§3: Linear Circuit Analysis
§3.0
§3.1
§3.2
§3.3
§3.4
§3.5
§3.6
§3.7
§3.8
§3.9
§3.10
§3.12
Introduction
Linear Systems
Superposition
Node-Voltage Analysis
Mesh-Current Analysis
Equivalent Circuits
Thevinin Equivalents
Norton Equivalents
Source Transformations
Maximum Power Transfer
Equivalent Resistance Revisited§3.11
Summary
Examples
§3.0 Introduction
When a carpenter works on a house or a mechanic tunes an engine, they each have many tools and various options for
completing their tasks. Most tasks are straightforward: the carpenter uses a hammer to pound in nails, and the mechanic uses
a specially designed wrench to remove the head gasket. Other tasks are slightly more ambiguous and require more information.
Consider the carpenter working on a doorframe. Whether he chooses to use a hand saw, a power saw, or a pocket knife depends
on the precision required, the number of doorframes to be built, and quite possibly the tools he has immediately available. The
mechanic may have even more trouble in determining whether to use a half-inch or 13mm wrench for tightening an assembly.
The analysis methods presented in this chapter will provide you with a toolbox capable of tackling almost any linear DC circuit.
The various methods will invariably terminate at the same answer, but some are clearly more efficient in solving particular circuits.
Our goal is to construct both the toolbox and an understanding for what problems each tool is effective. More importantly, these
tools will provide you with the necessary preperation to analyze more complex circuits: those containing diodes, transistors, and
digital logic.
Beginning with simple applications of Ohm’s and Kirchoff’s laws, we will use equivalent impedances and dividers to lay the
foundation for our remaining circuit analyses. Larger circuits using both current and voltage dividers will create the framework.
The properties of basic linear systems will allow us to perform analyses by superposition and Node/Mesh equations, further
fleshing out our options. The last analysis step will be to create simple circuit equivalents for any linear circuit, thus completing
our toolbox. Finally, we will address the various methods in a general setting and demonstrate a few tricks that can be used to
view the problems more easily.
Where possible, we will demonstrate multiple methods in analyzing the circuits: you need only pick the one with which you
feel most comfortable. The benefit of solving a circuit multiple ways lies in the ability to check your answers–you can be pretty
well certain that you are correct when you solve by completely different methods and obtain the same solution.
We will begin with a simple example to determine an output voltage given a DC input voltage as shown in Figure 3.1.
Example: Determine the current I50 traveling downward in the 50Ω resistor.
Figure 1
We know two methods so far to solve this problem:
1
2
1. We may determine the equivalent resistance looking to the right from the voltage source to then find the total current
entering the collection of resistors. Application of a current divider then gives the current through the 50 Ω resistor.
Method 1:
Req = 25 + 50k(30 + 20) = 50 Ω
2V
= 40 mA
50Ω
(30 + 20)
=
IT = 20 mA
(30 + 20) + 50
IT =
I50
Figure 2
2. We may combine the 20, 30, and 50 Ohm resistors as an equivalent and then find the voltage across the equivalent (same
as the voltage across the 50 Ω resistor). Simple application of Ohm’s law then gives us the current.
Method 2:
Rcomb = 50k(30 + 20) = 25 Ω
Vcomb =
I50 =
Rcomb
(2V ) = 1 V
Rcomb + 25
Vcomb
= 20 mA
50Ω
Figure 3
In many ways, this is a very typical problem to solve: a circuit containing one source and perhaps four or five resistors. Let’s try
another, more difficult, example that verifies the conservation of energy principle: all energy dissipated by a resistor is generated
by the source.
Example: Verify the conservation of energy in the circuit shown in Figure 3.4.
3
Figure 4
We use a combination of dividers and Ohm’s law applications to calculate the currents and the power dissipated.
Current calculations:
I10Ω =
I6Ω =
1V
= 100 mA
10Ω
30
· I15Ω = 41.67 mA
30 + 6
I30Ω =
I15Ω =
1V
1V
=
= 50 mA
(15 + 6k30) Ω
20Ω
6
· I15Ω = 8.33 mA
30 + 6
I1V = I10Ω + I15Ω = 150 mA
Power calculations:
P1V = (1V )(150mA) = 150 mW
P6Ω = I62 (6Ω) = 10.41 mW
2
P15Ω = I15
(15Ω) = 37.5 mW
2
P10Ω = I10
(10Ω) = 100 mW
2
P30Ω = I30
(30Ω) = 2.08 mW
A fully labeled circuit is shown in Figure 3.5.
Figure 5
To verify that the conservation of energy holds, compare the power generated by the voltage source to the sum of powers
dissipated in the resisters.
.
150 mW = (10.41 + 100 + 37.5 + 2.08) mW
Notice that there is an implicit assumption in this last step: we have shown that the powers are equal, but not the energy! Under
steady-state conditions, the rate of energy into a circuit (the power) is constant and therefore the conservation of instantaneous
power implies conservation of energy over any time-scale.
§3.1: Linear Systems
One of the most common models studied in engineering is the linear system. The generic linear system is characterized by the
property of linearity, which requires that the output is a weighted sum of the inputs. Figure 3.6 shows the block diagram of a
basic linear system: for inputs x1 , x2 , ..., xn there are corresponding outputs y1 , y2 , ..., yn .
4
Figure 6
While this definition sounds trivial, answer the following question: is f (x) = x + 1 a linear system? Actually, no! Plotting the
function on a set of axes as shown in Figure 3.7, you will see the answer jump out at you.
Figure 7
If we scale an input, say αx1 , then the output is the same multiple times the corresponding output, αy1 . Likewise, if we add
two inputs and process them simultaneously, say x1 + x2 , then the output is the corresponding sum y1 + y2 .
Now, let’s return to the question of whether f (x) = x + 1 qualifies as a linear system: consider the three values f (0) = 1,
f (1) = 2, and f (2) = 3. If f (x) was linear, then f (0 + 1) = f (0) + f (1), but 2 6= 1 + 2 = 3. Further, if f (x) were linear then
f (2 · 1) = 2 · f (1) which is again contradicted. The only conclusion is that f (x) = x + 1 is nonlinear, despite being a line; a line
need not be mathematically linear! If the line happens to pass through the origin, then it will be linear.
Combining the properties discussed, we obtain the system response
y k = c k · xk
∀k ∈ N
succinctly stated as: “the output is a weighted sum of the inputs.” Two important characteristics of the linear system to
notice are that a zero input always corresponds to a zero output and that the weighting terms, c k , do not depend on any of the
other inputs.
§3.2: Superposition
The bridge between mathematical definitions of linear systems and electrical circuits is a circuit technique called superposition.
We will consider many circuits in this chapter, all of which are linear systems: the inputs of the circuit as a linear system are
voltage and current sources, and the outputs are any of the node voltages or currents in the circuit elements. A process called
superposition entails activating one source at a time and zeroing all others, allowing us to determine the response of the linear
system (find each ck ). Adding the contributions from each of these voltage and current sources, we will then be able to obtain the
overall response of our linear circuits. The limitation will be that we can only use superposition to determine linear quantities:
power calculations (quadratic), diode calculations (exponential), and transistor analyses (mixture) will require additional methods.
Example: Use superposition to determine the current in the 5Ω resistor shown in Figure 3.8.
5
Figure 8
Up to now, we would have attacked this problem by some combination of Ohm’s law calculations or voltage and current
dividers. With two power sources, we have to try superposition. We stated above that if we “zero the other sources,” we may
calculate the response of each source independently. Stated more simply, we can figure out the contributions of each source to
the overall response I5 .
So, how do we zero a source? First consider an ideal voltage source: it has a fixed voltage and will allow any current to flow
through it. The only circuit element we have that will allow any current to flow through it, but is guaranteed to have zero voltage
across its terminals is a short circuit. To zero a current source, we follow the same process: a current source transmits a fixed
current independent of the voltage across it, so a “zero current source” would be identical to an open circuit, which is guaranteed
to stop the flow of current (therefore zero) independent of voltage. We may also obtain these zeroed sources from the internal
resistances of the ideal voltage and current sources as derived in Chapter 2: a voltage source has zero internal resistance and
thus zeros to a short circuit, while a current source has infinite internal resistance, corresponidng to an open circuit when zeroed.
Figure 3.9 shows a graphical representation of zeroed sources.
Figure 9
Returning to the example in Figure 3.8, we find that by zeroing the sources one at a time, we obtain two separate circuits,
shown in Figure 3.10, corresponding to two separate (and linear) contributions to current in the 5Ω resistor.
Figure 10
The contributions from each source, I53A and I59V , may then be added up to obtain the actual current, I5 .
(9V )
10
+
· (−3A) = (0.6A) + (−2A) = −1.4 A
(10 + 5)Ω
10 + 5
What about power calculations? Power is relative to the square of either voltage or current (which we know to both be linear
quantities). Prove to yourself that superposition may not be used to directly calculate the power in a circuit; that is, compare
I52 (5Ω) and (I523A + I529V )(5Ω). The correct calculation of power always depends on the net current, I 5 , not the contributions from
I5 = I59V + I53A =
6
each source.
Example: Use superposition to determine the voltage across the resistor R 1 in Figure 3.11.
Figure 11
Repeating the same process as before, we zero all but one source to calculate the contribution from each voltage and current;
Figure 3.12 shows the circuit with Vx zeroed.
Figure 12
VR1 , Ix = Ix · Req1 kReq2 ,
where
Req1 = R3 k(R4 + R5 + R6 )
and
Req2 = R1 kR2
Repeating the analysis for the contribution of Vx , we zero the current source and replace it with an open circuit as shown in
Figure 3.13.
Figure 13
VR1 , Vx = (−Vx ) ·
where
Req2
Req1 + Req2
Req1 = R3 k(R4 + R5 + R6 )
and
Req2 = R1 kR2
The total voltage, VR1 , can be written as the sum of contributions from each source.
7
VR1 = V R1 , Ix + V R1 , V x
= Ix · Req1 kReq2 + (−Vx ) ·
Req2
Req1 + Req2
= Ix · R1 kR2 kR3 k(R4 + R5 + R6 ) − Vx ·
R1 kR2
R1 kR2 + R3 k(R4 + R5 + R6 )
Example: Solve for the voltage across the 50 Ω resistor in Figure 3.14.
Figure 14
First consider the 20 V source.
Figure 15
V5020V = (+20 V ) ·
70k(20 + 50)
50
·
= 10 V
70k(20 + 50) + 15 20 + 50
Next consider the 2 A source.
Figure 16
V502A = (+2 A) · [50k(20 + 70k15)] = 39.29 V
and finally consider the 10 V source.
8
Figure 17
V5010V = (+10 V ) ·
50
= 6.07 V
50 + 15k70 + 20
Adding up the individual contributions, we obtain the total voltage, V50 .
V50 = V5020V + V502A + V5010V = 55.36 V
What if we throw in a twist to the last problem? We find out that the circuit we started with was drawn incorrectly: the
engineer specified a 25 V source in place of the 10 V source. Do we need to re-do the entire process? Not a chance! Instead of
25V
= 2.5) in the voltage source input and, by linearity, carry
recalculating any circuits, we can determine the ratio of change ( 10V
that through to the output. Therefore, the new response V5025V is 2.5 · V5010V = 15.18 V . This easy correction is one of the
benefits of superposition over other analysis methods.
The final example of superposition will have sources that can be combined (series voltage source and parallel current sources)
once another source in the circuit is zeroed. The analysis will work just as well considering each source individually, but will
proceed quicker by combining them in a smart fashion.
Example: Use superposition to solve for the voltage Vx across R1 in Figure 3.18.
Figure 18
If we zero both of the voltage sources, we end up with two parallel current sources, whose equivalent is simply the difference
of I1 and I2 .
9
Figure 19
Likewise, when we zero both current sources, there is a single series loop containing two voltage sources.
Figure 20
By solving the circuits in Figures 3.19 and 3.20, we reduce the workload by nearly half (two circuits instead of four). The
overall solution is:
R1
R1 + R 2 + R 3
To conclude the section, we state the general steps necessary to solve any circuit using superposition.
Vx = VxI1 −I2 + VxV2 −V1 = (I1 − I2 ) · R1 k(R2 + R3 ) + (V2 − V1 ) ·
1. Establish the unknown’s voltage polarity or current direction (and keep consistent throughout the analysis).
2. Redraw a circuit containing k voltage or current sources into k seperate circuits where only one source is active and all
others are zeroed. Remember: zero voltage sources are short circuits while zero current sources are open circuits.
3. Solve for each sources’ contribution to the overall voltage or current using Ohm’s law and dividers as appropriate.
4. Add the individual contributions to obtain the desired overall quantity.
A final comment on superposition: nearly every mainstream textbook on linear circuit analysis has a disclaimer that you cannot
perform superposition on circuits containing controlled sources. This notion is false! 1 We will routinely use superposition when
considering operational amplifier and transistor circuits, treating all controlled sources as independent (non-controlled sources)
and rectifying the algebra later.
Other Linear techniques
Another consequence of circuits as linear systems is that we may write systems of equations and employ linear algebra to solve.
~ = [Z]I~ or
The equations will ultimately fit into Ohm’s law: we will re-write the traditional matrix equation [A]~x = ~b as either V
~
~
[Y ]V = I, where [Z] is the matrix of impedances ([Y ] is the matrix of admittances) relating voltages to currents. The same basic
rules from linear algebra apply: the matrix [Z] or [Y ] must be invertible (have a non-zero determinant) for there to be a solution,
1Leach: something
10
and you need as many independent equations as unknowns to fully solve.
§3.3: Node-Voltage Analysis
Node-voltage analysis expands Kirchoff’s current law to write equations at each node in a circuit, and then solve with linear
algebra (ideally using a computer). We also draw on the fact that voltage is a conservative field, whose potentials can be taken
relative to an arbitrary reference point; thus we will be allowed to define a ground node as 0 V (typically, we choose the node
that has the most connections to other nodes in an attempt to simplify the equations).
Consider the first circuit that we analyzed using superposition.
Example: Use node-voltage analysis to solve for the current through the 5Ω resistor in Figure 3.21.
Figure 21
The most basic step is to define the nodes: we will first choose the ground node to be the one at the bottom since it is common
to the most other nodes, node A as a connecting point between the two resistors and the current source, and finally node B as
the interface between the voltage source and 10 Ω resistor.
Figure 22
Be extremely careful to notice that a single node may have many points and connections; oftentimes students will erroneously
define two different nodes in place of node A, one above the 3A source and another above the 5Ω resistor. Those two points are
directly connected by wire, which conducts current freely and, more importantly, will not allow a difference in potential across it.
The voltage at the point above the 3A source is always identical to the voltage above the 5Ω resistor, so we define it as a single
node. Similarly, the ground node has three connections, yet is still a single node.
Figure 23
11
We will write two independent equations relating the unknown voltages at nodes A and B to the input sources and passive
components. First, writing a KCL equation at node A (assume all currents to be leaving the node), we obtain:
VA − V B
VA − 0
+
=0
5Ω
10Ω
Normally, we would try to write a second equation at node B, but we find our first problem: we do not know the current into
an ideal voltage source (the potential is fixed completely independent of the current flowing through the source).
3A +
VB − V A
+ I9V = 0
10Ω
I9V =???
There is a way to bypass the indeterminate nature of the current through the voltage source: simply realize that KCL works
for arbitrary black boxes (linear) as well, and thus state that the current entering the voltage source is equal to the current leaving
the voltage source.
Figure 24
The actual name for this treatment of the voltage source is that of a supernode. Writing the new equation, we end up with
another KCL equation.
VB − V A
0 − VA
+
+ −3A = 0
10Ω
5Ω
Simplifying this equation shows another interesting false start to node-voltage analysis: this equation is exactly the same as
the one written at node A! Multiply both sides by a (−1) and you get the other equation. If the two equations are the same, then
they are also linearly dependent; to solve a linear system consisting of two unknowns, you must have two independent equations
(in general, the number of independent equations must equal the number of unknowns). We’ve exhausted the nodes at which to
write KCL equations, but we do know a very simple equation relating the two nodes having the voltage source between them.
VB − 0 = 9V
Thus, taking the voltage source as a fixed potential between two points, we obtain the final equation. Writing the two equations
together, we can form a linear system of equations (i.e. a matrix) to solve.
«
« „ « „
„ 1
1
1
−3A
VA
+ 10Ω
− 10Ω
5Ω
=
·
9V
VB
1
0
Notice that the elements of the matrix are simply the coefficients of the unknown node voltages. Further, you should check
that the units of each element in the expression make sense (matrix multiplication being across a row of the matrix and down the
column vector of the unknowns).
Once the equations are set into a matrix, we have a slew of choices how to solve: Gaussian elimination, Cramer’s rule, or
inverting the matrix. In general, you should be well versed in solving linear equations either by hand or with the aid of computation
software, so we will not cover those here.2 Even so, we are not quite finished. When a specific variable is requested (we asked for
the downward current in the 5 Ω resistor), we must solve for that quantity in terms of the, now known, node voltages.
„
„ «
«
~ = VA = [Y ]−1 · I~ = −7V
V
VB
9V
I5 =
VA
= −1.4 A
5Ω
2A great reference on linear algebra for solving linear systems of equations by hand is “Apostol: Linear Algebra”
12
Example: Use node-voltage analysis to solve for the voltage across R 1 in Figure 3.25.
Figure 25
We start out by defining the bottom left node as ground (since it is common to the most other nodes), and then by writing
the voltage equation at the supernode containing the voltage source.
VA + V x = V B
Before writing any current equations, we will define two equivalent resistances, and also collapse nodes C and D. The fact that
we are able to collapse the nodes means that we can write a dependent equation in terms of V A and VB for each node.
Req2 = R1 kR2
Req1 = R3 k(R4 + R5 + R6 )
R5 + R 6
VC =
· VB
R4 + R 5 + R 6
The simplified circuit is shown in Figure 3.26.
VD =
R6
· VB
R4 + R 5 + R 6
Figure 26
Using the reduced circuit (and a supernode in place of the voltage source), we can write the last equation by inspection.
VB
VA
+
= Ix
Req2
Req1
Putting everything together in matrix format, we have:
! „ « „
«
1
1
Ix
VA
Req2
Req1
=
·
−Vx
VB
1
−1
The requested voltage across R1 is the same as the voltage across the equivalent Req2 , or VA .
Req2
+ Ix · Req1 kReq2
Req1 + Req2
Example: Use node-voltage analysis again to solve for all of the node voltages in Figure 3.27. Specifically, determine the
voltage accross the 50Ω resistor.
VR1 = V A = V x ·
13
Figure 27
The first step is to write the easy equations between nodes having voltage sources between them, and then collect any
impedances into equivalents.
VA − 0 = 20 V
VD − VC = 10 V
Since we have four unknown voltages, and so far two equations, we need to write two independent KCL equations as follows.
VB − V A
VB − 0
VB − V C
+
+
=0
15 Ω
70 Ω
20 Ω
VD − 0
VC − V B
+
− 2A = 0
20Ω
50 Ω
We have applied the supernode condition in the second equation: the current leaving node C into the voltage source is the
same as the sum of the currents leaving node D. Combining the four equations into matrix format (we will now omit the units
~ = I.
~
since they cancel), we reduce the linear circuit to [Y ]V
0
1
B 0
B 1
@−
15
0
0
0
1
15
1
+ 70
+
1
− 20
1
20
0
−1
1
− 20
1
20
1 0 1 0 1
0
VA
20
BVB C B10C
1C
C·B C=B C
0 A @ VC A @ 0 A
1
VD
2
50
Not only does solving this system via nodal analysis result in the same node voltages that we obtained in the superposition
analysis, but the process finds all of the voltages at once!
0
1 0
1
VA
20 V
B C B
C
~ = BVB C = B 27.5 V C
V
@VC A @45.36 V A
VD
55.36 V
Trivially, we solve for the voltage across the 50Ω resistor,
V50Ω = VD = 55.36 V
Any circuit parameter that we wish to find is immediately available as a function of the node voltages. As an exercise, you
should verify the conservation of energy using the solution above (you will need to use the voltages across the 15 and 50 Ω resistors
to obtain the currents in the 20 and 10 V sources, respectively).
Example: Use node-voltage analysis to solve for the voltage across R 1 in Figure 3.28.
14
Figure 28
There are three independent nodes (two of which form supernodes) and one dependent node (B). The resulting set of equations
are:
VA − V C
VA − V D
+
+ I2 − I1 = 0
R1
R2 + R 3
R3
VD − 0 = V 2
VB =
· (VA − VD )
R2 + R 3
with the matrix format (we need only consider the independent equations).
VC − 0 = V 1
0
@
1
R1
+
1
R2 +R3
0
0
− R11
1
0
1 0 1 0
1
1
− R2 +R
VA
I1 − I 2
3
A · @ VC A = @ V1 A
0
VD
V2
1
The voltage across R1 is then the difference of VA and VC .
Summary of the steps for Node-Voltage analysis
The basic steps in solving a circuit via node-voltage analysis are:
1. Establish a ground or reference node.
2. Label remaining independent nodes – nodes are dependent if they can be written as a voltage divider times other node
voltages.
3. For each supernode, write an equation for the fixed potential and collapse the node.
4. Combine impedances where possible.
5. For each remaining independent node, write one KCL equation. If you always take currents as leaving the node, then the
entries in your matrix will in practice have a minimal number of negative signs.
~ = I~ and solve. Possible solution methods include Gaussian
6. Combine all KCL equations into the matrix format [Y ] · V
elimination by hand, plugging into a computer (ideal), or using Cramer’s rule.
~ of unknowns, solve for the specific voltages or currents requested.
7. After determining the vector V
§3.4: Mesh Current Analysis
Similar to node-voltage analysis, mesh-current analysis is a method used to completely solve a linear circuit, but by using mesh
or loop currents as the unknowns instead of node voltages. Everywhere that we wrote KCL equations before, we will now write
KVL equations based on these unknown currents. Consider the following examples.
15
Example: Use mesh-current analysis to solve for the current downwards in the 5Ω resistor.
Figure 29
The loop currents Ia and Ib represent imaginary vortices of current: while they are not measurable in a laboratory, they
are quite useful for theoretical analysis. The direction that the current flows (clockwise/counter-clockwise) is arbitrary, but a
suggestion is to take all loop currents to be oriented clockwise (the matrices will again have a minimal number of minus signs).
We will begin the analysis by seeing that the current Ia must be equal in magnitude, but opposite in sign, to the 3 A current
source.
Ia = −3 A
For the other equation, we apply KVL around the second loop. The net current through the 5Ω resistor is the difference of the
currents Ia and Ib ; if we are considering a clockwise loop going upwards across the 5Ω resistor, then the first sign of the voltage
across the resistor is positive when the net current is taken upwards, or I5Ω = Ib − Ia . Ib is the only current in the 10Ω resistor,
and we are going right to left for a clockwise loop, so the net current is just I b
Ib · (10 Ω) + 9 V + (Ib − Ia ) · (5 Ω) = 0
~.
Putting these two equations together in matrix format, we obtain a matrix equation of the form [Z] · I~ = V
„
« „ « „ «
1
0
Ia
−3
·
=
Ib
−5 10 + 5
−9
Which gives the solution:
«
„ « „
−3A
Ia
=
I~ =
−1.6A
Ib
We use this solution to calculate the current in the 5Ω resistor. As solved for previously, the current downwards in the resistor
is Ia − Ib .
I5Ω = Ia − Ib = (−3) − (−1.6) = −1.4 A
Example: Use mesh-current analysis to solve for the all of the currents in the circuit shown in Figure 3.30. Also, determine
the voltage across the 50Ω resistor.
Figure 30
We are unable to directly determine the voltage across a current source, so are unable to write a KVL equation containing the
2 Amp source. Instead, we write a current equation relating the current source to I b and Ic (notice that Ic has the same direction
as the 2 A source and thus contributes positively, while Ib is in the opposite direction and contributes negatively).
16
Ic − I b = 2 A
We then write our KVL equations such that they bypass the current source (also called creating a super-mesh).
−20 + 15 · Ia + 70 · (Ia − Ib ) = 0
70 · (Ib − Ia ) + 20 · Ib − 10 + 50 · Ic = 0
Combining these three equations into matrix format, we obtain another set of linear equations, whose solution is the vector of
unknown mesh currents.
0
1
0
1 0 1 0 1
2
−0.5A
0
−1
1
Ia
@15 + 70
−70
0 A · @ Ib A = @20A
=⇒
I~ = @−0.893AA
10
1.11A
Ic
−70
70 + 20 50
To compare this solution to our other methods, notice that V50 = 50 Ω · Ic = 55.36V just as before.
Example: Solve for the voltage across R1 using mesh-current analysis.
Figure 31
The first step is to determine how many equation we really need out of the four possible meshes: by combining the resistors
into appropriate equivalents as before, we can reduce the number of meshes to two.
Figure 32
Req1 = R3 k(R4 + R5 + R6 )
Writing the two independent equations,
Req2 = R1 kR2
Ia · Req2 − Vx + Ib · Req1 = 0
Ib − I a = I x
we can set up the appropriate matrix equation,
„
Req2
−1
Req1
1
« „ « „ «
Vx
Ia
=
·
Ix
Ib
17
and finally solve.
„ «
Ia
I~ =
=
Ib
Vx −Ix ·Req1
Req1 +Req2
Vx +Ix ·Req2
Req1 +Req2
!
Notice that this solution yields the same expressions as when we solved by superposition and nodal analysis.
Ix · Req1 − Vx
Req2
· Req2 = Ix · Req1 kReq2 − Vx ·
Req1 + Req2
Req1 + Req2
Example: Use mesh-current analysis to solve for the voltage across R1 in Figure 3.33.
VR1 = −Ia · R1 kR2 =
Figure 33
The two current equations will eliminate the left loop and combine the larger loops.
Ia = I 1
Ib − I c = I 2
The remaining equation is obtained by a clockwise loop about the current source I 2 .
Combining these equations
0
1
@ 0
−R1
0
1
R1
−V1 + R1 · (Ib − Ia ) + (R2 + R3 ) · Ic + V2 = 0
and solving yet another linear system, we obtain the currents in each of the meshes.
1
0
1 0 1 0
1
I1
0
Ia
I1
B I R +I2 ·(R2 +R3 )+V1 −V2 C
−1 A · @ Ib A = @ I2 A
=⇒
I~ = @ 1 1 R
A
1 +R2 +R3
V1 −V2 +(I1 −I2 )·R1
R2 + R 3
Ic
V1 − V 2
R +R +R
1
2
3
Summary of the steps for Mesh-Current analysis
The basic steps in solving a circuit via mesh-current analysis are:
1.
2.
3.
4.
5.
Draw and label as many mesh currents as necessary.
Write equations for any meshes containing current sources. Replace with super-meshes as appropriate.
Combine impedances where possible.
Write one independent equation for each of the remaining meshes. Solve the equations by hand, calculator, or computer.
Determine the desired quantities using the resulting vector of current values.
Comparison of Analysis Techniques
So far, we have seen three different analysis techniques for linear circuits. Voltage and current dividers work well, but only
when there is one power source. Superposition reduces the analysis of complicated circuits to simplified circuits (with zero-ed
18
sources) where we can use dividers to calculate the contribution of each source to a single desired quantity. Node-voltage analysis
uses KCL equations to determine all of the node voltages in a circuit, and mesh-current analysis uses KVL equations to determine
all the mesh currents, but neither solves directly for specific voltages or currents. Node/mesh analyses do however provide all
the information in the circuit at once. With this information, the question becomes: how do you choose which method to use in
solving a circuit?
We will explore various factors that affect which method we use: time, access to computing tools, analysis of perturbations,
and desired form of outputs are just a few to consider. Strictly speaking, analysis of circuits can take the form of any combination
of these methods. One remaining way to solve linear circuits is to create circuit equivalents (containing both impedances and
power sources), reducing the circuit as much as possible before analysing.
§3.5: Equivalent Circuits
Now that we know the basic steps to analyze a circuit, what happens if we want to use the circuit in a system, especially if that
system changes? Common devices as seemingly simple as an alarm clock may consist of hundreds of circuit elements, so we need
a modeling technique that allows us replace the large circuits by much smaller, and much easier to calculate, equivalent circuits.
The two types of representations we introduce now are called Thevinin and Norton equivalents. The Thevinin equivalent of any
linear circuit is defined as a voltage source in series with a single resistance (impedance) having the same terminal characteristics
as the original circuit, while a Norton equivalent is defined by a current source in parallel with the same Thevinin resistance
(impedance), still with identical terminal characteristics. Both the Thevinin and Norton equivalents could replace the original
circuit without us being able to tell the difference from outside the circuit.
Consider the proverbial black box that has two output terminals and consists of any unknown number of linear circuit
components (sources, resistances, open or short circuits, capacitors and inductors, etc.). From this box, we can determine
both the Thevinin and Norton equivalents by making the three measurements shown in Figure 3.34.
Figure 34
The first two measurements are with the black box connected to whatever sources that power the circuit elements. We measure the open-circuit voltage, Voc , as the terminal voltage with nothing connected to the output (hence the open circuit). The
short-circuit current, Isc , is obtained by placing a short circuit across the output terminals and measuring the current flowing
through it. The final measurement is to zero all of the sources inside the box (for a real device, unplug the box from the wall
socket) and then measure the equivalent resistance, Rth (impedance denoted Zth ), seen looking into the two terminals. These
values may then be used to construct each of our Thevinin and Norton equivalent circuits.
§3.6: Thevinin Equivalents
As stated previously, the goal of solving for a Thevinin equivalent is to reduce a large circuit into an externally equivalent
circuit consisting of only a single voltage source in series with an impedance.
19
Figure 35
The value of the voltage source is equal to the open-circuit or Thevinin voltage, V oc , and the impedance is the Thevinin
impedance, Zth .
Example: Calculate the Thevinin or open-circuit voltage and the Thevinin impedance to the left of the 50 Ω resistor in Figure
3.36. Verify that the equivalent circuit produces the same voltage V50 as before.
Figure 36
The first step is to designate what is inside the black box we are trying to model and what is outside the box. As designated
on the schematic, we are looking for the Thevinin equivalent to the left of the dashed line. This specifically means that the 50 Ω
resistor is not to be considered part of the circuit for modeling purposes, resulting in a true open circuit at the dashed lines. To
calculate this open-circuit voltage, we may use any of the circuit analysis techniques learned previously. The 10 V source does
not contribute any current to the rest of the circuit (due to its placement in series with an open circuit), and we are only looking
for one specific voltage: superposition will be the best choice.
The three circuits in Figure 3.37 show the decomposition of Figure 3.36 into sub-circuits with only one source active. We
calculate the contributions from each source to the voltage Voc and add these contributions to obtain the total Voc .
Figure 37
The trickiest of the three sub-circuits is the third one: since no current is able to flow through the open circuit, there is zero
current in any of the resistors. This results in a zero voltage across the resistors, whose potential is then added in series to the
potential difference of the 10V source. For similar reasons, there is no current flowing in the 20Ω resistor in the first circuit, and
thus Voc is the same as the voltage across the 70Ω resistor.
20
70
· (20V ) + (2A) · (20 + 15k70)Ω + (10V ) = 91.18 V
70 + 15
To calculate the Thevinin resistance (impedance), we zero all of the independent sources, and solve for the equivalent impedance
looking into the two terminals. Figure 3.38 shows the circuit with all sources zeroed.
Voc = Voc20V + Voc2A + Voc10V =
Figure 38
By inspection,
Rth = (20 + 15k70) Ω = 32.35 Ω
Putting these two parameters together, we form a series combination of V oc and Rth and replace everything to the left of the
original dashed line with our new equivalent.
Figure 39
How do we check if this new “equivalent” circuit truly has the same effect as the original? Reconnecting the 50 Ω resistor as a
load onto the new circuit, we can calculate the voltage V50 , which should be the same voltage we solved for in previous examples.
Figure 40
Using a single voltage divider,
50
= 55.36 V
50 + Rth
Now why in the world did we go through that much work to obtain the same answer as before? Consider a case where the
left side of our circuit (for which we obtained the Thevinin equivalent) is a stereo amplifier and the 50 Ω resistor is a collection
of speakers that we want to hook up. What happens if we then want to add/remove a speaker? Each of the other numerical
analyses performed would be useless once an impedance value is changed, whereas the Thevinin equivalent makes the calculation
easy in any case; simply recalculate the voltage divider at the end. Let’s try a few more examples, and then we will explore the
V50 = Voc
21
relationships between methods in more depth.
Example: Use a Thevinin equivalent circuit to solve for the current through the 5Ω resistor in Figure 3.41. Repeat the
solution with the 5Ω resistor changed to 25Ω.
Figure 41
Although smaller, this circuit is harder to conceptualize. We may redraw the circuit without the 5Ω resistor and stretch out
the nodes for better visualization.
Figure 42
Solving for Voc by superposition, and Zth by zeroing all of the sources, we obtain the values for the Thevinin equivalent.
Voc = 9V + (−3A) · 10Ω = −21V
Zth = 10Ω
To calculate the voltage across a load resistor connected to the output nodes, we attach an arbitrary load, R load , as shown in
Figure 3.43, and use Ohm’s law to get the current.
Figure 43
−21V
10Ω + Rload
For load resistors of 5Ω and 25Ω, we calculate −1.4A and −0.6A, respectively.
IRload =
Some circuits simply have no easy solution with the previous techniques discussed. Thevinin and Norton equivalents provide
a new method that works where others fail.
Example: Consider the bridge circuit in Figure 3.44: determine the voltage across R 5 .
22
Figure 44
Initially, the options look rather bleak; node-voltage analysis will work, but the solution will be extremely tedious. If we
remember that potentials are the same anywhere along a node, we can simplify the circuit by splitting the voltage source into
two parallel sources of the same value, shown in Figure 3.45.
Figure 45
With identical potentials at both the top and bottom nodes, we can see that there will be no current flowing at the top of R 1
and R2 or below R3 and R4 . Therefore, removing the connections will not affect our analysis of the circuit.
23
Figure 46
Now we’re in business: take Thevinin equivalents to the right and left of R5 , and then replace to solve for VR5 . Both sides
solve as in Figure 3.47.
Figure 47
The resulting circuit with Thevinin equivalents inserted shows a single series loop from which to calculate V R5 .
Figure 48
24
VR5 = V · [
R3
R4
R5
−
]·
R1 + R 3
R2 + R4 R1 kR3 + R5 + R2 kR4
§3.7: Norton Equivalents
A Norton equivalent is the other equivalent where we reduce any linear circuit to a single current source in parallel with an
impedance.
Figure 49
The current is termed either the Norton current or the short-circuit current, and the impedance is again called the Thevinin
impedance (for reasons we shall see). The short-circuit current, Isc , is calculated by externally applying a short circuit to the two
nodes and then measuring the current across the short. The Thevinin impedance is calculated the same way as before: zero all
independent sources and find the equivalent impedance lookinig into the two terminals.
Example: Use a Norton equivalent circuit to calculate the voltage across the 50Ω resistor in Figure 3.50.
Figure 50
Let’s find the Norton equivalent to the left of the 50Ω resistor. We will solve for the short circuit current (via superposition)
flowing between the two output nodes.
25
Figure 51
Some of the divider terms are tricky at first glance. In the first circuit, calculate the current into the 20Ω resistor as a divider
between the 20 and 70 Ω resistors; notice that the two resistors are in parallel due to the addition of the external short circuit.
When calculating the equivalent impedance for the total current out of the 20V source, we consider the impedance looking right
from the voltage source, not left into the equivalency nodes. The second circuit is easy: current takes the path of least resistance,
so all of the current goes through the short circuit. The third circuit is a single application of Ohm’s law.
Isc =
20V
10V
70
·
+ (2A) +
= 2.818 A
20 + 70 (15 + 20k70) Ω
(15k70 + 20) Ω
Now, we zero all of the independent sources and calculate the Thevinin impedance.
Zth = 20 + 15k70 = 32.35Ω
Putting these two terms together, we redraw as the Norton equivalent circuit.
Figure 52
To verify our calculations, we find the voltage across the 50Ω resistor.
V50 = Isc · (Rth k50) = 55.35 V
We again see an immediate benefit of the equivalent circuit: responses may be easily calculated for any load resistor!
Example: Calculate the Norton equivalent of the circuit shown in Figure 3.53, to the right of R 1 , and then us the equivalent
to determine the voltage VR1 .
26
Figure 53
We reduce the circuit by taking the equivalent resistance of the right side, and begin using superposition to solve for the short
circuit current, followed by the Thevinin impedance.
Figure 54
Solving for the Norton equivalent values:
Isc = Ix +
−Vx
Req1
Zth = R2 kReq1
The resulting Norton equivalent with R1 reattached is shown in Figure 3.55.
Figure 55
We will solve for the voltage VR1 in order to verify equivalency.
VR1 = Isc · (Rth kR1 ) = (Ix +
R1 kR2
−V x
) · (R1 kR2 kReq1 ) = Ix · Req1 kR1 kR2 + (−Vx ) ·
Req1
Req1 + R1 kR2
Example: Use a Norton equivalent circuit to solve for the current through the 5Ω resistor in Figure 3.56.
27
Figure 56
Replacing the 5Ω resistor by a short circuit in Figure 3.57, we may calculate our two parameters: I sc and Zth .
Figure 57
9V
= −2.1 A
Zth = 10 Ω
10Ω
Replacing the 5Ω resistor and calculating a current divider, we obtain the current I 5 .
Isc = (−3A) +
10
· (−2.1A) = −1.4A
10 + 5
Example: Use a Norton equivalent circuit to solve for the voltage across R 1 in Figure 3.58.
I5 =
Figure 58
Using superposition again (with a short circuit in place of R1 ), we obtain the expressions for short circuit current (oriented
downward) and Thevinin impedance.
Isc = I1 − I2 +
V2 − V 1
R2 + R 3
Zth = R2 + R3
28
And then finally solve for the voltage Vx .
Vx = Isc · R1 k(R2 + R3 ) = (I1 − I2 ) · R1 k(R2 + R3 ) + (V2 − V1 ) ·
R1
R1 + R 2 + R 3
§3.8: Duality of Equivalents: Source Transformations
We have so far derived two entirely different methods for modeling a large linear circuit as a simple “equivalent;” if both
methods produce circuits equivalent to the original, then they should also be equivalent to each other. Consider the two circuits
in Figure 3.59, which we will take to be the Thevinin and Norton equivalents of yet a third circuit, not pictured.
Figure 59
By taking the Thevinin equivalent of the Norton equivalent, and likewise taking the Norton equivalent of the Thevinin
equivalent, we shall find a direct relation between them.
Rth = Zth
Voc = Zth Isc
Isc =
Voc
Zth
Thus to find both equivalent circuits, we only need to find one of the equivalents and then convert to the other. In practice,
this property allows us to build in another verification step to tell us whether our analysis is correct. The easiest, and thereby
the least likely to generate errors, of the three quantities (Voc , Isc , and Zth ) is the equivalent impedance. As a result, the surest
oc
to the calculated value of Zth .
method is to calculate both Voc and Isc and then compare the ratio VIsc
Converting a single voltage source in series with an impedance to current source in parallel with the same impedance is called
a source transformation. This transformation can often simplify circuits dramatically; let’s again consider the circuit shown in
Figure 3.60 for which we found Thevinin and Norton equivalents.
Figure 60
We still want to find the Thevinin/Norton equivalent to the left of the dashed line, but we are going to systematically perform
source transformations from the far left until we have completed the equivalent. The first step will be a source transformation for
the series combination of 20 V voltage source and 15 Ω resistor, shown in Figure 3.61.
29
Figure 61
V
Taking the Norton equivalent of just the 20V source and the 15Ω resistor, we calculate a short-circuit current I sc = 20
=
15 Ω
1.333 A, and the same 15 Ω impedance. Substituting this equivalent into the circuit, we obtain the one shown in Figure 3.62.
Figure 62
We now have the 15 and 70 Ω resistors in parallel, creating an equivalent impedance of 12.35 Ω. Reversing the source transformation procedure with the new impedance (taking a Thevinin equivalent of the 12.35Ω resistor in parallel with 1.33 A source),
we obtain the circuit shown in Figure 3.63.
Figure 63
Absorbing the 20 Ω resistor and taking one more source transformation, we see a new simplification: two current sources in
parallel.
30
Figure 64
The two current sources add directly, so we are in effect absorbing the current source into our model. Transforming one last
time, we obtain two series voltage sources (which add directly) and a single impedance, which is the overall Thevinin equivalent
and the same equivalent circuit we obtained previously.
Figure 65
We shall consider one more example of using source transformations to reduce a circuit.
Example: Determine the Norton equivalent to the right of R1 in the circuit shown in Figure 3.66.
Figure 66
Using the same equivalent resistance, Req1 = R3 k(R4 + R5 + R6 ), as before and a single transformation to the right of the
current source, we nearly end up with the overall Norton equivalent.
31
Figure 67
Combining parallel current sources and resistances, we obtain the same equivalent as before, but in only two simple steps!
One word of caution: equivalent circuits as with any other “equivalent” are not the same thing as the original. The equivalent
circuits are wonderful mathematical tools for modeling a system, but this convenience does come at a cost: tracking variables
internal to the collapsed circuit is nearly impossible. The equivalent circuit derived at one two-port junction will be different than
another two elements away. By no means does this realization make Thevinin or Norton equivalents unimportant–we simply see
that Norton/Thevinin equivalents are suitable for some applications and not others.
In general, the steps used to find the Thevinin and Norton equivalents of a circuit are:
1. Separate those elements that are part of the equivalent and those that are external. Clearly label the polarity of V oc and
the direction of Isc , making sure they line up with the polarity/direction of the desired equivalent.
2. Solve for Voc and Isc using any method desired: superposition, nodal or mesh analysis, etc.
3. At any point, you may wish to employ a source transformation to reduce the circuit to a simpler form.
4. Replace the larger circuit by its calculated equivalent and proceed with any external load calculations.
§3.9: Maximum Power Transfer
We have stated that both Thevinin and Norton equivalents can be used to model a linear circuit for use in calculating responses
due to many loads. The most common goal in electronics is to transfer information with the least loss in the signal: in circuits,
this goal translates to delivering the maximum amount of power possible to a load. How do we achieve such maximum power
transfer?
Consider a loaded Thevinin equivalent, where both the Thevinin impedance and the load are purely real (resistors), as shown
in Figure 3.68.
Figure 68
The power transferred to the load is a product of its current and voltage (other forms give the same result). We may write
this power as:
Pload =
Vth · Rload
Vth
·
Rth + Rload Rth + Rload
32
To maximize this power relation, we differentiate with respect to Rload , and solve for the condition on Rload to make the
derivative zero.
d
Pload = 0
=⇒
Rload = Rth
dRload
We find that the maximum power transfer occurs when Rload = Rth ; or, equivalently, maximum power transfer occurs when
the load resistance to any device is equal to the Thevinin resistance of the driving circuit (for general impedances, maximum
∗
power transfer occurs when Zload = Zth
).
To verify this condition on Rload qualitatively, consider again the power delivered to Rload as the product of the voltage across
it and the current through it. If Rload is too small, then the voltage across it is also small, while if Rload is too large, the current
through both Rth and Rload will be made small. Somewhere in between we obtain the maximum of this product, as shown in
Figure 3.69.
Figure 69
The value of the maximum real power that a circuit can deliver may be written using the Thevinin resistance and either the
short-circuit current or the open-circuit voltage. (Notice that for Rload = Rth , the value of both the voltage divider for the voltage
across the load and the current divider for the current through the load are equal to 21 .)
2
Voc
I 2 Rth
= sc
4Rth
4
We will reconsider this limit on power transfer for general impedances later on.
Pmax =
Example: Solve for the load resistance that maximizes the power transfer from the circuit in Figure 3.70. What is the
maximum power that can be transferred?
Figure 70
Solution: To solve for the appropriate load resistance, we only need to find the Thevinin resistance of the circuit, and then
specify the load resistance to have the same value.
Rload = Rth = 15 + 6k30 = 20 Ω
To determine the amount of power delivered to the load, we need to calculate either the open-circuit voltage or the short-circuit
current. The expressions for both are shown.
33
30
+ (1 A) · (6k30) = 15 V
30 + 6
6k30
(12 V )
30
·
+ (1 A) ·
= 0.75 A
=
6 + 15k30 15 + 30
6k30 + 15
Voc = (12 V ) ·
Isc
With these values calculated, we determine the power delivered to a 20 Ω load resistor.
P20 Ω =
(15 V )2
(0.75 A)2 · 20 Ω
=
= 2.81 W
4 · 20 Ω
4
Example: What is the value of R in Figure 3.71, assuming that the 10 Ω load resistor was chosen for maximum power transfer?
How much power is delivered?
Figure 71
To determine R, find the Thevinin resistance and set it equal to the 10 Ω to ensure maximum power transfer.
Rth = R + 2 + 0 = 10 Ω
=⇒
R = 8Ω
Notice that zeroing the voltage sources also shorts out the 15 Ω resistor, resulting in the zero term above. Next, to find the
power delivered, find either the open-circuit voltage or the short circuit current,
Voc = (2 V ) + (1 A) · (2 Ω) = 4 V
Isc =
(2 V )
2
+ (1 A) ·
= 0.4 A
2+8
2+8
and calculate the power.
P10 Ω =
(4 V )2
(0.4 A)2 · 10 Ω
=
= 0.4 W
4 · 10 Ω
4
Exercise: Explain why the 15 Ω resistor in the previous circuit played no part in the analysis or solution. Consider the
properties of an ideal voltage source.
You will often hear electrical engineers speak of “impedance matching:” one of the most common errors in system design is
for subsystems to have an impedance mismatch, thereby wasting electrical power. The idea of maximum power transfer and
impedance matching drives good circuit design. We will come back to this idea in discussions of complex power and system
integration.
§3.10: Equivalent Resistance Revisited
We have seen that some configurations of resistances (impedances), for example the bridge configuration, have no series or
parallel components. To obtain the equivalent resistances of these circuits, we must try a new method besides successive parallel
and series equivalents. If we use our knowledge of Thevinin and Norton equivalent circuits, we may express the equivalent
resistance (impedance) as a quasi-ratio of a open-circuit voltage and short-circuit current. More exactly, we attach a 1V voltage
source or a 1A current source and test for the current or voltage, respectively.
34
Figure 72
The numerical value of the induced voltage, VT , or the numerical value of the inverse of the current IT will be the same as the
equivalent resistance (impedance), Req .
VT = (1A) · Req
=⇒
Req =
VT
= VT [Ω]
1A
IT =
(1V )
Req
=⇒
Req =
1V
1
=
[Ω]
IT
IT
The bracket notation attached to the final answer is to remind you of the appropriate units associated with the quantities.
Let’s begin by considering a simple example that could be solved using the original method.
Example: Use the more generalized solution method to determine the equivalent resistance of the circuit in Figure 3.73.
Figure 73
Solution: Choosing a 1A current source as an input to the circuit, we will test for the corresponding voltage across the two
terminals. With 1A through the 10 Ω resistor, we see a 10V voltage in series with the voltage across the rest of the elements,
6k(25 + 5) · 1 A = 5 V . This total voltage, 15V, is then numerically identical to the equivalent resistance of 15 Ω.
Clearly, the last example used a more round-about approach to solving for the equivalent resistance than introduced previously,
but let us consider again the resistance of the Wheatstone bridge circuit shown in Figure 3.74.
Figure 74
First, we attach a 1V test source to the terminals; to then determine the equivalent resistance, we must measure the current
flowing out of the source which is identical to the sum of the currents flowing into R 1 and R3 . The problem is that we cannot do
this directly: instead we will repeat our previous solution for the voltages at nodes C and D (using Thevinin equivalents to the
left and right of R5 ).
35
»
–
R5 + R3 kR4
R2
R4
R4
+ (1 V ) ·
−
·
R3 + R 4
R1 + R 2
R3 + R 4
R1 kR2 + R3 kR4 + R5
»
–
R3 kR4
R2
R4
R4
+ (1 V ) ·
−
= (1 V ) ·
·
R3 + R 4
R1 + R 2
R3 + R 4
R1 kR2 + R3 kR4 + R5
VC = (1 V ) ·
VD
Now, returning to the original circuit, we may calculate the currents down through R 1 and R3 ,
VA − V D
VA − V C
I R3 =
R1
R3
and then add them using a Kirchoff’s Current Law equation at node A to obtain the current through the voltage source and
our equivalent resistance.
I R1 =
R5 · (R1 kR3 ) + R1 R3
1
[Ω] = .... =
+ R2 kR4
IT
R1 + R 3 + R 5
We will not waste space with massaging the algebraic formulation into the closed form; rather, we will present one numerical
example, and show that both representations work.
I T = I R1 + I R3
Req =
Example: Determine the equivalent resistance of the bridge circuit in Figure 3.75.
Figure 75
Attaching a 1V test source, we calculate the voltages at nodes C and D.
3 + 2k5
5
4
5
V)+(
−
)V ·
= 0.787 V
2+5
1+4
2+5
1k4 + 3 + 2k5
2k5
4
5
5
V)+(
−
)V ·
= 0.738 V
VD = (
2+5
1+4
2+5
1k4 + 3 + 2k5
Returning to the original circuit, we calculate the current into the 1 Ω and 2 Ω resistors.
VC = (
(1 V ) − (0.738 V )
(1 V ) − (0.787 V )
= 0.213A
I2 Ω =
= 0.131A
1Ω
2Ω
We then calculate a total current from the voltage source of IT = 0.344 A. Finally, the overall equivalent resistance may be
calculated.
I1 Ω =
Req =
(1 V )
= 2.90 Ω
(0.344 A)
We compare this to the closed-form solution to verify.
3 · (1k2) + 1 · 2
+ 4k5 = 2.89 Ω
1+2+3
The two solutions are identical, but you should notice that roundoff errors due to truncating values CAN affect the final answer,
as in any engineering calculation...
Req =
36
Exercise: In many power systems, bridge circuit are used on a regular basis; explain how you would modify the previous
analysis for two bridge circuits in parallel (hint: think about the step where KCL is applied)?
Delta-Wye Transforms
To conclude our discussion on equivalent resistances (impedances), we state two well-known, but rarely used, results for
converting to and from the ∆ and Y configurations of generic impedances shown in Figure 3.76. We have derived all of the
analysis to come up with these results in previous examples, but have omitted the algebraic mess.
Figure 76
Z1 Z3
Z1 + Z 2 + Z 3
Z2 Z3
ZB =
Z1 + Z 2 + Z 3
Z1 Z2
ZC =
Z1 + Z 2 + Z 3
ZA ZB + Z A ZC + Z B ZC
ZB
ZA ZB + Z A ZC + Z B ZC
Z2 =
ZA
ZA ZB + Z A ZC + Z B ZC
Z3 =
ZC
Z1 =
ZA =
⇐⇒
§3.11: Linear Circuit Examples
Without a doubt, the prerequisite to truly understanding linear circuits is practice. The individual concepts are in practice
extraordinarily simple: Ohm’s law is just V = I R, power is P = I V , and Kirchoff’s laws are simple sums. Putting the concepts
together to solve any particular circuit is a more difficult task. We complete this rather long discourse on solving linear circuits
by giving two more basic circuits and solving with each of the methods presented.
The first set of examples uses a large, but relatively simple, linear circuit shown in Figure 3.77. We will see that fundamental
concepts like equivalent resistances as well as voltage and current dividers are used over and over.
37
Figure 77
Example: To start out, solve for the equivalent voltage Vx across the 6 Ω resistor.
We solve by zeroing all but one source at a time, calculating the contribution of each source to the voltage V xi , and finally
adding the contributions to determine the total voltage Vx . Figure 3.78 shows each of the four circuits with only one source
activated.
Figure 78
Each of the individual circuits reduces to a voltage or current divider, adding up to obtain the voltage V x .
Vx = Vx0.5 A + Vx15 V + Vx1 A + Vx22 V
6k(8 + 12)
12
· (6k50 Ω) + (15 V ) ·
12 + (8 + 6k50)
6k(8 + 12) + 50
6k50
8 + 12
+ (−1 A) ·
· (6k50 Ω) + (22 V ) ·
8 + 12 + 6k50
6k50 + 8 + 12
= (1.268 V ) + (1.268 V ) + (−4.225 V ) + (4.648 V )
= (0.5 A) ·
Example: Next, setup and solve the necessary node-voltage equations to obtain V x .
The redrawn circuit, with all nodes labeled, is shown in Figure 3.79.
Figure 79
= 2.958 V
38
We write the four equations necessary to solve,
VA
VA − V B
VB − V A
VC
VC − V D
+
= 0.5 A
VD = 15 V
+ (1 A) +
+
=0
12
8
8
6
50
and then compute the four node voltages. You should readily notice that the desired unknown, V X , is the same as node voltage
VC .
VC − VB = 22 V
VA = −9.028 V
VB = −19.046 V
VC = Vx = 2.954 V
VD = 15 V
We see that the node voltage VC is 2.954 V, as compared to the 2.958 V found previously (identical within roundoff).
Example: Now, setup and solve the necessary mesh-current equations to solve for V x .
The redrawn circuit, with all meshes labeled, is shown in Figure 3.80.
Figure 80
Writing the equations for the four meshes,
I1 = 0.5 A
(I2 − I1 ) · (12 Ω) + I2 · (8 Ω) − 22 V + (I3 − I4 ) · (6 Ω) = 0
I2 − I 3 = 1 A
(I4 − I3 ) · (6 Ω) + I4 · (50 Ω) + (15 V ) = 0
and solving to obtain Vx
I1 = 0.5 A
I2 = 1.2521 A
I3 = 0.2521 A
I4 = −0.2408 A
Vx = (I3 − I4 ) · (6 Ω) = 2.597 V
Example: Next, let’s find Vx using a Thevinin equivalent. To do so, we remove the 6 Ω resistor from the circuit and proceed
to find the open circuit voltage (via superposition) and the Thevinin equivalent resistance as shown in Figure 3.81.
Figure 81
The open circuit voltage is obtained by superposition,
(8 + 12)
50
50
+ (−1 A) · [(8 + 12)k50 Ω] + (22 V ) ·
+ (15 V ) ·
8 + 50
50 + (12 + 8)
50 + (8 + 12)
= (4.286 V ) + (−14.286 V ) + (15.714 V ) + (4.286 V )
Voc = (0.5 A) · [12k(8 + 50) Ω] ·
= 10 V
and the equivalent resistance is calculated by first zeroing all the sources and then finding the equivalent.
39
Zth = 50k(8 + 12) = 14.286 Ω
The final calculation is to use a voltage divider for the voltage Vx across the replaced 6 Ω resistor.
Vx = Voc ·
6
6
= (10 V ) ·
= 2.958 V
Zth + 6
14.286 + 6
Example: One more method available is to use the Norton equivalent circuit. Find the Norton equiavlent and then use Ohm’s
law to determine the voltage Vx .
Once again, we remove the 6 Ω resistor, but this time attach a short circuit in its place as shown in Figure 3.82. Care has
been taken to make sure that the positive direction of Isc corresponds to the proper polarity for Vx .
Figure 82
Using superposition one last time, we solve for the short circuit current,
(22 V )
(15 V )
12
+ (−1 A) +
+
8 + 12
(8 + 12) Ω
50 Ω
= (0.3 A) + (−1 A) + (1.1 A) + (0.3 A)
Isc = (0.5 A) ·
= 0.7 A
and then solve for the voltage across the parallel equivalent of Thevinin resistance (same as before) and the 6 Ω resistor.
Vx = Isc · (Zth k6) = (0.7 A) · (14.286k6 Ω) = 2.958 V
Example: The final method to solve this circuit is to use source transformations. Taking the original circuit, we iteratively
reduce the circuit to either a Thevinin or Norton equivalent. Performing one source transformation on both the left and the right
side of the circuit, we reduce it to the one shown in Figure 3.83.
Figure 83
Three more iterations shown in Figures 3.84, 3.85, and 3.86 give the same Norton equivalent as solve for previously.
40
Figure 84
Figure 85
Figure 86
With the final step, we calculate the voltage Vx .
Vx = (0.7 A) · (14.286k6 Ω) = 2.958 V
Exercise: What are the benefits/limitations for each of the methods presented in this section?
We will repeat the full analysis for a second circuit that contains a number of transparent tricks. The analysis for the various
methods will actually be less involved, but require you to see fundamental simplification.
Example: Solve for the voltage Vx across the 15Ω resistor in Figure 3.87 using each of the circuit methods.
Figure 87
41
Before we begin the analyses, we should take a moment to guesstimate what the overall characteristics will be:
1. The circuit has four independent sources, so superposition will require solving four separate circuits.
2. The circuit has 6 independent sources, but three voltage sources that will make for simple node equations.
3. The circuit has 5 independent meshes, yet only one current source for a simple equation.
4. The 3V, 12V, and 1A will contribute to current flowing upwards in the 15Ω resistor (creating a negative contribution to
Vx ), while the 22V source will contribute to a downward current and a positive V x .
5. The 4Ω resistor in series with the current source and the 8Ω resistor in parallel with the 3V voltage source will not affect
any of the other circuit variables; the voltage across a current source and the current through a voltage source are indeterminate.
Any value of resistance can be used for these two without changing other circuit variables.
6. Although non-trivial to determine, by far the easiest method to solve this circuit would be source transformations in combination with superposition.
We will comment on these observations in more details as we proceed.
Superposition: We repeat the same process as before to solve the circuit by superposition, but simplify by removing the
4Ω resistor (shorting to 0Ω) and the 8Ω resistor (opening to ∞ Ω) when redrawing the equivalent circuits. The four simplified
circuits with a single source active are shown in Figure 3.88.
Figure 88
To calculate the voltage Vx , we take each contribution and then add them up.
42
6
· (15k30 Ω) = −3.396 V
6 + 10k2 + 15k30
15k30
Vx3 V = (−3 V ) ·
= −1.698 V
15k30 + 10k2 + 6
15k(6 + 2k10)
= 3.182 V
Vx22 V = (22 V ) ·
30 + 15k(6 + 2k10)
2k(6 + 15k30)
15k30
Vx12 V = (−12 V ) ·
·
= −1.132 V
10 + 2k(6 + 15k30) 6 + 15k30
Vx = Vx1 A + Vx3 V + Vx22 V + Vx12 V = −3.04 V
Vx1 A = (−1 A) ·
Node-Voltage Analysis: We label the nodes as shown in Figure 3.89, choosing the bottom node as ground.
Figure 89
We have left the 4Ω and 8Ω resistors in the circuit, but can quickly show that they do not affect the remainder of the circuit.
The current through the 4Ω resistor is the same as the 1A source
VA − V B
+ (1 A) = 0
4Ω
while the potential difference between node E and ground is just the voltage source.
VE − 0 = (−3 V )
Changing the values of these two resistors will change either the current through (the 8Ω) or the voltage across (the 4Ω) the
resistors. Writing the equations to solve for the node voltages (taking all currents leaving the nodes),
VE − 0 = (−3 V )
VC − VD = (12 V )
VF − 0 = (22 V )
VB − V E
VB − V D
VB − V C
VD − V B
VC − V B
VD − 0
VD − V F
+
+
=0
+
+
+
=0
6
2
10
2
10
15
30
Solving 5 equations simultaneously is a pain, but the results are identical.
(1 A) +
VA = −6.77 V
VB = −2.77 V
VC = 8.96 V
VD = −3.04 V
Vx = VD = −3.04 V
Mesh Current Analysis: First, we label the meshes as shown in Figure 3.90,
VE = −3 V
43
Figure 90
Disregarding the 8Ω resistor, we end up with four meshes, resulting in the four equations:
I1 = −(1 A)
(I2 − I3 ) · 2 + I2 · 10 + (12 V ) = 0
(3 V ) + (I3 − I1 ) · 6 + (I3 − I2 ) · 2 + (I3 − I4 ) · 15 = 0
(I4 − I3 ) · 15 + I4 · 30 + (22 V ) = 0
Solving these equations, we obtain the four mesh currents,
I1 = −1 A
I2 = −1.173 A
I3 = −1.038 A
I4 = −0.835 A
and the overall solution for Vx .
Vx = (I3 − I4 ) · (15 Ω) = −3.044 V
Thevinin and Norton Equivalents
Now we will repeat the solution for the voltage Vx using both Thevinin and Norton equivalent circuits; since the two methods
are so close to one another, we will solve them together. We first remove the 15Ω resistor from the circuit and solve for the open
circuit voltage, Voc , across the terminals using superposition.
Figure 91
The solution of Voc in this circuit is identical to the analysis of the original circuit with the 15Ω resistor increased to ∞ Ω, or
an open circuit. To show this modification in a little more detail, first take the previous expression for V x using superposition
0
and including the 15Ω resistor (which we call Voc
).
44
0
Voc
= (−1 A) ·
1A
6
· (15k30 Ω) = −3.396 V
6 + 10k2 + 15k30
0
Voc
= (−3 V ) ·
3V
15k30
= −1.698 V
15k30 + 10k2 + 6
0
Voc
= (22 V ) ·
22 V
15k(6 + 2k10)
= 3.182 V
30 + 15k(6 + 2k10)
0
Voc
= (−12 V ) ·
12 V
2k(6 + 15k30)
15k30
·
= −1.132 V
10 + 2k(6 + 15k30) 6 + 15k30
0
Now we replace each 15Ω term with ∞ Ω to obtain Voc from Voc
. The equivalent resistance of an open circuit (∞ Ω) in series
with any finite resistor is the open circuit, in parallel with any finite resistor is just the resistor.
Voc1 A = (−1 A) ·
6
6
· (∞k30 Ω) = (−1 A) ·
· (30 Ω) = −4.779 V
6 + 10k2 + ∞k30
6 + 10k2 + 30
Voc3 V = (−3 V ) ·
∞k30
30
= (−3 V ) ·
= −2.389 V
∞k30 + 10k2 + 6
30 + 10k2 + 6
Voc22 V = (22 V ) ·
∞k(6 + 2k10)
(6 + 2k10)
= (22 V ) ·
= 4.479 V
30 + ∞k(6 + 2k10)
30 + (6 + 2k10)
Voc12 V = (−12 V ) ·
2k(6 + ∞k30)
∞k30
2k(6 + 30)
30
·
= (−12 V ) ·
·
= −1.593 V
10 + 2k(6 + ∞k30) 6 + ∞k30
10 + 2k(6 + 30) 6 + 30
Voc = Voc1 A + Voc3 V + Voc22 V + Voc12 V = −4.282 V
Before having the answer for Vx , we need to determine the Thevinin resistance as seen from the two terminals with all sources
zeroed: looking right, we see a 30Ω resistor and to the left, we see 6 + 10k2 Ω (the 4Ω is swamped by the open circuit of the zeroed
current source and the 8Ω resistor is killed by the zeroed 3V voltage source).
Rth = 30k(6 + 10k2) = 6.106 Ω
Finally, we determine the voltage Vx as a voltage divider with the Thevinin equivalent circuit.
15
15
= (−4.282 V ) ·
= −3.043 V
Rth + 15
6.106 + 15
Without a doubt, this circuit was much easier to solve with the original application of superposition! For completeness, we
will repeat to find the short circuit current of the circuit shown in Figure 3.92.
Vx = Voc ·
Figure 92
Again, the most straightforward method is to simply use superposition to add up the individual sources’ contributions,
45
6
−12 V
2
−3 V
22 V
+
·
+
+
6 + 10k2
10 + 2k6 2 + 6
2k10 + 6
30
= (−0.783 A) + (−0.261 A) + (−0.391 A) + (0.733 A)
Isc = (−1 A) ·
= −0.702 A
and then apply Ohm’s law to determine the voltage Vx across the parallel combination of same Rth calculated before and the
replaced 15Ω resistor.
Vx = Isc · (Rth k15) = (−0.702 A) · (6.106k15 Ω) = −3.046 V
Question: Why can you not modify the superposition solution for Vx to solve for Isc ?3 One hint is to consider the voltage
across a short circuit.
We found that the Norton equivalent method requires roughly the same amount of work as the original superposition solution.
After working many, many, of these circuits, you will begin to see how the methods can be intermixed for quicker solutions. Our
last example will be a hybrid method of source transformations (single-stage Thevinin or Norton equivalent) and superposition.
Hybrid Source Transformations and Superposition
First of all, take another look at the original circuit.
Figure 93
As previously discussed, the 4Ω and 8Ω resistors do not affect the rest of the circuit, so we throw them away (shorting the 4Ω
and opening the 8Ω). We then notice that we have three instances of single voltage sources in series with singular resistances; we
perform a source transformation on each of these terms simultaneously, obtaining the circuit in Figure 3.94.
Figure 94
3Although practically useless, you may divide the entire expression for V as derived using superposition by the 15Ω resistor and then take the
x
limit of the entire expression as the 15Ω → 0Ω, but will be forced to use L’Hopital’s rule and many, many extra steps.
46
We begin adding parallel current sources and parallel resistors to obtain the simplified circuit shown in Figure 3.95.
Figure 95
From this point forwards, superposition and source transformations will require roughly the same amount of effort; we choose
superposition since the expression can be written by inspection and without redrawing the circuit any further (you can easily
visualize the zeroed current sources by completely detaching them).
»
Vx = (10 Ω) · (−1.5 A) ·
–
(6 + 1.667)
1.667
6
+ (−1.2 A) ·
+ (0.733 A) ·
= −3.045 V
6 + (10 + 1.667)
1.667 + (10 + 6)
(6 + 1.667) + 10
Exercise: What characteristics of the circuit cause each of the methods to be easier or harder?
§3.12: Summary
In this chapter, we have seen all of the basic tools used to solve linear circuits: voltage and current dividers; superposition;
node-voltage and mesh-current analyses; Thevinin and Norton equivalents; and source transformations.
Voltage and current dividers give the fraction of a total voltage or current from a single voltage or current source.
Superposition allows you to decompose a circuit containing multiple sources into multiple circuits (having only a single source)
that make linear contributions to the overall circuit quantities.
Node-voltage and mesh-current analyses make possible simultaneous solution of all linear circuit variables by transforming
multiple applications of Kirchoff’s laws into linear systems of equations. Ohm’s law is reinterpreted in a matrix form: either
V̄ = [R] · I¯ or I¯ = [G] · V̄ . Note that the “ground node” is simply a or potential energy reference defined to be zero; all node
voltages are measured relative to this ground. A final comment is that mesh currents are not physically measurable quantities;
node voltages can be easily measured in a lab and solved by a computer.
Thevinin and Norton equivalents are a circuit modeling technique providing a simple circuit, either a single voltage source in
series with a resistor or a single current source in parallel with a resistor, that can be used to calculate a circuit’s effect on any
number of loads.
Source transformations are a special case of Thevinin and Norton equivalents, where a single voltage source in series with a
resistance is transformed into a current source in parallel with a resistor, or vice versa.
More often than not, the best method of solving a linear circuit will consist of a hodge podge of methods. Superposition is
perhaps the most anaytically straightforward, yet requires (redrawing and) solving multiple smaller circles. Node-voltage analysis
is the most computationally efficient way to solve an entire circuit, while superposition is often better for a single circuit variable.
Finally, the two circuit equivalents help determine how an entire circuit will contribute when attached to an external load; by
using these equivalents, we ease external calculations, but we lose the ability to calculate internal circuit quantities.
The bottom line in understanding how to solve a linear circuit is practice. No manual, textbook or professor can tell you
anything beyond V = IR, P = IV , or sums of currents and voltages being zero at a node or in a closed loop, respectively. Every
one of the time-domain, AC, analog, and digital circuits discussed throughout the remainder of the text relies on these four basic
concepts.
47
Problem 3.1 What is the Thevinin equivalent of a voltage source in parallel with a resistor? Why can you not find a Norton
equivalent?
Problem 3.2 What is the Norton equivalent of a current source in series with a resistor? Why can you not find a Thevinin
equivalent?
Problem 3.3: For the linear circuit shown in Figure 3.96, determine the voltage across the two output terminals.
Figure 96
Problem 3.4: For the linear circuit shown in Figure 3.97,
(a)
Use superposition to determine the voltage Vx .
(b)
Solve for Vx using node-voltage analysis.
(c)
Solve for Vx using mesh-current analysis.
Figure 97
Problem 3.5: For the linear circuit shown in Figure 3.98,
(a)
Use superposition to determine the voltage Vx .
(b)
Solve for Vx using node-voltage analysis.
(c)
Solve for Vx using mesh-current analysis.
(d)
Solve for Vx using the Thevinin equivalent (first remove the 84kΩ resistor).
(e)
Solve for Vx using the Norton equivalent (first remove the 84kΩ resistor).
(f)
Solve for Vx using source transformations.
Figure 98
Problem 3.6: For the linear circuit shown in Figure 3.99,
48
(a)
(b)
(c)
(d)
(e)
(f)
Use superposition to determine the current Ix .
Solve for Ix using node-voltage analysis.
Solve for Ix using mesh-current analysis.
Solve for Ix using the Thevinin equivalent (first remove the 11kΩ resistor).
Solve for Ix using the Norton equivalent (first remove the 11kΩ resistor).
Solve for Ix using source transformations.
Figure 99
Problem 3.7: For the linear circuit shown in Figure 3.100,
(a)
Use superposition to determine the current Ix .
(b)
Solve for Ix using node-voltage analysis.
(c)
Solve for Ix using mesh-current analysis.
(d)
Solve for Ix using the Thevinin equivalent (first remove the 93kΩ resistor).
(e)
Solve for Ix using the Norton equivalent (first remove the 93kΩ resistor).
(f)
Solve for Ix using source transformations.
Figure 100
Problem 3.8: For the linear circuit shown in Figure 3.101,
(a)
Use superposition to determine the current Vx .
(b)
Solve for Vx using node-voltage analysis.
(c)
Solve for Vx using mesh-current analysis.
(d)
Solve for Vx using the Thevinin equivalent (first remove the 55kΩ resistor).
(e)
Solve for Vx using the Norton equivalent (first remove the 55kΩ resistor).
Figure 101
49
Problem 3.9: For the linear circuit shown in Figure 3.102,
(a)
Use superposition to determine the voltage Vx .
(b)
Solve for Vx using node-voltage analysis.
(c)
Solve for Vx using mesh-current analysis.
(d)
Solve for Vx using the Thevinin equivalent (first remove the 62kΩ resistor).
(e)
Solve for Vx using the Norton equivalent (first remove the 62kΩ resistor).
(f)
Solve for Vx using source transformations.
Figure 102
Problem 3.10: For the linear circuit shown in Figure 3.103,
(a)
Use superposition to determine the voltage Vx .
(b)
Solve for Vx using node-voltage analysis.
(c)
Solve for Vx using mesh-current analysis.
(d)
Solve for Vx using the Thevinin equivalent (first remove the 43kΩ resistor).
(e)
Solve for Vx using the Norton equivalent (first remove the 43kΩ resistor).
Figure 103
Problem 3.11: For the linear circuit shown in Figure 3.104,
(a)
Solve for Ix using node-voltage analysis.
(b)
Solve for Ix using mesh-current analysis.
(c)
Solve for Ix using the Thevinin equivalents.
(d)
Solve for Ix using the Norton equivalents.
Figure 104
50
Problem 3.12: For the linear circuit shown in Figure 3.105,
(a)
Use superposition to determine the voltage Vx .
(b)
Solve for Vx using node-voltage analysis.
(c)
Solve for Vx using mesh-current analysis.
(d)
Solve for Vx using the Thevinin equivalent.
(e)
Solve for Vx using the Norton equivalent.
(f)
Solve for Vx using source transformations.
Figure 105
Problem 3.13: For the linear circuit shown in Figure 3.106,
(a)
Use superposition to determine the voltage Vx .
(b)
Solve for Vx using node-voltage analysis.
(c)
Solve for Vx using mesh-current analysis.
(d)
Solve for Vx using the Thevinin equivalent (first remove the 20kΩ resistor).
(e)
Solve for Vx using the Norton equivalent (first remove the 20kΩ resistor).
(f)
Solve for Vx using source transformations.
Figure 106
Problem 3.14: For the linear circuit shown in Figure 3.107,
(a)
Use superposition to determine the voltage Vx .
(b)
Solve for Vx using node-voltage analysis.
(c)
Solve for Vx using mesh-current analysis.
(d)
Solve for Vx using the Thevinin equivalent.
(e)
Solve for Vx using the Norton equivalent.
(f)
Solve for Vx using source transformations.
51
Figure 107
Problem 3.15: For the linear circuit shown in Figure 3.108,
(a)
Use superposition to determine the voltage Vx .
(b)
Solve for Vx using node-voltage analysis.
(c)
Solve for Vx using mesh-current analysis.
(d)
Solve for Vx using the Thevinin equivalent.
(e)
Solve for Vx using the Norton equivalent.
(f)
Solve for Vx using source transformations.
Figure 108
Problem 3.16: For the circuit shown in Figure 3.109, the unknown circuit element in the middle is nonlinear and that
Vx = 2 · I 2 where I is the current flowing through the device. Show that superposition is no longer a valid method for the
nonlinear element.
Figure 109
Problem 3.17: For the linear circuit shown in Figure 3.110,
(a)
Use superposition to determine the current Ix .
(b)
Solve for Ix using node-voltage analysis.
(c)
Solve for Ix using mesh-current analysis.
(d)
Solve for Ix using the Thevinin equivalent (first remove the 78kΩ resistor).
(e)
Solve for Ix using the Norton equivalent (first remove the 78kΩ resistor).
(f)
Solve for Ix using source transformations.
52
Figure 110
Problem 3.18: For the linear circuit shown in Figure 3.111,
(a)
Use superposition to determine the current Ix .
(b)
Solve for Ix using node-voltage analysis.
(c)
Solve for Ix using mesh-current analysis.
(d)
Solve for Ix using the Thevinin equivalent (first remove the 58kΩ resistor).
(e)
Solve for Ix using the Norton equivalent (first remove the 58kΩ resistor).
(f)
Solve for Ix using source transformations.
Figure 111
Problem 3.19: For the linear circuit shown in Figure 3.112,
(a)
Use superposition to determine the voltage Vx and the current Ix .
(b)
Solve for Vx and Ix using node-voltage analysis.
(c)
Solve for Vx and Ix using mesh-current analysis.
(d)
Solve for Vx using the Thevinin equivalent (first remove the 98kΩ resistor).
(e)
Solve for Ix using the Norton equivalent (first remove the 31V voltage source).
Figure 112
Problem 3.20: For the linear circuit shown in Figure 3.113,
(a)
Use superposition to determine the voltage Vx and the current Ix .
(b)
Solve for Vx and Ix using node-voltage analysis.
(c)
Solve for Vx and Ix using mesh-current analysis.
(d)
Solve for Vx using the Thevinin equivalent (first remove the 25kΩ resistor).
(e)
Solve for Ix using the Norton equivalent (first remove the 88kΩ resistor).
53
Figure 113
Problem 3.21: For the linear circuit shown in Figure 3.114,
(a)
Use superposition to determine the voltage Vx and the current Ix .
(b)
Solve for Vx and Ix using node-voltage analysis.
(c)
Solve for Vx and Ix using mesh-current analysis.
(d)
Solve for Vx using the Thevinin equivalent (first remove the 87kΩ resistor).
(e)
Solve for Ix using the Norton equivalent (first remove the 38kΩ resistor).
Figure 114
Problem 3.22: For the linear circuit shown in Figure 3.115,
(a)
Use superposition to determine the current Ix .
(b)
Solve for Ix using node-voltage analysis.
(c)
Solve for Ix using mesh-current analysis.
(d)
Solve for Ix using the Thevinin equivalent (first remove the 8kΩ resistor).
(e)
Solve for Ix using the Norton equivalent (first remove the 8kΩ resistor).
Figure 115
Problem 3.23: For the linear circuit shown in Figure 3.116,
(a)
Use superposition to determine the voltage Vx and the current Ix .
(b)
Solve for Vx and Ix using node-voltage analysis.
54
(c)
(d)
(e)
Solve for Vx and Ix using mesh-current analysis.
Solve for Vx using the Thevinin equivalent (first remove the current source).
Solve for Ix using the Norton equivalent (first remove the 29kΩ resistor).
Figure 116
Problem 3.24: For the linear circuit shown in Figure 3.117,
(a)
Use superposition to determine the voltage Vx and the current Ix .
(b)
Solve for Vx and Ix using node-voltage analysis.
(c)
Solve for Vx and Ix using mesh-current analysis.
(d)
Solve for Vx using the Thevinin equivalent (first remove the 25kΩ resistor).
(e)
Solve for Ix using the Norton equivalent (first remove the 16kΩ resistor).
Figure 117
Problem 3.25: For the linear circuit shown in Figure 3.118,
(a)
Use superposition to determine the voltage Vx and the current Ix .
(b)
Solve for Vx and Ix using node-voltage analysis.
(c)
Solve for Vx and Ix using mesh-current analysis.
(d)
Solve for Vx using the Thevinin equivalent (first remove the 12kΩ resistor).
(e)
Solve for Ix using the Norton equivalent (first remove the 6kΩ resistor).
(f)
Solve for Vx and Ix using source transformations.
Figure 118
55
Problem 3.26: For the linear circuit shown in Figure 3.119,
(a)
Use superposition to determine the voltage Vx and the current Ix .
(b)
Solve for Vx and Ix using node-voltage analysis.
(c)
Solve for Vx and Ix using mesh-current analysis.
(d)
Solve for Vx using the Thevinin equivalent (first remove the 23kΩ resistor).
(e)
Solve for Ix using the Norton equivalent (first remove the 19kΩ resistor).
Figure 119
Problem 3.27: For the linear circuit shown in Figure 3.120,
(a)
Use superposition to determine the voltage Vx and the current Ix .
(b)
Solve for Vx and Ix using node-voltage analysis.
(c)
Solve for Vx and Ix using mesh-current analysis.
(d)
Solve for Vx using the Norton equivalent (first remove the current source).
(e)
Solve for Ix using the Thevinin equivalent (first remove the voltage source).
Figure 120