Examples for Introduction to Electrical Engineering
Prepared by Professor Yih-Fang Huang, Department of Electrical Engineering
1BasicTerminologies
An electric circuit is a collection of interconnected multi-terminal electrical devices (also referred to as
circuit elements). Conventional circuit analysis focuses on devices that can be modeled as two-terminal
devices,namely,acircuitelementwithtwoleadscomingoutofit. Examples of two-terminal devices include
independent sources (e.g., a battery), simple resistors, capacitors, inductors and diodes. In practice, there are
also many devices that have more than two terminals. Typical such examples are transistors, transformers,
and operational amplifiers. In circuit analysis, those multi-terminal de vices are typically modeled as two-
terminal devices with dependent sources.
Before we proceed to show some circuit analysis examples, we shall define some terminologies first:
• Node:anodeisapointofconnectionbetweentwoormorecircuitelements.
• Branch:abranchisaportionofthecircuitthatconsistsofone circuit element and its two terminal
nodes.
• Loop:a(closed)loopisasequenceofconnectedbranchesthatbegin and end at the same node.
• Branch Current:abranchcurrentisthecurrentthatflowsthroughtheelectrical device of the
branch. The physical unit of current is ampere,namedaftertheFrenchmathematicianandphysicist,
Andr´e-Marie Amp`ere (1775-1836).
• Branch Voltage:abranchvoltageisthepotentialdifferencebetweenthetwoterminal nodes of the
electrical device of the branch. The physical unit of voltageisvolt,whichisnamedinhonorofthe
Italian physicist Alessandro Volta (1745-1827), who invented the voltaic pile, possibly the first chemical
battery.
• Node Voltage:anodevoltageisthepotentialdifferencebetweenadesignated node and the reference
node.
• Loop Current:aloopcurrentisthecurrentthatflowsthroughaclosedloop.Notethatloopcurrent
is different from branch current if that branch is common to twoormoreloops.
It is important to note that conventional circuit analysis assumes that all wires (or lead) are perfect conduc-
tors, namely, the lump ed parameter circuit model. Therefore, no energy is lost when currents flow through
the wires, and non-zero branch voltages occur only across a circuit element (a wire itself is not a circuit
element).
In circuit analysis , each branch has two branch variables, i.e., branch voltage and branch current. These are
algebraic variables,thustheyhavesignsassociatedwiththeirvalues.Acircuitiscompletelycharacterizedif
all the branch variables are known. Thus the objective of circuit analysis is to solve for all branch voltages
and branch currents.
2CircuitAnalysis
In principle, three s e ts of equations are sufficient to performcircuitanalysisforanycircuit,whichare
Kirchoff’s current law (KCL), Kirchoff’s voltage law (KVL), and branch element equations (e.g., Ohm’s
1
law). The KCL and KVL together characterize the circuit topology, i.e., how the circuit elements are
connected, while the branch element equation characterizesthephysicalproperty,specifically,thecurrent-
voltage relation, of the circuit element.
2.1 Kirchhoff Laws
The Kirchhoff Laws are results of conservation principles that must always be obeyed by any electric circuit.
These laws are named after Gustav Robert Kirchhoff (1824 -1887), a German physicist who contributed to
the fundamental understanding of electrical circuits and who formulated these laws in 1845 when he was
still a student at University of K¨onigsberg in Germany. Simply stated, KCL is concerned with the currents
entering a node, while KVL is stated with respect to the branchvoltagesaroundaloopinacircuit. Note
that these Kirchhoff laws hold regardless of the ph ysical properties of the circuit elements.
KCL:
The algebraic sum of all currents entering any node is zero.
KVL:
The algebraic sum of the branch voltages around any closed loop equals zero.
KCL is simply a statement that charges cannot accumulate at the nodes of a circuit. This principle is
identical to concepts found in fluid dynamics. Namely that if you look at the fluid flowing into one end of a
pipe, you expect the same amount of fluid to flow out the other end. If this did not occur, then fluid would
accumulate in the pipe and eventually cause the pipe to burst.KCLisnothingmorethananelectrical
equivalent of this intuitive physical idea from fluid mechanics. In a nutshell, KCL is a charge conservation
law, while KVL is an energy conservation law which states, in essence, that the total work done in going
around a loop will be zero.
To employ KVL and KCL, we must always keep in mind that, in circuit analysis, the currents and voltages
are algebr aic variables.Assuch,itishelpfultohaveasignconventionforthosevariables. To facilitate the
subsequent discussion, we adopt the following convention: (1) when applying KCL to a node, we designate
any current flowing into the node as p ositive and any leaving the node as negative; (2) when applying KVL
around a closed loop, we travel around the loop clo ckwise, andabranchvoltageispositiveifweenterthe
positive polarity of the voltage first, otherwise it is negative.
To explain how KCL works, let us consider the circuit shown in Figure 1. This figure shows a circuit
consisting of three branches, an independent voltage s ourceconnectedinparallelwithtworesistors.Thisis
also commonly known as a single-node circuit, for there is only one node a,apartfromthegroundnodeb.
The single node a of this circuit is shown in the righthand drawing of Figure 1. At this node, we see three
currents. Two of these currents i
1
and i
2
are leaving node a and the third current i
0
is entering node a.By
the sign convention discussed above and by KCL, we can write anequationasbelow:
i
0
− i
1
− i
2
=0
To illustrate applications of KVL, consider the single-loopcircuitshowninFigure2. Thisistheloopformed
from branches
(a, b) → (b, c) → (c, a)
The voltages obtained by traversing this loop are
v
ab
,v
bc
,V
In this figure, we start a t node a and begin tracing out our loop in a clockwise direction, we seethatthe
traverse of branch (a, b)goesfrom+to−.Thisisconsideredasanegativechangeinpotential(i.e.we
2
R
R
+
V
_
a
i
1
i
2
i
0
b
V
+
-
a
b
i
0
R
R
i
1
i
2
a
i
0
i
2
i
1
Parallel Circuit
Circuit Graph
KCL at Node a
Figure 1: KCL at node b
are decreasing the potential). The same is true for the voltage over branch (b, c). Note, however, that
in traversing branch (c, a)thatwearegoingfromanegativetopositivepolarity. Thechange in potential,
therefore, is positive. On the basis of our preceding discussion, we can see that KVL will lead to the following
equation:
−V + v
ab
+ v
bc
=0
R
R
+
V
_
a
i
1
i
2
i
0
b
V
+
-
a
b
R
R
Series Circuit
Circuit Graph
c
c
+
v
ab
_
+
v
bc
_
Figure 2: A Simple Resistive Circuit
2.2 Branch Element Equations
The third s e t of equations needed in performing circuit analysis is the set of branch element equations,which
are equations that characterize the relation between the branch voltage and branch current (i.e., the V-I
relation) and that relation is due to the physical propertiesoftheelectricaldeviceofthebranch. Inour
discussion, and in most two-terminal device modeling, the following three equations are most commonly
used:
1. For an ideal resistor, the Ohm’s Law states that the branch voltage is directly prop o rtional to the
branch current, namely,
v(t)=Ri(t) (1)
where the proportionality R is called the resistance and is measured in Ohms,denotedbyaGreek
symbol Ω, named after the German physicist Georg Simon O hm (1789-1854). Equation (1) is the
well-known Ohm’s Law.
3
2. For an ideal capacitor, the branch current is directly proportional to the derivative of the branch
voltage, namely ,
i(t)=C
dv(t)
dt
(2)
where the proportionality constant C is called the capacitance and is measured in far ads named in
honor of the English Chemist and Physicist Michael Faraday (1791-1867).
3. For an ideal inductor, the branch voltage is directly proportional to the derivative of the branch current,
namely,
v(t)=L
di(t)
dt
(3)
where the proportionality co ns tant L is called the inductance and is measured in henrys,namedafter
an American scientist Joseph Henry (1797-1878).
2.3 Nodal Analysis and Mesh Analysis
We have learned in previous sections that three sets of equations (KCL, KVL, and Ohm’s law) are sufficient
to perform circuit analysis for any circuits. To be more specific, in so lving a circuit that comprises B
branches (i.e., it has B cir cuit elements) and N nodes, we are solving for 2B variables, i.e., B branch
voltages and B branch currents. Accordingly, we need 2B independent equations to obtain unambiguous
solutions. For a circuit with B branches and N nodes, there will be N − 1independentKCLequations,
B − N +1 independent KVLequations,andB branch element equations. Thus there are, altogether, 2B
independent equations. However, since B is usually large (much larger than N ), solving 2B equations may
be an onerous task. The simpler alternatives are nodal analysis and mesh analysis.Bothmethodsarederived
from the aforementioned three sets of equations, however, they solve a smaller set of alternative algebraic
variables. The nodal analysis first defines node voltages for each node as the p otential difference between
the designated node and the reference (ground) node. Then, itemploysKCLtosetuptheso-callednode
equations, expressing every term in the equation in terms of node voltages. Thus each node equation is a
KCL equation, except that the unknown algebraic v ariables are node voltages. Once all node voltages are
solved (with the node equations), we can obtain all branch voltages and branch currents. In essence, instead
of directly solving for all branch voltages and branch currents, Nodal Analysis solves for node voltages and
then use those solutions to determine all branch voltages andbranchcurrents.Similarly,themeshanalysis
defines mesh currents for each mesh (a closed loop) as the current traversing around this closed loop. Then,
it employs KVL to set up the so-called mesh equations , expressing wherever possible every term in the
equation in terms of mesh currents. Thus each mesh equation isaKVLequation,exceptthattheunknown
algebraic variables are mesh currents. Once all mesh currents are solved (with the mesh equations), we can
obtain all branch voltages and branch currents. Since the number of nodes (or that of meshes) is significantly
less than the number of branches, those two methods involve many fewer variables to solve for. All circuits
can be solved either by nodal analysis or by mesh analysis.
To facilitate the employment of no dal analysis and mesh analysis, we need to introduce another sign conven-
tion, i.e., the so-called associated passive sign convention which is defined in accordance with the physical
properties of passive circuit elements, i.e., circuit elements like resistors that absorb energy. The associated
passive sign convention states that the branch current of a passive circuit element flows from high to low
potential. Accordingly, the sign of the branch current is positive if it flows from the positive (+) polarity of
the v oltage to the negative (-) polarit y. As such, unless otherwise specified, once the polarity of the voltage
is defined, the direction of associated branch current is defined. The converse is also true. In other words,
if the current flow direction is marked, then the voltage polarity is defined accordingly.
The following section presents some e x amples that use nodal analysis and mesh analysis for simple resistive
circuits. The mathematical tools involved are simply that ofsystemsoflinearequations. Thoseequations
can be solved by hand (if the number of unknowns is less than or equal to 3), using Cramer’s rule, or using
acomputerprogramlikeMatlab.
4
3ExamplesforLinearResistiveCircuitAnalysis
Example 1 (Nodal Analysis)
Figure 3: Example 1
In this example, we e mploy nodal analysis to find the branch voltages and branch currents in the circuit
shown. We begin by marking a reference node as shown. Other than the reference node, there are only
two nodes, namely , Node 1 and Node 2. We thus define two node voltages, e
1
and e
2
.Notethate
1
is
defined such that its positive polarity is at node 1 and negative polarity is at the reference node. This is
the convention that we use in defining all node voltages. Sincethereareonlytwonon-referencenodes,we
only need to set up two node equations to solve for those no de voltages. The node equations are simply
KCL equations. Three branches are connected to Node 1 - an independent current source of 6 Amps, a 40Ω
resistor and an 8Ω resistor. Accordingly, there are three branch currents in the KCL equation, which states
that the algebraic sum of those three branch currents is equaltozero,namely,
6 − i
1
− i
12
=0 (4)
We can see that four branches are connected to No de 2 - three resistor branches and a 1 Amp independent
current source. Thus the KCL eq uation for Node 2 is,
i
12
− i
2
− i
22
− 1=0 (5)
In writing the above equations, we have adopted the convention that currents entering a no de is positive
while cur rents leaving a node is negative. Now, we need to convert the above equations into node equations,
expressing all unknown currents as functions of the node voltages by using branch element equations (Ohm’s
Law in this case, since there are only resistors) and KVL equations wherever needed. In particular,
i
1
= e
1
/40
i
2
= e
2
/80
i
22
= e
2
/120
In addition to the above expr e ssions, we need to express i
12
in terms of the node voltages. Since i
12
is
marked as flowing from left to right on the 8Ω resistor branch, the associated branch voltage must be defined
to hav e positive polarity at Node 1 and negative polarit y at Node 2, according to the associated passive
sign convention described in Section 2.3. As such, the branch voltage is e
1
− e
2
,whichcanbeseeneasily
by applying KVL around the closed loop that encompasses the three branches - 40Ω, 8Ω, and 80Ω. Hence
i
12
=(e
1
− e
2
)/8. Now, we are ready to write down the two node equations. Substituting all the current
terms expressed in node voltages into Eqs. (4, 5) yields
6 −
e
1
40
−
(e
1
− e
2
)
8
=0 (6)
(e
1
− e
2
)
8
−
e
2
80
−
e
2
120
− 1=0 (7)
We now have just two algebraic variables, e
1
and e
2
to solve for, with two independent equations. Solving
those two equations, (6), (7), yields e
1
=120Voltsande
2
=96Volts. Now,youcanverifythatoncee
1
and
e
2
are known, we can find all the branch voltages and branch currents.
5
Example 2 (Nodal Analysis)
Figure 4: Example 2
In this example, we can see that there are four nodes in addition to the reference node, thus there are four
node voltages to solve for. Note that this reference node has been chosen intentionally so that one of the
node voltages is given, −40 Volts, negative of the independent voltage source, thereby reducing by one the
number of v ariables to solve for. T o solve for the circuit shown in Figure 4, we proceed similarly as in
Example 1, except now e
1
is already given. Thus the node equations are as follows:
At Node 1: e
1
= −40
At Node 2:
(e
1
−e
2
)
12
−
e
2
25
− 5 −
(e
2
−e
3
)
20
=0
At Node 3: 5 +
(e
2
−e
3
)
20
−
(e
3
−e
4
)
40
+7.5=0
At Node 4:
(e
3
−e
4
)
40
− 7.5 −
e
4
40
=0
(8)
Now, we can solve for e
2
, e
3
,ande
4
from the equations in (8) - three variables with three independent
equations. The solutions are e
2
= −10 Volts, e
3
=132Volts,ande
4
= −84 Volts. Again, you should verify
that once you know all the node voltages, you can easily determine all branch voltages and branch currents
by applying Ohm’s Law and KVL. Before we leave this example, you may notice that the solution process
may become more complicated if we assign the reference node differently. More explicitly, you will need to
solve for four unknown variables with four equations, instead of three variables with three equations, as we
shall see in the next example.
6
Example 3 (Nodal Analysis)
Figure 5: Example 3
There are five nodes, in addition to the reference node, in the circuit shown. The node equations ar e:
Combining Node 1 and Node 2:
−e
1
1
−
(e
2
−e
3
)
3
=0
At Node 3:
(e
2
−e
3
)
3
−
e
3
40
−
(e
3
−e
4
)
2
=0
At Node 4:
(e
3
−e
4
)
2
−
(e
4
−e
5
)
4
+28 = 0
At Node 5:
(e
4
−e
5
)
4
−
e
5
2
− 28 = 0
Additional equation: e
2
− e
1
=40
(9)
Solving for five variables with the five equations in (9) yields e
1
=5Volts,e
2
=45Volts,e
3
=60Volts,
e
4
=73Volts,ande
5
= −13 Volts. This problem can also b e solved with a Matlab program. In this
example, the direct application of Nodal analysis calls for solution of five unknown variables. However,
further inv estigation may lead you to see that you can simplify this problem to one of solving three variables
with three equations by exploring the first and the last equations in (9) to eliminate e
1
and e
2
.
7
Example 4 (Mesh Analysis)
Figure 6: Example 4
In mesh analysis, our goal is to solve fo r the mesh currents using mesh equations. The mesh equations are
KVL equations with the unknown branch voltages expressed in terms of mesh currents. In this example,
there are three meshes (also termed fundamental loops). Employing KVL equation in Mesh 1 yields,
v
1
− v
2
− v
3
=0 (10)
In writing (10), we have adopted the convention of transversing the loop clockwise, and if the polarity of
the branch voltage results in a branch current defined throughtheassociated passive sign convention flowing
in the same direction as the mesh current direction when it transverses through the branch, that branch
voltage is a positive variable. Now we need to convert (10) into a mesh equation. This is accomplished
by employing branch element equation (Ohm’s Law in this case,sincethereareonlyresistors). Notethat
the branch current through the 6Ω resistor branch is j
1
,thebranchcurrentthroughthe3Ωresistorbranch
is −j
1
+ j
3
,andthebranchcurrentthroughthe1Ωresistorbranchis−j
1
+ j
2
.Again,wehaveadopted
associated passive sign convention in defining those branch currents, following the polarity of the branch
voltages that had been marked. Applying Ohm’s Law, we have:
v
1
= j
1
6
v
2
=(−j
1
+ j
3
)3
v
3
=(−j
1
+ j
2
)1
Substituting the above expressions into (10) yields,
j
1
6 − (−j
1
+ j
3
)3 − (−j
1
+ j
2
)=0 (11)
By the same token, we can derive the mesh equations for Mesh 2 and Mesh 3, noting that the branches with
independent voltage sources have known branch voltages equal to the value of the voltage source. Hence the
three mesh equations are:
Mesh 1: j
1
6 − (−j
1
+ j
3
)3 − (−j
1
+ j
2
)=0
Mesh 2: − 230 + (−j
1
+ j
2
)+(j
2
− j
3
)2 + 115 + j
2
4=0
Mesh 3: − 115 − (j
2
− j
3
)2 + (j
3
− j
1
)3 + 460 + j
3
5=0
(12)
Here we have three independent equations to solve for three unknown variables, j
1
, j
2
,andj
3
.Thesolutions
are j
1
= −10.6Amps,j
2
=4.4Amps,andj
3
= −36.8Amps. Onceagain,youmayverifythatonceweknow
the values of all mesh currents, we can determine all branch voltages and branch currents.
8
Example 5 (Mesh Analysis)
Figure 7: Example 5
This example demonstr ates that when there is an independent current source in an isolated branch, i.e., a
branch that is not common to two or more meshes, we can reduce byonethenumberofmeshcurrentsto
be solved for. For the circuit shown in Figure 7, we see that j
3
=30Amps. Thusweonlyneedtosolvefor
j
1
and j
2
.FollowingthesameproceduresasinExample4,wecanwritethe mesh equations.
Mesh 1: − 600 + j
1
4+(j
1
− j
2
)16 + (j
1
− j
3
)5.6=0
Mesh 2: − (j
1
− j
2
)16 + j
2
3.2+(j
2
− j
3
)0.8+424 = 0
(13)
Solving for j
1
and j
2
with the equations in (13) yields j
1
=35Ampsandj
2
=8Amps.
9
Example 6 (Mesh Analysis)
Figure 8: Example 6
In this example, we show that when the circuit contains dependent sources
1
,wemayneedtoaddmore
equations for solve for all the unknown variables. In principle, we need the same number of independent
equations as the number o f unknown variables to o btain the unique solution, as we have learned in Linear
Algebra. The mesh equations for the circuit shown in Figure 8 are:
Mesh 1: j
1
7+(j
1
− j
3
)1 + (j
1
− j
2
)2 = 0
Mesh 2: − 125 − (j
1
− j
2
)2 + (j
2
− j
3
)3 + 75 = 0
Mesh 3: j
3
= −0.5v
∆
Additional equation: v
∆
=(j
1
− j
2
)2
(14)
In summary, there are four equations for solving four unknownvariables,j
1
, j
2
, j
3
,andv
∆
.Thesolutions
are j
1
=6Amps,j
2
=22Amps,j
3
=16Amps,andv
∆
= −32 Volts.
1
There are basically four types of dependent sources that we may encounter in those simple resistive circuits: voltage-
controlled voltage source, current-controlled voltage source, voltage-controlled curren t source and current-controlled current
source
10
