Note 02
DC Circuit Analysis
The simplest circuits to understand and analyze are those that carry direct current (DC). In this note we
continue our study of DC circuits with the topics of DC voltage and current sources, the idea of an
equivalent circuit, and various techniques of analyzing DC circuits.
Voltage and Current Sources
We have already encountered an example of an
energy source (battery) in Note 01. Any energy source
supplies a voltage and a current, of course. It is useful,
however, to describe a source specifically as a voltage
or current source if we place a special meaning on the
terms. It is also useful to distinguish between ideal
voltage and current sources and their real (that is, real
world) counterparts.
The Voltage Source
By an ideal voltage source (Figure 2-1) we shall mean
an energy source that is capable of delivering a constant voltage across a load regardless of the current
drawn by the load.
+
+
–
V
–
V
load resistor are at the same potential. In this state the
source behaves electrically as if it were shorted by a
connecting wire. Though it might seem trivial at this
stage the idea of a zeroed voltage source will, in fact,
prove useful in our study of Thévenin equivalent
circuits later in this note.
V
I
r
V = ε − rI
ε
V
ε
I
€
(a)
(b)
Figure 2-2. Beyond a certain current a real voltage source (a)
delivers a voltage V = ε – rI across a load (approximately),
where ε is the source’s intrinsic emf and r its internal
resistance.
Figure 2-1. Circuit symbols for an ideal voltage source.
We have already seen in Note 01 that a cell or a
battery is a more-or-less ideal voltage source so long
as the current drawn from it is small. As the current
drawn rises, however, beyond a certain point, the cell
begins to behave non-ideally, that is, the voltage appearing across its terminals decreases. This behavior
is attributed (at least approximately) to the existence
of an effective internal resistance r in series with the
battery’s emf. Thus in these notes (where we deem it
important) we shall represent a real voltage source by
the elements drawn in Figure 2-2a. An ideal voltage
source would be one whose internal resistance is zero.
r can be measured experimentally. The results of a
method for finding r of a real voltage source are
sketched in Figure 2-2b. (We shall take up this topic
again in Example Problem 2-8.) If the terminal voltage
V is plotted vs the output current I beyond the point
where the source behaves non-ideally, then the value
of r is (–1) times the slope of the graph.
Even a “dead” or a “zeroed” voltage source is of
some small interest. In Figure 2-3 a dead voltage
source is shown connected to a load resistor. Since the
source is dead then both sides of the source and the
0
Figure 2-3. If the output of a voltage source is zero then the
source can be “replaced” by a shorting wire with no change
occurring in its electrical behavior.
The Current Source
A current source is another special case of an energy
source. By ideal current source (Figure 2-4) we shall
mean an energy source that is capable of delivering a
constant current to a load regardless of the voltage
developed across the load.
I
I
Figure 2-4. Circuit Symbols for an ideal current source.
N2-1
Note 02
A real current source is much less common than a real
voltage source. A real current source can be made
from a voltage source and a large series resistor or it
can be constructed from a circuit employing a transistor or an IC chip. If you encounter a real current
source in this course at all it will likely be in the form
of a special power supply.
As you might expect the behavior of a real current
source is not exactly ideal either. The current delivered by a real current source decreases as the voltage
developed across the load increases beyond a certain
point. This means that a real current source behaves
like an ideal source element in parallel with an internal resistance (Figure 2-5). An ideal current source
would be one whose internal resistance is infinite.
Example Problem 2-1
An Example of an Internal Resistance
If a D cell develops 1.5V across its terminals when
unloaded but only 1.3V when connected to a 100 Ω
load, what is the internal resistance of the D cell?
Solution:
The fact that the D cell’s terminal voltage drops from
1.5 V (effectively the cell’s emf) to 1.3 V when the load
resistor is connected means that the cell has an internal resistance r. The circuit is shown in Figure 2-7.
I
I
€
I0
Ri
r
V
R = 100 Ω
V
I = I0 −
Ri
ε = 1.5 V
€
Figure 2-5. A real current source behaves (approximately)
€ source with an internal parallel
like €
an ideal €current
€
€
resistance.
Note finally from Figure 2-6 that a current source with
an output of zero can be replaced by an open circuit
with no change in electrical behavior. “Zeroed”
current and voltage sources are in a sense complementary.
Figure 2-7. A real D cell connected to a load resistor.
We can solve this problem a couple of ways. When R
is connected the current I flowing must be
I=
The voltage drop across r is ∆Vr = 1.5 – 1.3 = 0.2 V.
€
Therefore
0
Figure 2-6. A zeroed current source is one which can be
“replaced” by an open circuit with no change in electrical
behavior.
Let us consider a chemical cell as a real source
element.
N2-2
ΔVR 1.3 (V )
=
= 0.013 A .
R
100 (Ω)
r=
ΔVr
0.2 (V )
=
= 15.4 Ω .
I
0.013 (A)
The D cell has an internal resistance of 15.4 Ω.
€
Circuit analysis is essentially the activity of solving for
all unknown circuit parameters. The unknowns might
be the currents that flow in the various branches of
the circuit, the voltages that exist between the various
nodes of the circuit, the effective resistances between
the various nodes and/or combinations of these.
Kirchoff’s rules are useful in solving for these
unknowns. In the next section we consider the
statements of Kirchoff’s rules and apply them to a
number of examples.
Note 02
Kirchoff’s Rules
In symbols,
Consider the multibranch circuit drawn in Figure 2-8.
Suppose that all the circuit parameters are known
except for the currents I1, I2, and I3. You cannot solve
for these currents by means of resistor reductions
because of the existence of the voltage sources in the
two branches of the circuit. (Convince yourself of
this.) To solve this problem you need to apply€
Kirchoff’s rules.
A good first step in problems of this kind is to number the nodes and to redraw the circuit to make it as
recognizeable as possible (Figure 2-8b). You can number the nodes any way you like.
I1
1
2
I2
4Ω
I3
2Ω
3V
2Ω
5
4
2V
I1
3
4Ω
3V
2Ω
5
2V
I2
2Ω
4
around a closed loop 1
…[2-1]
i
Kirchoff’s Current Rule
Kirchoff’s current rule (abbreviated KCR) is commonly
stated in these words:
The sum of the currents flowing into a node is zero.
In symbols,
∑I
n
= 0 into a node 2
… [2-2]
n
Note the phrase into a node. The currents indicated in
Figures 2 are shown flowing into node 2. We assumed
these directions and they are completely arbitrary. If
€
in the subsequent analysis we find that the numerical
value of a current is negative then the actual direction
of the current is opposite to the direction assumed.
Example Problem 2-2
Applying Kirchoff’s Rules
Solve for the currents I1, I2 and I3 in Figures 2-8 using
Kirchoff’s rules.
I3
2
i
3
Figure 2-8a. An example of a multibranch circuit.
1
∑ ΔV = 0
4
Figure 2-8b. Figure 2-8a redrawn to make it more
recognizeable
NOTE: We are using here the terms branch, loop and
node. A branch is a physical section of a circuit comprising wire, voltage source, resistors, and so on. A
node is a point where two or more branches meet. A
loop is an imaginary closed path that we traverse in
the task of calculating potential changes. The path
defined by the nodes 1-2-4-5-1 in Figure 2-8 is a loop.
The path marked by the nodes 2-4 is a branch (and of
course the nodes are numbered).
Kirchoff’s Voltage Rule
Kirchoff’s voltage rule (abbreviated KVR) is commonly
stated in these words:
The sum of voltage changes around a closed loop, or
path, is zero.
Solution:
To apply Kirchoff’s rules correctly we must observe
the signs of the various voltage changes and current
directions consistently. Applying KVR around the
loop shown in Figure 2-8b defined by the nodes 5-1-24-5, we get
+3 − 4I1+ 2I2 = 0 .
…[2-3]
(The first voltage change we encounter is across the 3
V battery which is +ve, the change across the 4 Ω
resistor
€ in the branch 1-2 is negative, and so on).
Applying KVR around the loop defined by the nodes
2-3-4-2 (moving clockwise), we get
+2I3 + 2 − 2I2 = 0 .
…[2-4]
(The first voltage change is that across the 2Ω resistor
which is +ve, and so forth.)
Applying
KCR to node 2 we get
€
1
It can be shown that KVR around a closed loop follows from the
fact that the electric field within the conductor forming the loop is a
conservative field.
2
It can be shown that KCR is a consequence of charge
conservation.
N2-3
Note 02
I1+ I2 + I3 = 0 .
…[2-5]
imaginary current I1’, I2’ and I3’ (we use a prime to
remind us that these are not in general the true currents). For example, the true current flowing through
the top 3Ω resistor is I1’ – I2’ (or the reverse). Three
currents would mean ordinarily that we would need
to solve three simultaneous equations. But one of the
loop currents we know already, I1’ = 3 A, making
only two equations actually necessary in this example.
We can now solve the three simultaneous equations
(eqs[2-3] through [2-5]) in the usual way. The results
are €
I1 =
4
A,
5
I2 =
1
A,
10
I3 = −
9
A.
10
One of the currents (I3) turns out to be negative. This
means that I3’s true direction is opposite to the
€ direction we assumed. To check the consistency of
these results we can find the total potential change
around 1-2-4-5-1 (starting from node 1):
3Ω
I1'
4
1
− x4 + x2 + 3 = 0 .
5
10
3A
This is zero as required by KVR.
2Ω
I2'
4Ω
3Ω ➂
12 V
I3'
Figure 2-9. An example of a multibranch circuit to illustrate the method of loop or mesh currents.
€
We now continue with three methods that are based
on Kirchoff’s Rules: the methods of Loop currents, node
voltages and superposition.
WARNINGS: Make sure that when you move around
a loop you go continuously in the same clockwise or
counterclockwise direction. Take the current that is
flowing in the direction you travel as positive. Avoid
applying KVR to the loop containing a current source
as the voltage across a current source is usually not
known. 3
Applying KVR to loops “2” and “3” keeping the
warnings in mind we get
The Method of Loop Currents
The method of loop currents (also called m e s h
currents) is an advance on the basic Kirchoff’s rules.
As we have seen in the basic application of Kirchoff’s
rules we assign a true current to each branch of a
circuit and then apply KVR and KCR. The solutions of
the resulting simultaneous equations are the “true”
currents—the currents that are actually flowing.
But in the method of loop currents we assign an
imaginary current, numbered as usual, to each loop. In
a branch of the circuit that is shared by a number of
loops more than one loop current will be “flowing”.
We must find the true currents from the solutions of
the simultaneous equations; they are, in general, the
sum or the difference of adjacent loop currents. Used
carefully this method is superior to the usual method
in that we need apply only KVR and we need solve a
fewer number of equations. Let us consider an
example.
−4I2'+4I3'−3I2'+3I1'−2I2'= 0
and
…[2-7]
solve them in the usual way. The results are:
€
I1'= 3 A,
Solve for the currents in each branch of the circuit in
Figure 2-9 by the method of loop currents.
Solution:
We begin by naming the three loops “1”, “2” and “3”
as shown. To each of these loops we assign an
−3I3'+3I1'−4I3'+4I2'−12 = 0 .
…[2-6]
€ Substituting I1’ = 3A into eq[2-6] and [2-7] we can
€
Example Problem 2-3
The Method of Loop Currents
€
N2-4
①
➁
I2'=
51
A,
47
I3'=
63
A.
329
Having the loop currents, we can now solve for the
true currents. If we take the true (unprimed) current
I1 = I1’ = 3 A, I2 as the current through the 2Ω resistor,
I3 the current through the upper 3Ω resistor, I4 the
current through the 4Ω resistor and I5 the current
through the lower 3Ω resistor we have
I2 = I2'= 1.08 A, I3 = I1'−I2'= 1.92 A ,
3
A current source is conventionally labelled with the value of
current it is able to sustain through itself.
Note 02
I4 = I3'−I2'= −0.90 A, I5 = I1'−I3'= 2.82 A .
Note that the current through the 4Ω resistor is negative; this means that it is actually flowing to the left.
Solution:
Following the recipe we obtain these equations for
nodes 2 and 3:
€
Method of Node Voltages 4
and
The method of node voltages also goes by the name of
nodal analysis. The method is another variation on €
Kirchoff’s rules and has the advantage in some instances in requiring only KCR. The method consists of
first choosing an arbitrary (but convenient) node to €
serve as a “zero” voltage reference. The voltages at all
the other nodes are then assigned and referenced to
this zero voltage. KCR is then applied to each node
and the equations resulting are solved in the usual
way. In applying this method you may wish to follow
this four step “recipe”:
€
1
2
3
Choose a reference node whose voltage can be set
to zero. This node is arbitrary though it is usually
best to choose the node that is connected directly
to the negative side of the voltage source.
Assign voltages V1 , V2 , etc. to the various nodes.
Apply KCR to the various nodes to derive the
simultaneous equations.
Solve the equations.
V1− V 2 V 3 − V 2 0 − V 2
+
+
=0
3
4
6
0 − V 3 V1− V 3 V 2 − V 3
+
+
= 0.
4
1
4
…[2-8]
…[2-9]
Note V1 = 3V is obvious. Substituting V1 = 3V results
in two equations in two unknowns. The results are:
V1 = 3 V, V 2 =
36
40
V, V 3 =
V.
17
17
CHALLENGE: How would you solve this problem
any other way?
Method of Superposition
The method of superposition is also a variation of
Kirchoff’s rules, or at the very least, involves Kirchoff’s rules. Voltages and currents obey a “principle of
superposition” which can be stated as follows:
Let us consider an example.
The current in any branch of a circuit equals the sum of
the currents produced in that branch by each individual
independent source with all others zeroed.
Example Problem 2-4
The Method of Node Voltages
The voltage at any node of a circuit equals the sum of the
voltages produced at that node by each individual
independent source with all others zeroed.
4
Solve for the voltages indicated in the circuit in Figure
2-10 by the method of node voltages.
V1
3V
2Ω
2Ω
1Ω
4Ω
0
V2
4Ω
V3
6Ω
Figure 2-10. A multi-branch circuit for nodal analysis.
Let us consider an example.
Example Problem 2-5
The Method of Superposition
Solve for the current I1 in the circuit in Figure 2-11a
using the method of superposition.
Solution:
This method requires that the sources be zeroed.
Recall that zeroing a voltage source amounts to replacing the source with a short circuit (Figure 2-3) and
zeroing a current source amounts to replacing the
source with an open circuit (Figure 2-6). The circuit of
Figure 2-11a with the sources zeroed results in the
circuits in Figures 2-11b and c. From Figure 2-11b we
get
I1' =
4
This method lends itself well to computer adaptation. It is used
in the program Spice and others.
4
A,
3
N2-5
€
Note 02
1Ω
Thevenin’s Theorem
1Ω
Thévenin’s theorem can be stated in these words: 5
+
4V
2A
zero
current source
I1
zero
voltage source
P 1Ω
(a)
1Ω
1Ω
+
4V
1Ω
The theorem states, in effect, that the circuit in Figure
2-12a is equivalent to the one in Figure 2-12b.
network
1Ω
–
Any two terminal network is composed of an ideal
voltage source VTH and a series resistor RTH.
1Ω
–
I1'
I1'
2A
(b)
1Ω
(c)
Figure 2-11. The process of zeroing the current and voltage
sources in (a) results in (b) and (c) respectively.
R TH
+
2
A,
3
–
by applying the method of node voltages to node P.
The total current is the sum of the two currents:
€
I1 = I1' + I1'' = 2A.
Clearly, the methods of loop currents, node voltages
and superposition
are quite equivalent. You should
€
try to master all of these methods as each has its own
advantages in particular situations.
Equivalent Circuits
Thus far we have considered techniques for analyzing
circuits whose circuit diagram or internal structure is
known. Sometimes, we are forced to deal with a
circuit whose internal structure is not known. Such a
circuit is often called a “black box”. Fortunately, in
these cases we can usually extract information about
the circuit using the Thévenin and Norton theorems.
These theorems (which are really statements given
without proof) enable us to express a network in
terms of equivalent sources and resistors. The
theorems are useful even when the network’s circuit
diagram is known, for they represent one more way
of reducing a circuit’s compexity. You must
remember, however, that they are applicable only to
networks consisting of linear elements (like voltage
sources and resistors).
N2-6
B
(a)
by applying Ohm’s law, and from Figure 2-11c we get
I1'' =
A
composed of
unknown
voltage and/or
current sources,
resistors etc.
VTH
(b)
A
A
RN
IN
B
(c)
B
Figure 2-12. According to the theorems the circuit in (a) is
electrically the same as the Thévenin equivalent in (b) and
the Norton equivalent in (c).
VTH and R TH may be found as follows:
1
Calculate (or measure with a suitable instrument)
the open circuit voltage VOC . Then VTH = VOC .
The method of finding R TH depends on whether the
circuit diagram is known or not. If unknown then
2a
Short the output and measure the short-circuit
current ISC with a suitable instrument. Then R TH
= VTH/ISC .
If the circuit diagram is known then
2b
5
Zero the voltage and/or current sources and
from the result calculate the effective resistance
R AB between the output terminals AB. Then R TH
= R AB.
In some places in the documentation for this course you will
find RTH and V TH written Req and V eq.
Note 02
Norton’s Theorem
r
Norton’s theorem can be stated in these words:
Any two terminal network is equivalent to an ideal
current source IN and a parallel resistor R N .
The meaning of the theorem is sketched in Figures 2€
12a and c. I N and R N may be found as follows:
1
Calculate (or measure with a suitable instrument)
the short circuit current ISC . Then IN = ISC .
As in the case of Thévenin’s theorem, the method of
finding RN depends on whether the circuit diagram is
known or not. If known or unknown
2a Calculate (or measure with a suitable instrument)
the open circuit voltage, i.e., V TH. Then R N =
VTH/ISC .
If the circuit diagram is known, the option is to
2b Zero the voltage and/or current sources and from
the result calculate the effective resistance R AB
between the output terminals AB. Then R N = RAB.
ε
r
ε
€ 2-13. Thevenin and Norton equivalents of a real
Figure
€
energy source (Figure 2-2a).
€
Let us now consider a more complex example.
Example Problem 2-7
Thévenin and Norton Theorems: An Example with
Voltage and Current Sources.
Find the Thévenin and Norton equivalents of the two
terminal network shown within the dashed outline of
Figure 2-14a.
1Ω
It follows from this and the previous section that R N =
R TH.
It should be emphasized here that the theorems are
most useful when the circuit is a complicated one or
when the circuit diagram is unknown. These cases we
explore in Example Problems 2-7 and 2-8. But first, a
simple equivalent.
P
+ ①
4V
–
(a)
Example Problem 2-6
A Simple Equivalent
R TH
RTH = r = RN .
Now shorting the output and calculating the short
circuit current we have
ε
ISC = IN = .
r
Thus the equivalents are as drawn in Figure 2-13.
€
2A
short output
P 1Ω
1Ω
A
= 2Ω
IN – 2
B
Draw the Thevenin and Norton equivalents of the real
energy source of Figure 2-2a.
€
1Ω
(b)
2Ω
+
–
2A
IN
(d)
A
A
6V
(c)
A
+
VTH
–
B
➁
Vo
–
I
1Ω
1Ω
+
zero sources
Solution:
The open-circuit voltage of Figure 2-2a is automatically VOC = VTH = ε.
Zeroing the voltage source and calculating the effective resistance between the output terminals we get
r
3A
B
2Ω
(e)
B
Figure 2-14. The Thévenin and Norton equivalent circuits of
(a) are shown in (c) and (e), respectively.
Solution:
The steps involved in finding the Thévenin equivalent
are sketched in Figures 2-14a, b and c, for the Norton
equivalent in Figures 2-14a, d and e.
N2-7
Note 02
Thévenin Equivalent
Applying KCR to point P in Figure 2-14a when AB is
open we get
I + 2 = 0, so that I = −2 A.
Applying KVR around loops “1” and “2” we get
€
+4 + (2)1− V0 = 0
and
V0 + 0 − VTH = 0 .
Solving€these equations we get
VTH = 6V .
€
Example Problem 2-8
Finding a Thévenin Equivalent Experimentally
Explain how the Thévenin equivalent of an active DC
black box might be found experimentally.
Solution:
This is done using a voltmeter with a large internal
resistance (DMM) and a variable resistor (Figure 2-15).
With RL disconnected (or of very large value) use the
voltmeter to measure the open-circuit voltage, V TH.
Connect the load resistor and vary it until the output
voltage Vo read on the meter drops to 1/2 VTH. The
corresponding value of R L then equals R TH. You will
follow this procedure in Experiment 01.
Now zeroing the sources of Figure 2-14a we get the
circuit in Figure 2-14b. Thus
€
DC
Black
Box
RTH = 2Ω .
Thus the Thévenin equivalent is as given in Figure 214c.
€
Norton Equivalent
By short-circuiting the output of Figure 2-14a we get
the circuit in Figure 2-14d. The short circuit current is
the Norton current IN. Therefore applying KCR to
node P we see that the current delivered by the
battery is IN – 2. Applying KVR around the loop (the
periphery of Figure 2-14d) we get
RL
V
DMM
Figure 2-15. A setup for finding the Thévenin equivalent of a
DC black box experimentally.
The procedure just described applies equally well to
any kind of “active” black box—from a battery to a
power supply. It does not work for a passive element
like a d’Arsonval meter movement. 6
4 − (IN − 2)1− ( IN )1 = 0 .
From which
Practice Problems
IN = 3 A.
From the Thévenin equivalent R N = RTH = 2Ω. The
€ equivalent is as shown in Figure 2-14e.
Norton
Note that you can derive the Thévenin equivalent
from the Norton equivalent and vice versa. For
example, in Figure 2-14c you can find the Norton
current by calculating the current when AB is shorted.
Conversely, the Thévenin voltage is the voltage between A and B across the 2 Ω resistor in Figure 2-14e.
You should keep both approaches in mind when
solving problems of this kind; you can often find a
Norton equivalent easier by first solving for the
Thévenin equivalent and vice versa.
6
A method of doing this is described in Experiment 01, “DC
Circuits and Measurements”.
N2-8
N2-9