Circuit Analysis – Engineering3821

Course Notes

Fall 2013

by

E.W. Gill, Ph.D., P.Eng.

Unit 1

Things Old and New

Here we will review the material of Engineering 1040 which is relevant to this course.

We will not consider these topics in depth here, but will do so as required in the

concepts to be developed later in the course. A general overview of the course is

given on the course outline. Some of the lecture material will consist of printed

notes such as these and some will require in-class note-taking. It is suggested that

a convenient method of keeping the course as orderly as possible would be to use a

three-ring binder with class handouts interspersed with class notes where appropriate.

1.1 Variables, Conventions and Elementary Cir-

cuit Elements

Here we consider the elements, basic terminology and sign conventions that will be

used in this course. It is essentially review and all variable names, definitions, etc.

closely follow those given in the first six chapters of the text. At this stage you

should review the Electric Circuits Module from Engineering 1040. That material is

prerequisite to what follows in this Unit. Also, read Chapters 1 and 2 of the text.

In particular, pay careful attention to the assumptions which allow us to use circuit

theory rather than electromagnetic field theory in the electric circuits which we will

in encounter in this course. The question, “How large can a circuit be in relation to

the wavelength of the signal existing in it so that we can consider it to be still ‘small

enough’ so that circuit theory applies?”, is a REALLY important one, but it is not

addressed in this course.

Current (i)

Electric current, i, is defined as the time, t, rate at which electric charge, q ,passes

a given point. The relationship given in ENGI1040 was for the case of the rate being

1

constant and could be represented simply as

i =q

t(1.1)

where q is measured in coulombs (C), t in seconds (s) and i in C/s. The coulomb

per second is named the ampere (A). To account for the very real possibility that the

rate of charge flow is not a constant, we consider a differential amount of charge, dq,

passing a given point in a differential time, dt, and write the current as a simple time

derivative

i =dq

dt. (1.2)

Of course, all units are the same as in equation (1.1). Equation (1.2) immediately

suggests that charge q may be found using a time-integral of the current, an idea

which is in perfect harmony with the simple case encountered in ENGI1040 in which

the charge passing a given point in a circuit could be found from the area under the

‘current versus time’ graph. That is

or

q =

∫i dt , (1.3)

it being understood that i is a function of time – i.e., i ≡ i(t).

Voltage (v)

Voltage, v, or potential difference between two points measured in volts (V) may

be defined as the energy, w, per unit charge, q, expended in moving the charge from

one position to the other in an electric field. If we think in terms of differential

amounts of energy and charge (i.e., dw and dq), then we may write

v =dw

dq. (1.4)

Because energy is measured in joules (J), the volt is a joule per coulomb (J/C).

Analogous to equation (1.3), we may also write

w =

∫v dq , (1.5)

which implies that energy could be determined from a ‘voltage versus charge’ graph

(not that we will commonly use this in circuit analysis).

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Ideal Basic Circuit Element

The ideal basic circuit element is an two-terminal device that can be described in

terms of the current through it and/or voltage across it and which does not consist of

sub-elements. Consider such a device as illustrated (note that in this illustration, the

current is passing through the element in the direction of an assigned voltage drop

from + to − :

Examples include the resistor, encountered in ENGI1040, and the inductor and ca-

pacitor, which we will study in this course. The circuit symbols for all three are:

If the current flows in the direction of the voltage drop, then the current is consid-

ered to be positive. Equivalently, in any expression relating voltage and current we

use a positive sign/negative sign whenever the reference direction for the current is

in the direction of the reference voltage drop/voltage rise. This is illustrated below

in our initial consideration of power. This is the so-called passive sign convention.

Obviously, the electric charges (electrons) flow from negative to positive, but we are

assigning the positive direction for the current to be from positive to negative – i.e.

we assume the current to be in the direction in which positive charges would flow,

even though it is not positive charges which are actually flowing. This current, which

is in the direction opposite to electron flow, is referred to as conventional current and

we will use this concept throughout the course.

Electric Power (p)

Power, p, in general, is defined as the rate of change of energy. Using the differential

forms dw, dt, and dq, for energy, time, and charge, respectively, and considering

equations (1.2) and (1.4), we therefore have from this definition

p =dw

dt=

dw

dq

dq

dt= vi . (1.6)

Using the passive sign convention we note that the configuration

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indicates that the element generates power (i.e. is a source of electric power) while

the configuration

indicates that the element absorbs or dissipates power – like a resistor. Since w is

measured in joules (J) and t is measured in seconds (s), the power unit is the J/s

or watt (W). Additionally, since the size of the power is vi for the cases illustrated,

the watt appears to be equivalent to volts × amperes. In more ‘complex’ situations

encountered later in the course we will not be able to strictly write this equivalence

– more later.

Ideal Voltage and Current Sources

The ideal voltage source is a circuit element which maintains a fixed voltage across

its terminals irrespective of the current flow between them. This ideal voltage source

may be (i) “independent” meaning that v is fixed and does not depend on the rest

of the elements connected to it (this is the type we saw in ENGI1040) or it may be

(ii) “dependent” indicating that its value depends on a voltage or current elsewhere

in the circuit (this is something NEW). The symbols are as shown:

Analogous definitions are associated with current sources and the symbols are

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An ideal dependent voltage source may be ‘voltage-controlled’ – i.e. it’s value

may depend on a voltage elsewhere in the circuit – or it may be ‘current-controlled,

meaning that it may depend on a particular current in the circuit. Likewise, an ideal

dependent current source may be voltage-controlled or current-controlled.

Example:

It should be noted that if we restrict our discussion to ideal sources then knowing

only the value of a voltage or current source eliminates the possibility of specifying

that voltage or current as a function one of the other. For example, if we know the

value of an (ideal) voltage source, then the current it delivers depends on the load to

which it is attached – i.e., more information than just the source voltage is required

in order to relate the voltage to the current.

Non-ideal Sources

While ideal sources have no internal resistance – and this may be a good ap-

proximation in some instances – real sources certainly do have internal resistance.

Consider the following source where it is indicated that the source itself has an inter-

nal resistance, Rin, and the source terminals are at a and b:

Obviously, the size of the load resistance, RL, will affect the value of the voltage, vs,

across RL. This is true because part of the source voltage will be dropped across

Rin. Since this voltage drop across Rin depends on the size of the current, which in

turn will depend on both Rin and RL, the voltage across RL will be different for each

RL. To deduce this we have implicitly applied Kirchhoff’s voltage law which we will

review shortly.

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Resistance (Ohm’s Law)

An element of pure electrical resistance, R, which opposes the flow of current

(i.e. it always dissipates energy) is by definition the ratio of the voltage v across that

element (resistor) to the current i through it.

Illustration:

That is

R =v

iOR v = iR (Ohm’s law). (1.7)

The resulting unit of V/A is called the ohm and is symbolized as Ω.

From equations (1.6) and (1.7), the power in watts dissipated by a resistor is

clearly

p = vi =v2

R= i2R . (1.8)

For a resistor, p > 0 and resistors must therefore always dissipate energy.

Closely related to the concept of resistance is that of conductance G. By definition,

G =1

R(1.9)

and is measured in units of siemans (S) or, equivalently, mhos (0).

We saw in ENGI1040, that by using Kirchhoff’s laws (as reviewed in Section 1.2

following,) resistors “in series” as indicated by

add to give a total equivalent resistance of

Req =n∑

i=1

Ri . (1.10)

It was similarly shown that resistors “in parallel” as indicated by

produce a total equivalent resistance of

1

Req=

n∑i=1

1

Ri

. (1.11)

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1.2 Kirchhoff’s Laws

At this point, it would be a really good idea to know the definitions of the following

words as they appear in Table 4.1 of the text: (1) node, (2) essential node, (3) path,

(4) loop, (5) mesh, (6) branch, (7) essential branch and (8) planar circuit. Most of

them will be used as we discuss particular circuit examples.

To completely analyze a circuit it is required that the voltage across and current

through each element be determined. For example, suppose that in the following

simple illustration the source voltage vs and resistance values are known. There

remain to be determined the current through the source and the voltage across each

resistance and the current through each. It thus appears initially that there are 5

unknowns:

The 5 unknowns are is, i1, i2, v1, and v2. Strictly speaking, 5 linearly-independent

equations are needed to obtain the solution. Two of these,

v1 = i1R1 . . . (1) and v2 = i2R2 . . . (2) ,

come from Ohm’s law while the remaining equations result from application of Kirch-

hoff’s laws as seen below.

1.2.1 Kirchhoff’s Current Law (KCL)

Kirchhoff’s current law states that the algebraic sum of all the currents at any node

in a circuit equals zero. We use the convention that currents entering a node are

negative and those leaving a node are positive. If we allow i1, i2, etc. to be the size of

the currents, then for the situation illustrated

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−i1 + i2 + i3 − i4 + i5 = 0

We could equally say that “the sum of the currents entering a node equals the sum

of the currents leaving the node” – i.e.

i1 + i4 = i2 + i3 + i5 .

This law is nothing more than a special case of the statement of continuity of current.

Applying KCL to the above circuit we consider three nodes, A, B, and C, as shown

and write

is = i1 . . . (3) i1 = i2 . . . (4) and i2 = is . . . (5) ,

Of these three equations, there are only two linearly-independent equations. In

fact, in any circuit with n nodes, there will be only n−1 linearly independent equations

from Kirchhoff’s current law. To get the remaining equation required to “solve” the

circuit we must use Kirchhoff’s voltage law (KVL).

1.2.2 Kirchhoff’s Voltage Law (KVL)

Kirchhoff’s voltage law states that “the sum of all voltages around any closed path in

a circuit equals zero”. If we assign a positive sign to voltage drops and a negative sign

to voltages rises as a loop is traversed in a fixed direction (see direction of arrows in

the above circuit) then

−vs + v1 + v2 = 0 . . . (6) .

Clearly this could be restated as the “sum of the voltage rises equals the sum of the

voltage drops”’ around any closed loop and we may write

vs = v1 + v2 .

Clearly, from the 5 equations associated with our illustrative circuit all unknowns

may be written in terms of vs and the resistance values.

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1.2.3 Multiple Sources

Circuits may contain more than one voltage source or more than one current source

and these sources may be independent or dependent. With the theory we have so far,

such circuits may still be solved using KVL, KCL and Ohm’s Law as required.

Example 1: Find i1, i2, and i3 in the following circuit:

A summary example for Sections 1.1 and 1.2 appears on Tutorial 1.

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1.3 A Few Important Resistive Circuits

Here, we consider a few examples of simple resistive circuits which require use of

the concepts presented in the previous sections. Most of these ideas were studied in

ENGI1040 and are simply summarized for future reference here.

Voltage Division

Example 1: Consider the circuit:

It is easily shown (DO IT) from Kirchhoff’s voltage law and Ohm’s law that

v1 = vs

(R1

R1 + R2

)and v2 = vs

(R2

R1 + R2

)′

or, if there were n resistors in series, the voltage vj across the jth resistor would be

vj = vs

(Rj

Req

)(1.12)

where Req is the equivalent resistance of the n resistors as given by equation (1.10).

Remember, the voltage on the right hand side of these equations is the voltage which

is being divided – depending on the problem, it won’t necessarily be the source voltage.

Example 2: Consider the following configuration, in which the circuit in Example

1 is attached to a load. We wish to show that the output voltage v0, which is some

fraction of the source voltage vs, is essentially independent of the load resistance RL

if RL R2.

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Current Division

Consider the following circuit:

We have seen in ENGI1040 that the currents for this two parallel-resistor case may

be determined as

i1 = is

(R2

R1 + R2

)and i2 = is

(R1

R1 + R2

)or for n resistors in parallel, the current, ij, through the jth branch made be deduced

as

ij =Req

Rj

is (1.13)

where Req is simply the parallel sum of the n resistors. Of course, for the two-resistor

case depicted above,

Req =R1R2

R1 + R2

.

Remember that in equation (1.13) and the two equations preceding it is is the current

which is being divided. Here this happens to be the source current, but in more

complicated circuits it may not be the source current which is involved in the formula

– it will, however, always be the current which is being divided.

**It is easy to show from equation (1.13) that the ratio of any two branch currents

equals the inverse ratio of the resistances through which they flow. For the above

case,

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Wheatstone Bridge

The circuit shown below, referred to as a Wheatstone bridge, is used to measure

resistance. We use the term “balanced” to mean that the current ig in the galvanome-

ter is zero. R3 is a variable resistor and is adjusted to balance the bridge. For this

balanced case, KCL and KVL may be used to show that the value of the unknown

resistance Rx is given by

Rx =

(R1

R2

)R3 . (1.14)

(Note: The text example has different labels on the resistors, but the analysis is the

same even though the equation in the text appears to be different from equation

(1.14) above. The difference arises only because of the labels on the resistors and the

currents.)

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Delta-to-Wye Equivalent Circuits

The configuration of elements (here, resistors) shown below is referred to as a

delta (∆) or pi (Π) interconnection – for obvious reasons.

The configuration of elements (here, resistors) shown below is referred to as a wye

(Y) or tee (T) interconnection.

These structures (which may contain elements other than resistors) appear in a variety

of useful circuits – especially in three-phase power applications to be studied later in

the programme – and it is thus useful to consider them here in the context of resistance

configurations. We wish to show that a ∆ configuration may be transformed to an

equivalent Y configuration (or vice versa). This may be done using series and parallel

simplifications as follows:

First consider the equivalent resistance Rab between points a and b in each configu-

ration:

If we proceed similarly for the resistances between b and c and between c and a we

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get

Rbc =Ra(Rb + Rc)

Ra + Rb + Rc

= R2 + R3 . . . (2)

Rca =Rb(Rc + Ra)

Ra + Rb + Rc

= R1 + R3 . . . (3) .

These 3 equations may be solved trivially for R1, R2, and R3 or for Ra, Rb, and Rc

to give the ∆-to-Y or Y-to-∆ transformations, respectively. For example,

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1.4 Circuit Analysis Techniques

In this section we review some of the circuit analysis issues encountered in Chapter

4 of the text. Since the details were extensively covered in Term 1, we will carry out

our review largely by way of example.

1.4.1 Node-Voltage Method

Recall that an essential node is a point in a circuit where three or more circuit elements

join. In the circuit shown below there are only 3 essential nodes. This is emphasized

by labelling one node twice – eg., electrically, the two c’s are the same point. We will

now use the node-voltage method to determine the (i) voltages across the 1 Ω and

12 Ω resistors, (ii) v as shown, and (iii) subsequently, we’ll use the results to find the

current through the 12 Ω resistor.

•

•

•

•

Step 1: Note that since there are 3 essential nodes, we may describe the circuit with

2 (i.e. one less than the number of essential nodes) node-voltage equations.

Step 2: Choose a reference node (indicated by the solid arrow) – it is usually a good

idea to pick this as the node with the most branches.

Step 3: Label the remaining essential nodes with their voltages with respect to the

reference. Here we have used v1 and v2.

Step 4: Use KCL at each of the nodes in Step 3. In doing this the text’s convention

is to sum the currents leaving each of these nodes to zero. Here, we get

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Step 5: Use the node-voltages to determine any required branch currents and powers.

(NOTE: An example involving dependent sources and supernodes will be completed

in Tutorial 2.)

1.4.2 Mesh-Current Method

Recall that a mesh is a circuit loop that does not enclose any other loops. In the

circuit shown below there are 3 meshes.

• • •

•

Step 1: Assign a mesh current to each of the meshes as indicated by the arrowed

curves labelled i1, i2, and i3.

Step 2: Apply KVL to each mesh to obtain a system of equations involving the mesh

currents. Solve these for the mesh currents.

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Step 3: Obtain any necessary branch currents from the mesh currents. This allows

any voltages and powers of interest to be calculated. Here we will find (a) the power

delivered by the 80 V source and (b) the power dissipated in the 8 Ω and 26 Ω

resistors. (It is also easy to check that “power delivered” = “power absorbed”.)

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Extra notes on mesh currents:

(1) If there is a dependent source in the circuit, the mesh equations will have to be

supplemented by an equation involving that source.

Illustration:

• • •

•

(2) When a branch contains a current source, the voltage across it is not known. Then

we may “ignore” that branch and create a supermesh which avoids it. The supermesh

must use the same mesh currents as originally chosen and a separate equation relating

the previously “ignored” source to these currents will be required also.

Illustration:

• • •

• • •

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1.4.3 Source Transformations and Equivalent Circuits

Source Transformations

It is possible to replace a voltage source in series with a resistor with a current

source in parrallel with the same resistor (or vice versa) so that the voltage vab across

a load, say RL, remains the same. Consider the following in which we require vab to

be the same in each case:

•

•

•

•

•

•

In the first circuit,

vab = . . . (1)

while in the second,

vab = . . . (2)

From (1) and (2), it is thus required that

This means that

is =vsR

. . . (3).

As far as the behaviour at the terminal a and b is concerned, both circuits are equiv-

alent if (3) holds.

Thevenin Equivalent Circuits

Consider a resistive network possibly containing both dependent and independent

sources. We wish to reduce the circuit to a single equivalent independent voltage and

resistance such that the current and voltage associated with a load connected across

a and b will be the same in the simplified circuit as it was in the original network.

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Illustration:

When this is the case, the voltage and resistance are referred to as the Thevenin

voltage and resistance, symbolized as VTh and RTh, respectively. This equivalence

must hold for all possible values of load resistance RL. It must therefore hold in the

two extreme cases: (i) RL → ∞ (i.e. an open circuit) and (ii) RL → 0 (i.e. a short

circuit). In the first of these cases,

vab = VTh = voc . . . (4)

where voc is the open-circuit voltage. Since, by definition, the Thevenin voltage is the

same for any load, equation (4) indicates we may get it by simply open-circuiting the

load. In the second case, the short-circuit current isc must be given by

isc =VTh

RTh

. . . (5)

again since both VTh and RTh are required to be unchanged for any load. From (5)

RTh =VTh

isc. . . (6)

and equations (4) and (6) thus define the Thevenin equivalent circuit.

Norton Equivalent Circuit

Before elaborating on Thevenin’s equivalent circuits we note the following source

transformation:

Illustration:

From our discussion on source transformations at the beginning of this subsection,

we know that the above circuits are equivalent as long as

is =

In this case the circuit on the right is referred to as the Norton equivalent circuit.

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Preliminary Example on Thevenin/Norton Equivalent Circuits

Find the Thevenin and Norton equivalent circuits for

•

•

•

•

•

•

Thevenin Equivalent Circuit:

Using source transformation, the circuit may be represented as

The Thevenin voltage and resistance may calculated simply as follows:

Norton Equivalent Circuit:

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More on Thevenin Equivalent Circuits

Depending on the type of circuit, reduction to its Thevenin equivalent may involve a

variety of techniques:

(1) In general, voc, isc, and/or RTh may be determined using the results of node-

voltage and mesh-current analysis.

(2) As above, sometimes use of equivalent sources is helpful.

(3) If the circuit contains only independent sources, RTh may be obtained easily by

(i) short-circuiting the voltage supplies and (ii) open-circuiting the current supplies.

This works only because RTh does not depend on the values of the independent voltage

and current supplies in the original circuit. Consider this for the above example:

(4) If the circuit contains dependent sources, RTh may be determined by (i) short-

circuiting the independent voltage supplies, (ii) open-circuiting the independent cur-

rent supplies and (iii) connecting a “test” source across the terminals and determining

the equivalent resistance of the revised circuit from its response to the test source.

Consider Drill Exercise 4.24, p. 148 of the text:

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1.4.4 A Preliminary Look at Maximum Power Transfer

Later we shall be concerned with maximum power being delivered from a circuit which

has elements other than resistors to a load which is not purely resistive. Here, however,

we review the idea of maximum power transfer from a purely resistive network to a

purely resistive load. The purely resistive network may be represented by its Thevenin

equivalent and is depicted below as being attached to a load resistance RL. We wish

to establish the value of RL that permits maximum power transfer from the network

to the RL.

Illustration:

We begin by noting that power delivered to the load is a function of RL as given by

p = i2RL = = . . . (1)

Since we wish to determine the value of RL which maximizes p, we may take the first

derivative of p in (1) with respect to RL, set the result to 0 and solve for RL:

dp

dRL

= = . . . (2)

Clearly, RL = RTH produces an extreme value. The second derivative of p with respect

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to RL is negative (SHOW THIS) for this value. Thus,

RL = RTH . . . (3)

is the value of RL which maximizes the power delivered to the load. From (1), this

power is given by

pmax = . . . (4)

Example: Problem 4.75, page 181 of text.

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1.4.5 Superposition

The principle of superposition states that whenever a linear system is driven by more

than one independent source of energy, the total response of the system equals the

sum of the responses from each of the sources acting individually. The application of

superposition to the problems encountered thus far, in which all independent sources

are dc, does not offer an improvement over the techniques already used. However, for

illustrative purposes we’ll consider superposition briefly.

In applying superposition: (i) to find the effects of independent voltage sources

(one at a time) the independent current sources are deactivated by open-circuiting

them; (ii) to find the effects of independent current sources (one at a time) the in-

dependent voltage sources are deactivated by short-circuiting them; (iii)dependent

sources are never deactivated.

Example: Drill exercise 4.28, page 156 of the text.

We wish to find the voltage v as shown.

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