fig 15
The parallel tuned circuit is used extensively in radio receivers and transmitters, and many
other types of equipment.
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2.31.12 Results Table
frequency
(Hz)
V1
(Vp–p)
V2
(Vp–p)
V1–V2
(Vp–p)
I
(mA p–p)
Z
()
150
200
250
300
350
400
450
500
550
600
700
800
900
1000
Fig 9
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Notes
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2.32 The Transformer Assignment
2.32.1 Objectives

To investigate the effects when two coils are wound in proximity on the same
former.

To understand the Transformer Principle.

To determine the relationships between currents and voltages in the two coils.

To see how the turns ratio of the two coils determines the current and voltage
relationships.
2.32.2 Prerequisite Assignments

Electromagnetic Induction

Inductance
2.32.3 Knowledge Level

See Prerequisite Assignments.
2.32.4 Equipment Required
Qty
Apparatus
1
Basic Electricity and Electronics Module 12-200-A
2
Multimeters
OR
Feedback Virtual Instrumentation may be used in place of one
of the multimeters
1
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Function Generator, 100 Hz – 5 kHz 20 V pk-pk sine
(eg, Feedback FG601)
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2.32.5 Background
In the Electromagnetic Induction assignment you saw that, if you have two coils wound on
the same core and a current was passed through one coil, then a current would flow in the
second coil whenever the current in the first coil was varied.
This was called Electromagnetic Induction.
This phenomenon was explained by showing that a current carrying conductor has around
it a magnetic field which is proportional to the magnitude of the current flowing. Thus as
the current changes so does the magnetic field surrounding the conductor.
The second coil was in the influence of this magnetic field, and it was shown that an emf is
generated in a conductor which is within the influence of a changing magnetic field.
In the Inductance assignment you explored the idea of this induced emf being a BackEMF and that its magnitude was proportional to the rate of change of the magnetic flux
linking the conductor.
ie, for a conductor,
e=
-d
dt
where Ф is magnetic flux.
When a conductor is wound into a coil each turn of that coil will cut the magnetic field, thus
the emf induced by each turn will be
e=
-d
dt
however, if there are N turns in the coil, then the total emf induced will be N times that for
one turn.
e total = - N
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Consider the circuit of fig 1.
Fig 1
If an alternating voltage is applied to the primary coil P of the circuit in fig 1, then an
alternating current will flow in the coil.
This alternating current will give rise to an alternating magnetic field in the core, shown by
the dotted lines.
The alternating magnetic field will link with the turns of the secondary coil S, and cause an
emf to be induced in them.
This induced alternating emf will cause an alternating current to flow in the load in the
secondary circuit.
This principle is know as the Transformer Principle.
2.32.5.1 Relationship between voltages and turns ratio
The voltage across the primary is Vp volts. If there are Np turns on the primary winding, the
Vp
voltage per turn on the primary will be
volts.
Np
Now a magnetic field will be set up by the primary winding. Assume all the magnetic flux
set up by the primary cuts the secondary turns. Thus an emf will be induced in the
secondary turns by this flux.
As an emf of
Vp
volts per turn in the primary circuit gives rise to a flux, say Ф, and all this
Np
flux cuts the secondary turns, then the flux Ф will induce an emf of
Vp
volts per turn in the
Np
secondary circuit.
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If there are Ns turns on the secondary winding, the total secondary
induced emf will thus be Ns times the voltage per turn.
ie,
V s = Ns .
Vp
.
Np
Thus:
Vp
Vs
=
Np
Ns
ie, the ratio of primary voltage to secondary voltage is equal to the ratio of primary turns to
secondary turns.
It is impossible to design a transformer in which all the flux generated by the primary
winding cuts the secondary winding. This expression is therefore only approximate but
nevertheless it is a very useful and widely used expression.
2.32.5.2 Relationship between currents and turns ratio
With the resistive load connected, the power in the secondary circuit P s, is given by:
Ps = Is Vs
A primary current flows to supply this load, and the power in the primary circuit, P p, is
given by:
Pp = Ip Vp
Normally the power to supply the load is very much greater than the small amount of
power that is needed to set up the magnetic flux in the core, thus generally you can say:
Ps = P p
 Is Vs = Ip Vp
Vp
Vs
=
Is
Ip
Np
Is
=
Ns
Ip
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2.32.5.3 Relationship between impedance and turns ratio
For the circuit of fig 2 the input resistance of the transformer is given by:
Rin =
Vp
Ip
Fig 2
Rin =
Vp
Ip
Vs .
=
Np
Ns
N
Is . s
Np
 Np 
V
= s .  
Is
 Ns 
2
But:
Vs
= RL
Is
 Np 
Rin = RL .  
 Ns 
2
Thus the transformer changes the impedance as seen by the source.
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2.32.6 Practical 1
The circuit that you are going to investigate is shown in fig 2, below.
Fig 2
The component designated by P and S is a transformer. The winding of the transformer
designated P is the primary of the transformer. The winding designated S is the
secondary of the transformer.
You will apply a number of different magnitudes of input voltage to the transformer and
you will measure the primary and secondary voltages and the primary currents that result.
You will use Ohm’s Law to calculate the secondary currents flowing in each case.
Connect up the circuit as shown in the Patching Diagram for this Practical.
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Practical 1 Patching Diagram
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2.32.6.1 Perform Practical
Ensure that you have connected up the circuit as shown in the Patching Diagram and that
it corresponds with the circuit diagram of fig 3.
Fig 3

Set the generator output to 2.0 V rms, 3.5 kHz and measure the voltage across the
primary coil, P, and the primary current.

Transfer the voltmeter to the secondary coil, S, and measure the voltage across
this.
Knowing the secondary voltage and the load resistance, calculate the secondary current.
Copy the results table as shown in fig 4, found in the Results Table section of this
assignment, and tabulate your results.

Repeat these measurements and calculations for input voltages of 3 V rms,
4 V rms and 5 V rms.
Record all of your results.
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2.32.6.2 Questions
1. What can you say about the ratio
2. What happens to the ratio
Ip
Is
Vp
Vs
for varying primary voltages?
?
3. Can you see any relationship between
4. From your calculations of
Vp
Vs
Vp
Vs
and
Ip
Is
?
, find the average voltage ratio, and thus the turns ratio of
the transformer. Does this agree with the 1:1.68 ratio of the transformer on the
module?
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2.32.7 Results Required
When you have performed this Assignment you should have:

measured the currents and voltages in the circuit for a number of input voltages,

determined the voltage ratio of the transformer,

determined the current ratio for the transformer,

determined the turns ratio of the transformer.
Your report should contain:

the circuits that you investigated,

the results that you achieved,

the calculations for turns ratio,

conclusions on your findings in the Assignment.
To produce your report you should use a word processing package.
To achieve the calculated values you could use a spread sheet package.
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2.32.8 Practical Considerations and Applications
A transformer consists essentially of two or more windings of wire inductively coupled;
these are called the primary and secondary windings. The windings may be wound on a
magnetic core, which ensures that there is high magnetic flux linkage between the
windings.
An alternating voltage across one winding will induce an alternating voltage in the other
winding due to the changing flux linkages. The number of turns on the winding determines
the induced voltage.
The secondary voltage being less than the primary voltage if the number of turns on the
secondary is less than the number of turns on the primary.
The presence of the core intensifies the magnetic flux. The ratio of the flux density
produced with a certain core material, to the flux density produced with an air core under
the same conditions, is known as the relative permeability of the core material.
In addition to the transformation of voltages, a transformer may be used as a means of
isolating parts of a circuit from each other as far as dc is concerned, whilst allowing the
parts to behave as a single circuit with respect to the alternating voltages.
Also, the impedance across one winding may be transformed to appear as a different
value across other windings, i.e. impedance matching may be effected using a
transformer.
The main parts of the transformer are the magnetic core, coil former, wire for the windings,
and the materials used as wire and winding insulation. The transformer may be housed in
a metal case which, in addition to affording mechanical protection for the windings, acts as
an electromagnetic or electric screen.
The relationship between the various circuit conditions is as follows:
In the ideal transformer, if
N1 = Number of turns on primary
N2 = Number of turns of secondary
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V1
= Voltage across primary
V2
= Voltage across secondary
I1
= Current in primary
I2
= Current in secondary
Z1
= Impedance across primary
Z2
= Impedance across secondary
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Then:
I
N1
V
= 1 = 2=
I1
N2
V2
Z1
Z2
2.32.8.1 The Core
Two types of core in common use are made up of thin wafers of magnetic material
insulated from each other known as laminations, and are described as being of shell-type
or core-type construction. The shapes are illustrated in fig 5.
Fig 5
The reason for the use of a laminated core instead of a solid magnetic core is to reduce
the eddy current losses to a low value. Eddy currents are circulating currents induced in
the core material itself and are the result of the varying magnetic linkage between the
primary winding and the core; the core behaving as a secondary winding.
These eddy currents reduce the efficiency of the transformer. The use of insulated
laminations prevents the currents passing across the core and limits them to the
comparatively high resistance path of each lamination.
Most transformers in telecommunications circuits are comparatively small and the shelltype which requires only one coil former is extensively used. The magnetic circuit is
divided into two parallel paths, half the flux enclosing one side of the coil, and half the
other side. The centre limb on which the coil is mounted is twice as wide as each of the
outer limbs.
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For some applications, such as high power transformers, the core-type construction is
preferable. In this type of construction there is only one magnetic path, with two coils, one
on each limb of the core.
The cores are made up of a set of laminated strips, usually varnished on one side, and
clamped firmly together.
Shell-type transformers may be made up of laminations of the form shown in fig 6.
Fig 6
A partly assembled core using E and I laminations is shown in fig 7.
Fig 7
Alternate stacking of the lamination pairs may be used and this gives a good interleaved
joint between laminations. For instance, if the arrangement of the first layer is as fig 7(b)
then the subsequent odd numbered layers will be arranged likewise. The even numbered
layers will be arranged as fig 7(c).
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By this method it is possible to increase the efficiency of the magnetic path whilst still
preserving the conditions for low eddy current losses.
Illustrated in fig 8 are typical shapes of laminations used in the core-type transformer.
Alternate stacking of pairs may again be used when these laminations are built up to form
a core.
Fig 8
In recent years cores wound from continuous steel strip have come into use, and are
known as C-type cores. The steel strip is wound on a mandrel to the required size then
annealed, impregnated with bond to prevent the coils splaying apart, and cut in two to
permit assembly with the coil.
The faces of the core at the point of cutting are machine finished to ensure that when
assembled they abut with the minimum air gap between them. If a gap is required, the
requisite thickness of insulating material is inserted between the faces. A steel band is
fitted around each loop to hold it together.
In the shell-type construction shown in fig 9 four core elements are required to make up
the complete core. Other sizes of core elements suitable for smaller transformers are also
illustrated.
Fig 9
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2.32.8.2 Core Materials & Windings
Core Materials
The cores of transformers are made from a large variety of steels depending upon the
purpose for which the transformer is intended. Some of the steels and their chief uses are
as follows:
Silicon steel has a permeability of 8000 and is used on most power and audio-frequency
transformers.
High permeability nickel alloy steel and Mumetal, with relative permeabilities of up to
80,000 and 200,00 respectively, are used on small wide-band audio-frequency
transformers.
Radiometal, which is a nickel-iron alloy with a maximum relative permeability of 20,000, is
used for audio and lower radio-frequency transformers.
Rhometal, a nickel-iron alloy with high resistivity and a maximum relative permeability of
5000, is also used for lower radio-frequency transformers.
Ferrites are magnetic ceramic materials usable up to several MHz.
Windings
The size of wire used for the coil winding is often estimated in terms of the current it is to
carry. For instance, some manufacturers use the rule, 3 A/mm2 as a basis for the first
choice of wire. Temperature rise, and regulation determine the final choice.
The space occupied by the wire, especially with small wires sizes, depends upon the type
of insulation used as well as the cross-sectional area of the conductor.
Space can be saved by using, whenever possible, enamelled wire instead of cotton or silk
covering.
In the winding process the start lead is placed on the coil former, suitable insulation placed
over it, and the first layer of turns for the first winding is wound over the insulation.
Dependent upon the design requirements insulating material may be placed over this layer
of turns and between subsequent layers. Individual windings are separated as required
with insulating material, and if necessary with electric screens. The coil leads to each
winding are carefully insulated, and one or more layers of insulating material are wound
over the entire coil and lead insulation.
As stated earlier, a transformer in its simplest form has two windings: known as the
primary and secondary windings respectively. For high efficiency of operation it is
essential that the magnetic linkage between the two windings should be as great as
possible. In order to achieve this result the primary and secondary windings can be
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disposed on the core in one of several ways depending upon the operating frequency and
circuit requirements.
For a mains transformer operating at a low frequency and high flux density, high efficiency
of operation can be achieved by simply winding the secondary coil over the primary.
It is not absolutely necessary to have two separate windings on a transformer although
from a safety point of view it is desirable when one is a high voltage winding. In the autotransformer one winding only is used which acts as both primary and secondary.
The principle of the auto-transformer is illustrated in fig 10.
Fig 10
It consists of a single winding which is tapped at several points to provide a fraction of the
primary voltage across the secondary load. (A tapping on a winding consists in bringing
out an external lead from a point on the winding other than at its ends).
In fig 10 a step-down transformer is shown (the secondary voltage E2 is less than the
primary voltage E1), but by simply interchanging E1 and E2 a step-up transformer is
obtained.
When the difference between the primary and secondary voltages is slight the size of an
auto-transformer is very much less than that of a two-winding transformer handling the
same power. Where however the voltage difference is considerable there is little
advantage in using an auto-transformer.
One disadvantage of the auto-transformer is the direct electrical contact between the
primary and secondary windings, which under certain fault conditions may result in
damage to equipment.
For instance when used as a step-down transformer with E1 much greater than E2, a break
in the winding at point 'X' would result in Practically the whole of the mains voltage being
applied to the low voltage side.
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2.32.8.3 Impregnation and Insulating Materials
Immediately a coil is wound it is the best practice to impregnate it in chemically neutral
mineral wax, or a good varnish. This affords protection against mechanical damage,
prevents the ingress of moisture, and also improves the dielectric strength of the insulating
materials. For low operating temperatures wax is used, but for higher operating
temperatures it is necessary to use varnish.
The main group of insulating materials used in transformers consists of such materials as
bakelite, paper, cotton, silk, varnish, and wire enamel. Coil formers are of bakelite and the
wire insulation is either cotton, silk or enamel. Paper or varnish is used as insulation
between coil layers when require and as a final covering for the complete winding.
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2.32.9 Results Table
primary
voltage
Vp(rms)
primary
current
Ip(mA)rms
secondary
voltage
Vs(Vrms)
secondary
current
Is(mA rms)
Vp
Vs
Ip
Is
2
3
4
5
Fig 4
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