Lecture 2: Resistor Networks
This lecture will serve to recall and formally set down our approach to analysing electrical
circuits. The main focus will be on resistor networks and our interest will typically be on
the determination of input-output voltage relationships, but power transfer will also be
considered. In effect, we will be treating the electrical circuit as if it were a signal
processing block and determining what the block does to the input voltage. Two
important circuit analysis tools, Thevenin’s theorem and the superposition principle, are
covered.
Learning Outcomes:
On completing this lecture, you will be able to:



2.1
Analyse the voltage relationships in resistor networks of moderate complexity;
Recognise when and how to apply Thevenin’s theorem and the superposition
principle;
Carry out power transfer calculations.
Notation
The following symbols denote ideal voltage and current sources:
Voltage source
Current source
i or I
v or V
A lower case letter will denote instantaneous signal value, such as
v  10 sin( 2 50 t ) V
while an uppercase letter will denote a DC or rms quantity. Note that the voltage
arrowhead points in the reference direction of the voltage rising while the current
arrowhead gives the reference direction for positive current flow.
A particular voltage source is the DC power source or battery, denoted by
DC power source
E
A voltage quantity is the difference in electrical potential between two nodes. In most
electrical and electronic circuits, where our interest lies in voltage signals, we tend to
assign one node a reference potential of 0V (zero volts) and designate all other voltages
2-1
with respect to this 0V reference node. Such a 0V reference is termed a ground and
carries the symbol
The following diagram illustrates two ways of working with a 5V battery, itself shown in
Fig (a):
0V
+5 V
5V
5V
5V
0V
(a)
-5 V
(b)
(c)
In Fig (b) the lower node is grounded resulting in a potential of +5V on the upper node;
in Fig (c) the upper node is grounded resulting in a potential of -5V on the lower node.
2.2
Device Characterisation
I
I
D
V
R
D
I
V
R
V
The electrical operation of any electronic device is captured by means of the relationship
between the current I flowing through the device in response to a voltage V across the
device. For most electronic devices this relationship is non-linear, such as illustrated for
device D above. For the device known as a resistor, the relationship (at a given
temperature) is linear. This linear relationship is expressed by Ohm’s Law:
I
1
V
R
where R is the resistance value given in ohms.
2-2
Where we have a circuit with multiple devices — resistors for the present — we make use
of Kirchhoff’s laws to analyse the circuit. In building up these multi-device circuits, there
are only two possibilities for configuring the devices: putting them in series or putting
them in parallel.
(i) Resistors in Series:
I
R1
V1
R2
V2
VS
Noting that the same current I flows in each resistor, the voltage drops are given
V1  R1 I
and
V2  R2 I
Applying Kirchhoff’s Voltage Law to the loop, we get
VS  V1  V2
 I R1  R2 
Noting that VS is the voltage across the series resistor combination and I is the current
through the series combination, we have
RSERIES 
VS
 R1  R2
I
Resistors in series add.
(ii) Resistors in Parallel:
I
I1
VS
R1
I2
V1
R2
Noting that the same voltage is dropped across each resistor
2-3
V2
VS  V1  V2
the individual currents can be expressed
I1 
V1 VS

R1 R1
and
I2 
V2 VS

R2 R1
Applying Kirchhoff’s Current Law we get
I  I1  I 2

VS VS

R1 R2
 VS
R1  R2
R1 R2
Again noting that VS is the voltage across the parallel resistor combination and I is the
current flowing through the combination, we can express the effective resistance value by
RPAR 
2.2
VS
RR
 1 2
I
R1  R2
Voltage Dividers
Consider applying an input voltage Vi to a series resistor combination R1 and R2, and
taking as output Vo the voltage across R2:
V1
I1
R1
Vi
As before
I1  I 2  I
With
Vi  I R1  R2 
I2
R2
and
Vo  I R2 
The output-input relationship is given by
2-4
Vo
Vo
R2 I
R2


Vi R1  R2 I R1  R2
This basically states that the output is a scaled replica of the input with the scale factor
being a fraction determined by the resistor ratio R2/(R1+R2). The following example
illustrates a practical application of this circuit model.
Example 2.1:
microphone
amplifier
RS
Vo
VS
RL
Vo
Consider the electrical connection between a microphone and an amplifier. The
microphone is modelled by a non-ideal voltage source (comprising an ideal voltage source
VS and a source resistance RS) while from the point of view of the microphone the
amplifier appears as a resistive load RL. The voltage signal actually going in to the
amplifier is given by
VO 
RL
VS
RS  R L
Usually we would wish to get as much signal as possible into the amplifier and this clearly
requires
RL  RS
2.3
Thevenin’s Theorem
Consider the following circuit.
R1
VS
R2
Vo
RL
Noting that the two resistors R2 and RL can be combined in parallel into an equivalent
resistor
RPAR 
R2 R L
R2  R L
we can express the output voltage by
2-5
RPAR
R1  RPAR
VO 
R2 R L
R2  R L
VS 
VS
R2 R L
R1 
R2  R L

R2 RL
VS
R1 R2  R1 RL  R2 RL

RL R2
VS
R1 R2  RL R1  R2 

RL
R2
VS
R1 R2
R1  R2
 RL
R1  R2
If we make the following substitutions:
Let
RT 
R1 R2
R1  R2
and
VT 
R2
VS
R1  R2
We can express the output as
VO 
RL
VT
RT  RL
It is as if we had replaced all the circuit to the left of RL by a voltage source VT and a
resistor RT:
RT
Vo
VT
RL
This is the essence of what is known as Thevenin’s theorem which states:
Any part of a linear electrical circuit can be replaced by a single voltage source V T
and a source resistor RT.
The Thevenin voltage source VT can be determined by calculating the open circuit
voltage for that part of the overall circuit being replaced.
The Thevenin resistance RT can be determined by calculating the overall resistance
looking back into that part of the overall circuit being replaced and putting any
voltage (or current) source to zero.
Looking again at our original circuit
2-6
R1
R2
VS
Vo
RL
what we want to do is to replace that part of the circuit to the left of the V O arrow:
R1
VT
R2
VS
RT
For this sub-circuit, clearly the open circuit voltage is given by
VT 
R2
VS
R1  R2
If we now put VS to zero and look back into the sub-circuit we get
R1
R2
RT
showing that RT is indeed the parallel combination of R1 and R2
RT 
2.4
R1 R2
R1  R2
The Superposition Principle
Consider the following voltage divider:
V1
V2
I1
I2
R1
R2
Vo
VS1
2-7
VS2
The problem is to determine the dependency of the output voltage V O on the two voltage
sources VS1 and VS2. The designated currents I1 and I2 may be expressed
I1 
V1 VS1  VO

R1
R1
I2 
From Kirchhoff’s current law, we have
V2 VS 2  VO

R2
R2
I1   I 2
giving
VS1  VO
V  VO
  S2
R1
R2
VS1 VS 2 VO VO



R1 R2
R1 R2
VO
R2  R1 VS1 R2  VS 2 R1

R2 R1
R2 R1
VO 
R2
R1
VS 1 
VS 2
R1  R2
R1  R2
An alternative means of arriving at this result is based on the superposition principle.
Basically the principle states:
In a linear circuit, if a node voltage VO depends on a number of independent
voltage sources, say VS1 and VS2, then
Calculate the node voltage due to the first source with all other sources zeroed;
Calculate the node voltage due to the second source with all other sources zeroed;
etc.
The overall node voltage is then the sum of the individual values as calculated
above.
In the case of our example,
VO  VO VS1 , VS 2  0  VO VS 2 , VS1  0
 VO VS1   VO VS 2 
for short.
VO(VS1) may be determined from the following diagram:
R1
Putting VS2 = 0V
R2
VO(VS1)
VS1
From our consideration of voltage dividers we have
2-8
VO VS1  
R2
VS 1
R1  R2
VO(VS2) may be determined from
R1
R2
VO(VS2)
Putting VS1 = 0V
VS2
Again voltage divider consideration shows
VO VS 2  
R1
VS 2
R1  R2
The superposition principle now gives
VO  VO VS1   VO VS 2 

2.5
R2
R1
VS1 
VS 2
R1  R2
R1  R2
as before.
Power Transfer
Sometimes when connecting a non-ideal signal source (ie a signal source with a non-zero
source resistance) to a resistive load we are primarily interested in delivering as much
power as possible to the load rather than trying to keep the signal level as high as
possible. We now investigate the requirements for maximising the transfer of average
power.
RS
Vs
RL
Vl
Assume the signal quantities are rms quantities to facilitate average power calculations.
The average power delivered to the load is
Pl 
Vl 2
RL
2-9
1

RL
 RL


Vs 
 RS  R L


RL
R S  R L 
2
2
Vs2
The question now is: for a given source (ie known Vs and RS), what value of RL maximises
Pl?
Differentiating Pl with respect to RL we obtain


dPl
3
2
 RL  2RS  RL   RS  RL  1 Vs2
dRS
 Vs2 RS  RL   2RL  RS  RL 
3
 Vs2 RS  RL  RS  RL 
3
0
 R L  RS
for a maximum
for maximum power transfer.
That is, the load resistance should be chosen to equal the source resistance to maximise
the power transfer from source to load.
2.6
Concluding Remarks
In this lecture, necessary and important aspects of electrical circuit theory have been
reviewed and a consistent system of notation has been set out. The particular methods of
analysis used here are not the only methods to arrive at the desired solution or
relationship — in many practical problems, there will be different possible approaches to
the solution; what has been provided here is a set of approaches.
2-10