Basic Concepts electric circuit, and each component of the circuit … 2015 - PreMidterm... ·...

CHAPTER 1

Basic Concepts

In electrical engineering, we are often interested in communicating ortransferring energy from one point to another. To do this requires aninterconnection of electrical devices. Such interconnection is referred toas an electric circuit, and each component of the circuit is known as anelement.

Definition 1.0.1. An electric circuit is an interconnection of electri-cal elements.

• Examples of circuit elements are discussed in Section 1.3.

1.1. International System of Units

1.1.1. As engineers, we deal with measurable quantities. Our mea-surement must be communicated in standard language that virtually allprofessionals can understand irrespective of the country. Such an inter-national measurement language is the International System of Units(SI).

5

DinText BoxECS 203 (CPE2)Asst. Prof. Dr.Prapun Suksompong

6 1. BASIC CONCEPTS

1.1.2. In this system, there are six principal units from which the unitsof all other physical quantities can be derived.

Quantity Basic Unit SymbolLength meter mMass kilogram kgTime second s

Electric Current ampere ATemperature kelvin K

Luminous Intensity candela cd

1.1.3. One great advantage of SI unit is that it uses prefixes based onthe power of 10 to relate larger and smaller units to the basic unit.

Multiplier Prefix Symbol1012 tera T109 giga G106 mega M103 kilo k10−2 centi c10−3 milli m10−6 micro µ10−9 nano n10−12 pico p

Example 1.1.4. Change of units:

600, 000, 000 mA =

1.2. Circuit Variables

1.2.1. Charge: The concept of electric charge is the underlying prin-ciple for all electrical phenomena. Charge is an electrical property of theatomic particles of which matter consists, measured in coulombs (C). Thecharge of an electron is −1.602× 10−19 C.

• The coulomb is a large unit for charges. In 1 C of charge, thereare 1/(1.602 × 10−19) = 6.24 × 1018 electrons. Thus realistic orlaboratory values of charges are on the order of pC, nC, or µC.• A large power supply capacitor can store up to 0.5 C of charge.

1.2. CIRCUIT VARIABLES 7

1.2.2. Law of Conservation of Charge: Charge can neither be creatednor destroyed, only transferred.

Definition 1.2.3. Current: The time rate of change of charge, mea-sured in amperes (A). Mathematically,

i(t) =d

dtq(t)

Note:

• 1 ampere (A) = 1 coulomb/second (C/s).• The charge transferred between time t1 and t2 is obtained by

∆q =

∫ t2t1

i(t)dt

1.2.4. Representing current in circuit (schematic) diagram:

• To talk about current, we need to specify its direction and amount.• These are conveyed by the direction of the arrow (referred to as the

reference direction) and the labeled value (which can be negative ifthe current actually flows in the opposite direction.)

1.2.5. Two types of currents:

(a) A direct current (dc) is a current that remains constant with time.

• Produced by sources such as batteries, power supplies.

8 1. BASIC CONCEPTS

(b) A time-varying current is a current that varies with time.• An alternating current (ac) is a type of time-varying current

that varies sinusoidally with time.

• Such ac current is used in your household, to run the air con-ditioner, refrigerator, washing machine, and other electric ap-pliances.

1.2.6. By convention the symbol I is used to represent such a constantcurrent. A time-varying current is represented by the symbol i.

Definition 1.2.7. Voltage (or potential difference) across an element:the energy required to move a unit charge though an element, measured involts (V).

• 1 volt (V) = 1 joule/coulomb = 1 newton-meter/coulomb

vab

+

–

a

b

• Representing voltage in circuit (schematic) diagram: To talk aboutvoltage, we need to specify(a) its polarity via the plus (+) and minus (-) symbols at the two

positions (points a and b in the picture above)

(b) its value (can be a negative number)


• The voltage between two points a and b in a circuit is denoted byvab and can be interpreted in two ways:

9 V

+

–

a

b

-9 V

+

– a

b

(a) point a is at a potential of vab volts higher than point b, or(b) the potential at point a with respect to point b is vab.

• vab = −vba• Mathematically,

vab =dw

dqwhere w is the energy in joules (J) and q is charge in coulombs (C).

1.2.8. Like electric current, a constant voltage is called a dc voltageand is represented by V , whereas a time-varying voltage is representedby v. A time-varying voltage that is sinusoidal is called an ac voltage.

Example 1.2.9. A dc voltage is commonly produced by a battery; acvoltage is produced by an electric generator.

1.2.10. Current and voltage are the two basic variables in electric cir-cuits. The common term signal is used for an electric quantity such asa current or a voltage (or even electromagnetic wave) when it is used forconveying information. Engineers prefer to call such variables signalsrather than mathematical functions of time because of their importance incommunications and other disciplines.

For practical purposes, we need to be able to find/calculate/measuremore than the current and voltage. We all know from experience that a100-watt bulb gives more light than a 60-watt bulb. We also know thatwhen we pay our bills to the electric utility companies, we are paying forthe electric energy consumed over a certain period of time. Thus powerand energy calculations are important in circuit analysis.

Definition 1.2.11. Power: time rate of absorbing (or expending) en-ergy, measured in watts (W). Mathematically, the instantaneous power

(1.1) p =dw

dt=dw

dq

dq

dt= vi

10 1. BASIC CONCEPTS

Definition 1.2.12. Passive Sign Convention (PSC): Sign of power

(a) Plus sign: Power is absorbed (consumed) by the element. (resis-tor)• For resistor, we say the power is dissipated.

(b) Minus sign: Power is supplied by the element. (battery, genera-tor)

1.2.13. To comply with the convention, we have to be careful whenapplying formula (1.1).

(a) If the current “enters” the element through the “positive terminal”of the voltage across this element, p = vi.

v

+

–

i

p = +vi

i

v

+

–

p = –vi

(b) If the current “enters” the element through the “negative terminal”of the voltage across this element, p = −vi.

Note that, here, to choose which version of the formula to apply, weonly consider the direction of the current’s arrow and the polarity of thevoltage. We do not have to care whether the variables i and v are positiveor negative.

Example 1.2.14.

4 V

+

–

3 A

4 V

+

–

3 A


4 V

+

–

3 A

4 V

+

–

3 A

Example 1.2.15. Light bulb or battery: which is which?

1.2.16. Law of Conservation of Energy : Energy can neither be creatednor destroyed, only transferred.

• For this reason, the algebraic sum of power in a circuit, at anyinstant of time, must be zero.• The total power supplied to the circuit must balance the total power

absorbed.

Example 1.2.17. The circuit below has five elements. Let Pk be thepower “absorbed” by element k. (Although we use the word “absorbed”,if Pk is negative, then element k is actually supplying power.)If P1 = −205 W, P2 = 60 W, P4 = 45 W, P5 = 30 W,calculate the power P3.


1.2.18. Energy: the energy absorbed or supplied by an element fromtime t = t1 to t = t2 is

w =

∫ t2t1

p(t) dt =

∫ t2t1

v(t)i(t) dt.

• Integration suggests finding area under the curve.• Need to be careful with negative area.

Example 1.2.19. Electricity bills: The electric power utility compa-nies measure energy in kilowatt-hours (kWh), where 1 kWh = 3600 kJ.

1.3. Circuit Elements

Definition 1.3.1. There are 2 types of elements found in electricalcircuits.

1) Active elements (is capable of generating energy), e.g., generators,batteries, and operational amplifiers (Op-amp).

2) Passive element, e.g., resistors, capacitors and inductors.

Definition 1.3.2. The most important active elements are voltage andcurrent sources:

(a) Voltage source provides the circuit with a specified voltage (e.g.a 1.5V battery)

(b) Current source provides the circuit with a specified current (e.g.a 1A current source).

Definition 1.3.3. In addition, we may characterize the voltage or cur-rent sources as:

1) Independent source: An active element that provides a specifiedvoltage or current that is completely independent of other circuit elements.

v DC V+

–

1.3. CIRCUIT ELEMENTS 13

i

2) Dependent source: An active element in which the source quantityis controlled by another voltage or current.

v i+ –

1.3.4. The key idea to keep in mind is that a voltage source comes withpolarities (+ -) in its symbol, while a current source comes with an arrow,irrespective of what it depends on.

Example 1.3.5. Current-controlled voltage source

5 V+

–

A

10i

B

+ –

i

C

Example 1.3.6. Draw the general form of a voltage-controlled currentsource.


Example 1.3.7. Indicate the type of the dependent source in each ofthe circuit in the figure below.

1.3.8. Remarks:

• Dependent sources are useful in modeling elements such as transis-tors, operational amplifiers and integrated circuits.• Ideal sources

– An ideal voltage source (dependent or independent) will pro-duce any current required to ensure that the terminal voltageis as stated.

– An ideal current source will produce the necessary voltage toensure the stated current flow.

– Thus an ideal source could in theory supply an infinite amountof energy.

• Not only do sources supply power to a circuit, they can absorb powerfrom a circuit too.• For a voltage source, we know the voltage but not the current sup-

plied or drawn by it. By the same token, we know the currentsupplied by a current source but not the voltage across it.

CHAPTER 2

Basic Laws

Here we explore two fundamental laws that govern electric circuits(Ohm’s law and Kirchhoff’s laws) and discuss some techniques commonlyapplied in circuit design and analysis.

2.1. Ohm’s Law

Ohm’s law shows a relationship between voltage and current of a resis-tive element such as conducting wire or light bulb.

2.1.1. Ohm’s Law: The voltage v across a resistor is directly propor-tional to the current i flowing through the resistor.

v = iR,

where R = resistance of the resistor, denoting its ability to resist the flowof electric current. The resistance is measured in ohms (Ω).

• To apply Ohm’s law, the direction of current i and the polarity ofvoltage v must conform with the passive sign convention. This im-plies that current flows from a higher potential to a lower potential

15


16 2. BASIC LAWS

in order for v = iR. If current flows from a lower potential to ahigher potential, v = −iR.

+

– Rv

il

Cross-sectional

area A

Material with

resistivity r

2.1.2. The resistance R of a cylindrical conductor of cross-sectional areaA, length L, and conductivity σ is given by

R =L

σA.

Alternatively,

R = ρL

Awhere ρ is known as the resistivity of the material in ohm-meters. Goodconductors, such as copper and aluminum, have low resistivities, whileinsulators, such as mica and paper, have high resistivities.

2.1.3. Remarks:

(a) R = v/i(b) Conductance :

G =1

R=i

vThe unit of G is the mho1 (f) or siemens2 (S)

1Yes, this is NOT a typo! It was derived from spelling ohm backwards.2In English, the term siemens is used both for the singular and plural.

2.1. OHM’S LAW 17

(c) The two extreme possible values of R.(i) When R = 0, we have a short circuit and

v = iR = 0

showing that v = 0 for any i.

R = 0v = 0

+

–

i

(ii) When R =∞, we have an open circuit and

i = limR→∞

v

R= 0

indicating that i = 0 for any v.

R = ∞ v

+

–

i = 0

2.1.4. A resistor is either fixed or variable. Most resistors are of thefixed type, meaning their resistance remains constant.

18 2. BASIC LAWS

A common variable resistor is known as a potentiometer or pot forshort

2.1.5. Not all resistors obey Ohms law. A resistor that obeys Ohmslaw is known as a linear resistor.

• A nonlinear resistor does not obey Ohms law.

• Examples of devices with nonlinear resistance are the lightbulb andthe diode.• Although all practical resistors may exhibit nonlinear behavior un-

der certain conditions, we will assume in this class thatall elements actually designated as resistors are linear.

2.1. OHM’S LAW 19

2.1.6. Using Ohm’s law, the power p dissipated by a resistor R is

p = vi = i2R =v2

R.

Example 2.1.7. In the circuit below, calculate the current i, and thepower p.

30 V DC+

– v

i

5 kΩ

Definition 2.1.8. The power rating is the maximum allowable powerdissipation in the resistor. Exceeding this power rating leads to overheatingand can cause the resistor to burn up.

Example 2.1.9. Determine the minimum resistor size that can be con-nected to a 1.5V battery without exceeding the resistor’s 14-W power rating.

20 2. BASIC LAWS

2.2. Node, Branches and Loops

Definition 2.2.1. Since the elements of an electric circuit can be in-terconnected in several ways, we need to understand some basic conceptof network topology.

• Network = interconnection of elements or devices• Circuit = a network with closed paths

Definition 2.2.2. Branch: A branch represents a single element suchas a voltage source or a resistor. A branch represents any two-terminalelement.

Definition 2.2.3. Node: A node is the “point” of connection betweentwo or more branches.

• It is usually indicated by a dot in a circuit.• If a short circuit (a connecting wire) connects two nodes, the two

nodes constitute a single node.

Definition 2.2.4. Loop: A loop is any closed path in a circuit. Aclosed path is formed by starting at a node, passing through a set of nodesand returning to the starting node without passing through any node morethan once.

54 C H A P T E R T W O r e s i s t i v e n e t w o r k s

illustrative circuit. We will then apply the same systematic method to solvemore complicated examples, including the one shown in Figure 2.1.

2.1 T E R M I N O L O G Y

Lumped circuit elements are the fundamental building blocks of electronic cir-cuits. Virtually all of our analyses will be conducted on circuits containingtwo-terminal elements; multi-terminal elements will be modeled using combi-nations of two-terminal elements. We have already seen several two-terminalelements such as resistors, voltage sources, and current sources. Electronicaccess to an element is made through its terminals.

An electronic circuit is constructed by connecting together a collection ofseparate elements at their terminals, as shown in Figure 2.2. The junction pointsat which the terminals of two or more elements are connected are referred to asthe nodes of a circuit. Similarly, the connections between the nodes are referredto as the edges or branches of a circuit. Note that each element in Figure 2.2forms a single branch. Thus an element and a branch are the same for circuitscomprising only two-terminal elements. Finally, circuit loops are defined to beclosed paths through a circuit along its branches.

Several nodes, branches, and loops are identified in Figure 2.2. In the circuitin Figure 2.2, there are 10 branches (and thus, 10 elements) and 6 nodes.

As another example, a is a node in the circuit depicted in Figure 2.1 atwhich three branches meet. Similarly, b is a node at which two branches meet.ab and bc are examples of branches in the circuit. The circuit has five branchesand four nodes.

Since we assume that the interconnections between the elements in a circuitare perfect (i.e., the wires are ideal), then it is not necessary for a set of elementsto be joined together at a single point in space for their interconnection to beconsidered a single node. An example of this is shown in Figure 2.3. Whilethe four elements in the figure are connected together, their connection doesnot occur at a single point in space. Rather, it is a distributed connection.

F IGURE 2.2 An arbitrary circuit.

Nodes

Loop

Branch

Elements

Definition 2.2.5. Series: Two or more elements are in series if theyare cascaded or connected sequentially and consequently carry the samecurrent.

Definition 2.2.6. Parallel: Two or more elements are in parallel ifthey are connected to the same two nodes and consequently have the samevoltage across them.

2.2. NODE, BRANCHES AND LOOPS 21

2.2.7. Elements may be connected in a way that they are neither inseries nor in parallel.

Example 2.2.8.

Example 2.2.9. How many branches and nodes does the circuit in thefollowing figure have? Identify the elements that are in series and in par-allel.

10 VDC 4 Ω

5 Ω

2 Ω 1 Ω

2.2.10. A loop is said to be independent if it contains a branch whichis not in any other loop. Independent loops or paths result in independentsets of equations. A network with b branches, n nodes, and ` independentloops will satisfy the fundamental theorem of network topology:

b = `+ n− 1.Definition 2.2.11. The primary signals within a circuit are its currents

and voltages, which we denote by the symbols i and v, respectively. Wedefine a branch current as the current along a branch of the circuit, anda branch voltage as the potential difference measured across a branch.

2.2 Kirchhoff’s Laws C H A P T E R T W O 55

Elements

Distributed nodeIdeal wires

A

BC

D

A

B

C

D

F IGURE 2.3 Distributedinterconnections of four circuitelements that nonetheless occurat a single node.

Branch current

+

-

Branchvoltage

vi

F IGURE 2.4 Voltage and currentdefinitions illustrated on a branch ina circuit.

Nonetheless, because the interconnections are perfect, the connection can beconsidered to be a single node, as indicated in the figure.

The primary signals within a circuit are its currents and voltages, which wedenote by the symbols i and v, respectively. We define a branch current as thecurrent along a branch of the circuit (see Figure 2.4), and a branch voltage as thepotential difference measured across a branch. Since elements and branches arethe same for circuits formed of two-terminal elements, the branch voltages andcurrents are the same as the corresponding terminal variables for the elementsforming the branches. Recall, as defined in Chapter 1, the terminal variables foran element are the voltage across and the current through the element.

As an example, i4 is a branch current that flows through branch bc in thecircuit in Figure 2.1. Similarly, v4 is the branch voltage for the branch bc.

2.2 K I R C H H O F F ’ S L A W S

Kirchhoff’s current law and Kirchhoff’s voltage law describe how lumped-parameter circuit elements couple at their terminals when they are assembledinto a circuit. KCL and KVL are themselves lumped-parameter simplificationsof Maxwell’s Equations. This section defines KCL and KVL and justifies thatthey are reasonable using intuitive arguments.1 These laws are employed incircuit analysis throughout this book.

1. The interested reader can refer to Section A.2 in Appendix A for a derivation of Kirchhoff’s lawsfrom Maxwell’s Equations under the lumped matter discipline.

22 2. BASIC LAWS

2.3. Kirchhoff’s Laws

Ohm’s law coupled with Kirchhoff’s two laws gives a sufficient, powerfulset of tools for analyzing a large variety of electric circuits.

2.3.1. Kirchhoff’s current law (KCL): the algebraic sum of currentsdeparting a node (or a closed boundary) is zero. Mathematically,∑

n

in = 0

KCL is based on the law of conservation of charge. An alternativeform of KCL is

Sum of currents (or charges) drawn as entering a node

= Sum of the currents (charges) drawn as leaving the node.

Example 2.3.2.

i5

i4

i3

i2

i1

Example 2.3.3.

2.3. KIRCHHOFF’S LAWS 23

Example 2.3.4. A simple application of KCL is combining currentsources in parallel.

I1 I2 I3

IT

a

b(a)

IT = I1 – I2 + I3

IT

a

b

(b)

Note that KCL also applies to a closed boundary. This may be regardedas a generalized case, because a node may be regarded as a closed surfaceshrunk to a point. In two dimensions, a closed boundary is the same as aclosed path. The total current entering the closed surface is equal to thetotal current leaving the surface.

Closed boundary

Example 2.3.5.

24 2. BASIC LAWS

A Kirchhoff’s voltage law (KVL): the algebraic sum of all voltagesaround a closed path (or loop) is zero. Mathematically,

M∑m=1

vm = 0

KVL is based on the law of conservation of energy. An alternativeform of KVL is

Sum of voltage drops = Sum of voltage rises.

Example 2.3.6.

v1

+ – v2 v3+ –

v4

v5 + –

Example 2.3.7. When voltage sources are connected in series, KVLcan be applied to obtain the total voltage.

VS = V1 + V2 – V3Vab

a

b

+

–

Vab

a

b

+

–

V1

V2

V3

2.3. KIRCHHOFF’S LAWS 25

Example 2.3.8. Find v1 and v2 in the following circuit.

10 V

+ – v2

v1+ –

8 V

4 Ω

2 Ω

Example 2.3.9.

Example 2.3.10 (HRW p. 725). In the figure below, all the resistorshave a resistance of 4.0 Ω and all the (ideal) batteries are 4.0 V. What isthe current through resistor R?

725QU E STION SPART 3

HALLIDAY REVISED

Pr at which energy is dissipated as thermal energy in the battery is

Pr � i2r. (27-16)

The rate Pemf at which the chemical energy in the battery changes is

Pemf � i�. (27-17)

Series Resistances When resistances are in series, they havethe same current. The equivalent resistance that can replace a se-ries combination of resistances is

(n resistances in series). (27-7)

Parallel Resistances When resistances are in parallel,they have the same potential difference. The equivalent resistancethat can replace a parallel combination of resistances is given by

(n resistances in parallel). (27-24)1

Req � �

n

j�1

1Rj

Req � �n

j�1 Rj

RC Circuits When an emf � is applied to a resistance R and ca-pacitance C in series, as in Fig. 27-15 with the switch at a, the chargeon the capacitor increases according to

q � C �(1 � e�t/RC) (charging a capacitor), (27-33)

in which C � � q0 is the equilibrium (final) charge and RC � t isthe capacitive time constant of the circuit. During the charging, thecurrent is

(charging a capacitor). (27-34)

When a capacitor discharges through a resistance R, the charge onthe capacitor decays according to

q � q0e�t/RC (discharging a capacitor). (27-39)

During the discharging, the current is

(discharging a capacitor). (27-40) i �dqdt

� �� q0RC �e�t/RC

i �dqdt

� � �R �e�t/RC

1 (a) In Fig. 27-18a, with R1 � R2, is the potential differenceacross R2 more than, less than, or equal to that across R1? (b) Is thecurrent through resistor R2 more than, less than, or equal to thatthrough resistor R1?

Fig. 27-18 Questions 1 and 2.

(a)

+–

R1 R2

R3(b)

+–

R3

R1R2

(d)(c)

R2R1+–

R3

R3

+–

R1

R2

R

Fig. 27-21 Question 6.

2 (a) In Fig. 27-18a, are resistors R1 and R3 in series? (b) Are resistors R1 and R2 in parallel? (c) Rank the equivalent resistancesof the four circuits shown in Fig. 27-18, greatest first.

3 You are to connect resistors R1 and R2, with R1 � R2, to a bat-tery, first individually, then in series, and then in parallel. Rankthose arrangements according to the amount of current throughthe battery, greatest first.

4 In Fig. 27-19, a circuit consists ofa battery and two uniform resistors,and the section lying along an x axisis divided into five segments ofequal lengths. (a) Assume that R1 �R2 and rank the segments accordingto the magnitude of the averageelectric field in them, greatest first. (b) Now assume that R1 � R2

and then again rank the segments. (c) What is the direction of theelectric field along the x axis?

5 For each circuit in Fig. 27-20, are the resistors connected in series, in parallel, or neither?

6 Res-monster maze. In Fig. 27-21, all the resistors have aresistance of 4.0 � and all the (ideal) batteries have an emf of 4.0V. What is the current through resistor R? (If you can find theproper loop through this maze, you can answer the question with afew seconds of mental calculation.)


+ –

R1 R2

a b c d e

x

+–

+– +–

(a) (b) (c)


7 A resistor R1 is wired to a battery, then resistor R2 is added inseries. Are (a) the potential difference across R1 and (b) the cur-

halliday_c27_705-734v2.qxd 23-11-2009 14:35 Page 725

26 2. BASIC LAWS

2.4. Resistor Combination

2.4.1. Series resistors:

(a) Meaning: From the point of view of “circuit 1”, the two resistorsR1 and R2 which are connected in series can be replaced by anequivalent resistor Req where Req = R1 +R2.

(b) The use of the symbol “≡” means that the two circuits are “equiv-alent” with respect to (wrt.) the two terminals a-b.

2.4.2. Parallel resistors:

Example 2.4.3. 6‖3 =

Example 2.4.4. R‖R =

Example 2.4.5. (R)‖(nR) =

Example 2.4.6. (mR)‖(nR) =

2.4. RESISTOR COMBINATION 27

2.4.7. Generalization to n resistors:

Example 2.4.8. Find Req for the following circuit.

Req

4 Ω

8 Ω

1 Ω

5 Ω

2 Ω

3 Ω 6 Ω

2.4.9. Caution: Equivalence is always with respect to some two termi-nals.

• To reduce confusion, it is important to specify these terminals.• However, in many cases, the terminals are not explicitly specified

because it is quite “clear” where they are in the circuit.

Example 2.4.10. Find the equivalent resistance of the following circuit.

28 2. BASIC LAWS

2.4.11. Later in this class, we will have to deal with many resistor-combination scenarios which have “hanging” branches or “dangling” branches.This is demonstrated in the next example.

When working on such problem, some resistors can be eliminated fromthe circuits. When it is not clear whether a particular resistor can beremoved, we can “try connecting a 1A current source” across the two ter-minals and see whether there is any current flowing through that resistor.

Example 2.4.12. Find Req of the following circuit

2.4.13. An advanced-level trick for finding Req: Connect a 1A currentsource across the two terminals and (use other techniques to) solve forthe value V of the voltage across the two terminals (under the 1A currentsource), then

the Ohm’s law implies that we must have

Req =V

1= V.

In other words, the voltage will be exactly the same as the equivalentresistance. Later on, we will study a seemingly impossible-to-solve resistor-combination question using such observation.

2.4. RESISTOR COMBINATION 29

Example 2.4.14. A revisit to Example 2.4.12:

2.4.15. Consider the circuit below.

(a) Find the equivalent resistance with respect to terminals a-c.(b) Find the equivalent resistance with respect to terminals c-d.

30 2. BASIC LAWS

2.5. Dividers

2.5.1. Voltage divider: When two resistors are connected in series,the “total” voltage across them is divided among the resistors in directproportion to their resistances.

Caution: Before applying this formula, make sure that the same currentpasses though both resistors. (No extra current leaks in or out at node c.)

• This means the formula still works when there is/are “hanging”branch(es) at node c.

Example 2.5.2. Find V0 in the following circuit.

Example 2.5.3. Find V0 in the following circuit.

2.5. DIVIDERS 31

2.5.4. Generalization:

2.5.5. Current divider: When two resistors are connected in parallel,the “total” current through them is divided among the resistors in inverseproportion to their resistances.

Example 2.5.6.

2.5.7. Generalization:

32 2. BASIC LAWS

Example 2.5.8. Find io, vo, po (power dissipated from the 3Ω resistor).

12 V

i a

b

+

–

v0 3 Ω 6 Ω

4 Ω i0

Example 2.5.9. Three light bulbs are connected to a 9V battery asshown below. Calculate: (a) the total current supplied by the battery, (b)the current through each bulb, (c) the resistance of each bulb.

I2

R2

I I1

R115 W

10 W

20 W9 V

9 V

R3

+

–

V2

+

–

V3

+

–

V1

2.7. MEASURING DEVICES 33

2.6. Practical Voltage and Current Sources

An ideal voltage source is assumed to supply a constant voltage. Thisimplies that it can supply very large current even when the load resistanceis very small.

However, a practical voltage source can supply only a finite amount ofcurrent. To reflect this limitation, we model a practical voltage source asan ideal voltage source connected in series with an internal resistance rs,as follows:

Similarly, a practical current source can be modeled as an ideal currentsource connected in parallel with an internal resistance rs.

2.7. Measuring Devices

Ohmmeter: measures the resistance of the element.Important rule: Measure the resistance only when the element is discon-nected from circuits.

Ammeter: connected in series with an element to measure currentflowing through that element. Since an ideal ammeter should not restrictthe flow of current, (i.e., cause a voltage drop), an ideal ammeter has zerointernal resistance.

Voltmeter:connected in parallel with an element to measure voltageacross that element. Since an ideal voltmeter should not draw current awayfrom the element, an ideal voltmeter has infinite internal resistance.

ECS 203 2014: Quiz 1

Instructions i. Separate into groups of no more than three persons.

ii. Only one submission is needed for each group. Late submission

will not be accepted.

iii. Write down all the steps that you have done to obtain your

answers. You may not get full credit even when your answer is

correct without showing how you get your answer.

iv. Do not panic.

1. Find I1 when IS = 10 A, R1 = 3 kΩ and R2 = 2 kΩ.

R2R1IS I2I1

2. Consider the circuit below.

12

6

a

b3

c

a. Find the equivalent resistance with respect to terminals a-b

b. Find the equivalent resistance with respect to

terminals b-c

c. Find the equivalent resistance with respect to terminals a-c

Name ID

CHAPTER 3

Methods of Analysis

Here we apply the fundamental laws of circuit theory (Ohm’s Law &Kirchhoff’s Laws) to develop two powerful techniques for circuit analysis.

1. Nodal Analysis (based on KCL)2. Mesh Analysis (based on KVL)This is the most important chapter for our course.

3.1. Nodal Analysis

Here, we analyze circuit using node voltages as the circuit variables.

Example 3.1.1. Back to Example 2.5.8

12 V

i a

b

+

–

v0 3 Ω 6 Ω

4 Ω i0

37


38 3. METHODS OF ANALYSIS

3.1.2. Steps to Determine Node Voltages:

Step 0: Determine the number of nodes n.Step 1: Select a node as a reference node (ground node). Assign

voltages v1, v2, · · · , vn−1 to the remaining n− 1 nodes.• The voltage are now referenced with respect to the reference

node.• The ground node is assumed to have 0 potential.

• Recall that voltages are measured between two points. Fornode voltages, the second point is always the ground node.

Step 1s: If a voltage source is connected between the reference node anda nonreference node, we simply set the voltage at the nonref-erence node “equal” to the voltage of the voltage source.

Step 2: Apply KCL to each of the remaining nonreference nodes.(a) Use Ohm’s law to express the branch currents in terms of the

node voltages.

(b) Current source automatically gives current value.Caution: Watch out for the direction of the arrow.

Step 3: Solve the resulting simultaneous equations to obtain the un-known node voltages.

Step 4: After all the node voltages are determined, it is easy to findeverything else.

3.1. NODAL ANALYSIS 39

3.1.3. Remarks: There are multiple methods to solve the simultaneousequations in Step 3.

• Method 1: Elimination technique (good for a few variables)• Method 2: Write in term of matrix and vectors (write Ax = b),

then use Cramer’s rule.• Method 3: Use

– computer (e.g., MATLAB) to find A−1 and then find x = A−1b– calculator (fx-991MS can solve simultaneous linear equations

with two or three unknowns.)

Example 3.1.4. Calculate the node voltages in the circuit below.

12

10 A3 Ω 6 Ω

4 Ω

5 A

2

Example 3.1.5. Calculate the node voltages in the circuit below.

12

0

3 A 4 Ω

2 Ω 3

4 Ω

8 Ω ix

2ix


The next example shows that the steps given in 3.1.2 are not sufficientto solve all interesting circuit-analysis problem.

Example 3.1.6. Find v and i in the circuit below.

21 V

i

9 V

+

– v 6 Ω 2 Ω

4 Ω

3 Ω

3.1.7. An extra step should be added to 3.1.2:

Step 1sn: If there is a voltage source connected between two nonref-erence nodes, the two nonreference nodes form a supernode. Weapply both KCL and KVL to determine the node voltages.

3.1.8. Note the following properties of a supernode:

(a) The voltage source inside the supernode provides a constraint equa-tion needed to solve for the node voltages.

(b) A supernode has no voltage of its own.(c) We can have more than two nodes forming a single supernode.(d) The supernodes are treated differently because nodal analysis re-

quires knowing the current through each element. However, there isno way of knowing the current through a voltage source in advance.

3.2. MESH ANALYSIS 41

3.2. Mesh Analysis

Mesh analysis provides another general procedure for analyzing circuits,using mesh currents as the circuit variables.

Definition 3.2.1. Mesh is a loop which does not contain any otherloop within it.

3.2.2. Steps to Determine Mesh Currents:

Step 0: Determine the number of meshes n.Step 1: Assign mesh currents1 i1, i2, . . ., in, to the n meshes.

• The direction of the mesh current is arbitrary–(clockwise orcounterclockwise)–and does not affect the validity of the solu-tion.• For convenience, we define currents flow in the clockwise (CW)

direction.Step 2a: From the current direction in each mesh, define the voltage

drop polarities.Step 2b: Apply KVL to each of the n meshes.

Use Ohm’s law to express the voltages in terms of the mesh current.• Tip (for combining Step 2a and 2b): Go around the loop in the

same direction2 as the mesh current (of that mesh). When wepass a resistor R, the voltage drops by I × R where I is thebranch current (algebraic sum of mesh currents) through thatresistor in the mesh-current direction.

Step 3: Solve the resulting n simultaneous equations for the meshcurrents.

Step 4: Other quantities related to the circuit can be found from themesh currents.

1Using mesh currents instead of element currents as circuit variables is convenient and reduces thenumber of equations that must be solved simultaneously.

2Note that according to our agreement above, this is in the CW direction.


Example 3.2.3. Back to Example 2.5.8

12 V

i a

b

+

–

v0 3 Ω 6 Ω

4 Ω i0

Example 3.2.4. Find the branch currents I1, I2, and I3 using meshanalysis.

15 V 4 Ω

6 Ω 5 Ω

10 Ω

I1 I2

I3

i1 i2

10 V

3.3. REMARKS ON NODAL ANALYSIS AND MESH ANALYSIS 43

3.3. Remarks on Nodal Analysis and Mesh Analysis

3.3.1. Nodal analysis applies KCL to find unknown (node) voltages ina given circuit, while mesh analysis applies KVL to find unknown (mesh)currents.

3.3.2. Mesh analysis is not quite as general as nodal analysis becauseit is only applicable to a circuit that is planar.

• A planar circuit is one that can be drawn in a plane with no branchescrossing one another; otherwise it is nonplanar.

3.3.3. Nodal Analysis vs. Mesh Analysis: Given a network to beanalyzed, how do we know which method is better or more efficient?

Suggestion: You should be familiar with both methods. Choose themethod that results in smaller number of variables or equations.

• A circuit with fewer nodes than meshes is better analyzed usingnodal analysis, while a circuit with fewer meshes than nodes is betteranalyzed using mesh analysis.

You can also use one method to check your results of the other method.

3.3.4. Nodal analysis and mesh analysis can also be used to find equiv-alent resistance of a part of a circuit.

This becomes extremely useful when the techniques that we studied inthe previous chapter cannot be directly applied (, e.g., when we can’t findresistors that are in parallel or in series; they are all connected in some“strange” configuration.)

The are two approaches to this kind of problems.

(a) Apply 1 V voltage source across the terminals, find the correspond-ing current I through the voltage source. Then,

Req =V

I=

1

I.

(b) Put 1 A current source through the terminals, find the correspondingvoltage V across the current source. Then,

Req =V

I=V

1= V.


Example 3.3.5. Find the equivalent resistance for the following circuit

1 Ω

1 Ω

1 Ω

1 Ω

1 Ω

Req

CHAPTER 4

Circuit Theorems

The growth in areas of application of electrical circuits has led to anevolution from simple to complex circuits. To handle such complexity, en-gineers over the years have developed theorems to simplify circuit analysis.These theorems (Thevenin’s and Norton’s theorems) are applicable to lin-ear circuits which are composed of resistors, voltage and current sources.

Definition 4.0.6. System:

4.1. Linearity Property

Definition 4.1.1. A linear system is a system whose output is lin-early related (or directly proportional) to its input1. In particular, whenwe says that the input and output are linearly related, we mean they needto satisfies two properties:

(a) Homogeneous (Scaling): If the input is multiplied by a constant k,then we should observed that the output is also multiplied by k.

(b) Additive: If the inputs are summed then the output are summed.

Example 4.1.2. Is the function f(x) = x2 + 1 linear?

1The input and output are sometimes referred to as cause and effect, respectively.

45


46 4. CIRCUIT THEOREMS

Example 4.1.3. Is the function f(x) = 3x+ 1 linear?

4.1.4. A one-dimensional linear function is a function of the form

y = ax

for some constant a.

• For a system, we may call it a single-input single-output (SISO)system.• In radio it is the use of only one antenna both in the transmitter

and receiver.

4.1.5. A multi-dimensional linear function is a function of the formy1y2...ym

= Ax1x2...xn

for some matrix A.

• For a system, when both m and n are greater than one, we may callit a multiple-input multiple-output system (MIMO) system.

• When m = n = 1, we are back to the one-dimensional case in 4.1.4.

Example 4.1.6. A resistor is a linear element when we consider thecurrent i as its input and the voltage v as its output.

4.1.7. For us, we are considering linear circuit.

(a) The xj’s will be all of the (independent) voltage and current sourcesin the circuit.

(b) Each yj will correspond to the current or voltage value under inves-tigation.

4.1. LINEARITY PROPERTY 47

Example 4.1.8. The circuit below is excited by a voltage source vs,which serves as the input. Assume that the circuit is linear.

vs

i

RLinear circuit

The circuit is terminated by a load R. We take the current i through Ras the output. Suppose vs = 10 V gives i = 2 A. By the assumed linearity,vs = 1V will give i = 0.2 A. By the same token, i = 1 mA must be due tovs = 5 mV.

Example 4.1.9. For the circuit below, find vo when(a) is = 15 and (b) is = 30.

4 Ω

12 Ω

is 8 Ω

+

– v0

4.1.10. Because p = i2R = v2/R (making it a quadratic function ratherthan a linear one), the relationship between power and voltage (or cur-rent) is nonlinear. Therefore, the theorems covered in this chapter are notapplicable to power.


4.2. Superposition

Example 4.2.1. Find the voltage v in the following circuit.

4 Ω

8 Ω

6 V+

– v 3 A

From the expression of v, observe that there are two contributions.

(a) When Is acts alone (set Vs = 0),

(b) When Vs acts alone (set Is = 0),

Key Idea: Find these contributions from the individual sources and thenadd them up to get the final answer.

Definition 4.2.2. Superposition technique is a way to determine cur-rents and voltages in a circuit that has multiple independent sources byconsidering the contribution of one source at a time and then add themup.

4.2.3. The superposition principle states that the voltage across (orcurrent through) an element in a linear circuit is the algebraic sum of thevoltages across (or currents through) that element due to each independentsource acting alone.

• So, if the circuit has n sources, then we have n cases: “source 1acting alone”, “source 2 acting alone”, . . . , “source n acting alone”.

4.2. SUPERPOSITION 49

4.2.4. To apply the superposition principle, we must keep two thingsin mind.

(a) We consider one independent source at a time while all other inde-pendent source are turned off.2

• Replace other independent voltage sources by 0 V (or shortcircuits)• Replace other independent current sources by 0 A (or open

circuits)This way we obtain a simpler and more manageable circuit.

(b) Dependent sources are left intact because they are controlled bycircuit variable.

4.2.5. Steps to Apply Superposition Principles:

S1: Turn off all independent sources except one source.Find the output due to that active source. (Here, you may use anytechnique of your choice.)

S2: Repeat S1 for each of the other independent sources.S3: Find the total contribution by adding algebraically all the contri-

butions due to the independent sources.

Example 4.2.6. Back to Example 4.2.1.

4 Ω

8 Ω

6 V+

– v 3 A

2Other terms such as killed, made inactive, deadened, or set equal to zero are often used to conveythe same idea.


Example 4.2.7. Using superposition theorem, find vo in the followingcircuit.

2 Ω

3 Ω

4 A

+

– v0 10 V

5 Ω

4.2.8. Remark on linearity: Keep in mind that superposition is basedon linearity. Hence, we cannot find the total power from the power dueto each source, because the power absorbed by a resistor depends on thesquare of the voltage or current and hence it is not linear (e.g. because52 6= 12 + 42).

4.2.9. Remark on complexity: Superposition helps reduce a complexcircuit to simpler circuits through replacement of voltage sources by shortcircuits and of current sources by open circuits.

However, it may very likely involve more work. For example, if the cir-cuit has three independent sources, we may have to analyze three circuits.The advantage is that each of the three circuits is considerably easier toanalyze than the original one.

4.3. SOURCE TRANSFORMATION 51

4.3. Source Transformation

We have noticed that series-parallel resistance combination helps sim-plify circuits. The simplification is done by replacing one part of a circuitby its equivalence.3 Source transformation is another tool for simplifyingcircuits.

4.3.1. A source transformation is the process of replacing a voltage sourcein series with a resistor R by a current source in parallel with a resistor Ror vice versa.

R

Ra

b

vs is

a

b

Notice that when terminals a − b are short-circuited, the short-circuitcurrent flowing from a to b is isc = vs/R in the circuit on the left-handside and isc = is for the circuit on the righthand side. Thus, vs/R = is inorder for the two circuits to be equivalent. Hence, source transformationrequires that

(4.2) vs = isR or is =vsR.

3Recall that an equivalent circuit is one whose v − i characteristics are identical with the originalcircuit.


4.3.2. Source transformation is usually used repeatedly in combina-tion with the “source combination” and “resistor combination” techniquesstudied in earlier chapter. In our class, when we say “use source transfor-mation”, we actually mean to use all the three techniques above repeatedlyto simplify the circuit. At the end, the unknown current or voltage valuecan usually be obtained by the current divider formula or the voltage di-vider formula, respectively.

Example 4.3.3. Use source transformation to find v0 in the followingcircuit:

8 Ω

2 Ω

3 A+

– v0 12 V

3 Ω

4 Ω

4.3. SOURCE TRANSFORMATION 53

4.3.4. Cautions:

(a) The “R” in series with the voltage source and the “R” in parallelwith the current source are not the same “resistor” even thoughthey have the same value. In particular, the voltage values acrossthem are generally different and the current values through themare generally different.

(b) Keep a circuit variable as a fixed point in the circuit. Do not blindly“transform” and “combine”.

(c)

(d) From (4.2), an ideal voltage source with R = 0 cannot be replacedby a finite current source. Similarly, an ideal current source withR =∞ cannot be replaced by a finite voltage source.


4.4. Thevenin’s Theorem

4.4.1. It often occurs in practice that a particular element in a circuitor a particular part of a circuit is variable (usually called the load) whileother elements are fixed.

• As a typical example, a household outlet terminal may be connectedto different appliances constituting a variable load.

Each time the variable element is changed, the entire circuit has to be ana-lyzed all over again. To avoid this problem, Thevenins theorem provides atechnique by which the fixed part of the circuit is replaced by an equivalentcircuit.

4.4.2. Thevenin’s Theorem is an important method to simplify a com-plicated circuit to a very simple circuit. It states that a circuit can be re-placed by an equivalent circuit consisting of an independent voltage sourceVTh in series with a resistor RTh, whereVTh: the open circuit voltage at the terminal.RTh: the equivalent resistance at the terminals when the independent

sources are turned off.

VTh

(a)

(b)

Ia

Linear

two- terminal

circuit

Load

b

V

+

–

I a

b

V

+

–

Load

RTh

This theorem allows us to convert a complicated network into a simplecircuit consisting of a single voltage source and a single resistor connectedin series. The circuit is equivalent in the sense that it looks the same fromthe outside, that is, it behaves the same electrically as seen by an outsideobserver connected to terminals a and b.

4.4. THEVENIN’S THEOREM 55

4.4.3. Steps to Apply Thevenin’s theorem.

S1: Find RTh: Turn off all independent sources. RTh is the equivalentresistance of the network looking between terminals a and b.

S2: Find VTh: Open the two terminals (remove the load) which youwant to find the Thevenin equivalent circuit. VTh is the open-circuit voltage across the terminals.

VTh = voc

a

Linear

two- terminal

circuit

b

voc

+

–

Linear circuit with

all independent

sources set equal

to zero

Rin

a

b

RTh = Rin

S3: Connect VTh and RTh in series to produce the Thevenin equivalentcircuit for the original circuit.

Example 4.4.4. Find the Thevenin equivalent circuits of each of thecircuits shown below, to the left of the terminals a-b.


Example 4.4.5. Find the Thevenin equivalent circuit of the circuitshown below, to the left of the terminals a-b. Then find the current throughRL = 6, 16, and 36 Ω.

12 Ω

4 Ω

2 A

a

RL

1 Ω

32 V

b

Solution:

12

4

2 A

a1

32 V

b

+

–

VTh

(b)

RTh

a

b

1 4

12

(a)

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4.4. THEVENIN’S THEOREM 57

Example 4.4.6. Determine the current I in the branch a-b in the circuitbelow.

3.6 Thévenin’s Theorem and Norton’s Theorem C H A P T E R T H R E E 163

The current I2 can be quickly determined from the network in Figure 3.66 as

I2 = vTHRTH + 10 �

.

We know that vTH = 4 V and RTH = 2 �, and so I2 = 1/3 A.

e x a m p l e 3.22 b r i d g e c i r c u i t Determine the current I in thebranch ab in the circuit in Figure 3.67.

There are many approaches that we can take to obtain the current I. For example, wecould apply the node method and determine the node voltages at nodes a and b andthereby determine the current I. However, since we are interested only in the current I,a full blown node analysis is not necessary; rather we will find the Thévenin equivalentnetwork for the subcircuit to the left of the aa′ terminal pair (Network A) and forthe subcircuit to the right of the bb′ terminal pair (Network B), and then using thesesubcircuits solve for the current I.

Let us first find the Thévenin equivalent for Network A. This network is shown inFigure 3.68a. Let vTHA and RTHA be the Thévenin parameters for this network.

We can find vTHA by measuring the open-circuit voltage at the aa′ port in the networkin Figure 3.68b. We find by inspection that

vTHA = 1 V

Notice that the 1-A current flows through each of the 1-� resistors in the loop containingthe current source, and so v1 is 1 V. Since there is no current in the resistor connectedto the a′ terminal, the voltage v2 across that resistor is 0. Thus vTHA = v1 + v2 = 1 V.We find RTHA by measuring the resistance looking into the aa′ port in the network inFigure 3.68c. The current source has been turned into an open circuit for the purpose

1 A

a

1 Ω

1 Ω

1 Ω

1 Ω

1 A

1 Ω 1 Ω 1 Ω

1 ΩI b

Sub-circuit A Sub-circuit B

a′ b′

F IGURE 3.67 Determining thecurrent in the branch ab.

There are many approaches that we can take to obtain the current I. For example, we could

apply nodal analysis and determine the node voltages at nodes a and b and thereby determine

the current I by Ohm’s law. However, here, we will find the Thvenin equivalent circuit for the

subcircuit to the left of the aa′ terminal pair (Subcircuit A) and for the subcircuit to the right

of the bb′ terminal pair (Subcircuit B), and then using these equivalent subcircuits to find the

current I.

164 C H A P T E R T H R E E n e t w o r k t h e o r e m s

F IGURE 3.68 Finding theThévenin equivalent for Network A.

1 A

a

1 Ω

1 Ω1 Ω

1 Ω

(a)

vTHA

RTHA

(b)

(c)

1 A

a

1 Ω

1Ω 1Ω1 Ω

vTHA+

-v2+ -

v1+

-

a

1 Ω

1 Ω1 Ω

1 ΩRTHAa′

a′

a′

F IGURE 3.69 Finding theThévenin equivalent for Network B.

1 A

1 Ω 1 Ω

1 Ωb

vTHB

RTHB

(b)

(c)

1 A

1 Ω 1 Ω

1 ΩbvTHB+

-

1 Ω 1 Ω

1 ΩbRTHB

(a)

b′

b′

b′

of measuring RTHA. By inspection, we find that

RTHA = 2 �.

Let us now find the Thévenin equivalent for Network B shown in Figure 3.69a. LetvTHB and RTHB be the Thévenin parameters for this network.

vTHB is the open-circuit voltage at the bb′ port in the network in Figure 3.69b. Usingreasoning similar to that for vTHA we find

vTHB = −1 V.

Prapun

Prapun

Prapun

Prapun

Prapun

Prapun

Prapun

Prapun

Prapun

Prapun

Prapun

Prapun

Prapun

Prapun


4.5. Norton’s Theorem

Norton’s Theorem gives an alternative equivalent circuit to Thevenin’sTheorem.

4.5.1. Norton’s Theorem: A circuit can be replaced by an equivalentcircuit consisting of a current source IN in parallel with a resistor RN ,where IN is the short-circuit current through the terminals and RN isthe input or equivalent resistance at the terminals when the independentsources are turned off.

Note: RN = RTH and IN =VTHRTH

. These relations are easily seen via

source transformation.4

IN

(a) (b)

a

Linear

two-terminal

circuit

b

a

b

RN

4.5.2. Steps to Apply Norton’s Theorem

S1: Find RN (in the same way we find RTH).S2: Find IN : Short circuit terminals a to b. IN is the current passing

through a and b.

a

Linear

two- terminal

circuit

b

isc = IN

S3: Connect IN and RN in parallel.

4For this reason, source transformation is often called Thevenin-Norton transformation.

Prapun

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4.5. NORTON’S THEOREM 59

Example 4.5.3. Back to the circuit in Example 4.4.5. Directly findthe Norton equivalent circuit of the circuit shown below, to the left of theterminals a-b.

12 Ω

4 Ω

2 A

a

RL

1 Ω

32 V

b

Remark: In our class, to “directly find the Norton equivalent circuit”means to follow the steps given in 4.5.2. Similarly, to “directly find theThevenin equivalent circuit” means to follow the steps given in 4.4.3.

Suppose there is no requirement to use the direct technique, then it isquite easy to find the Norton equivalent circuit from the derived Theveninequivalent circuit in Example 4.4.5.

Prapun

Prapun

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Prapun

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Prapun


Example 4.5.4. Directly find the Norton equivalent circuit of the fol-lowing circuit at terminals a-b.

R

a

IS

b

VS

Example 4.5.5. Directly find the Norton equivalent circuit of the circuitin the following figure at terminals a-b.

4 Ω

8 Ω

a8 Ω

2 A

b

12 V

5 Ω

4.6. MAXIMUM POWER TRANSFER 61

4.6. Maximum Power Transfer

In many practical situations, a circuit is designed to provide power toa load. In areas such as communications, it is desirable to maximize thepower delivered to a load. We now address the problem of delivering themaximum power to a load when given a system with known internal losses.

4.6.1. Questions:

(a) How much power can be transferred to the load under the most idealconditions?

(b) What is the value of the load resistance that will absorb the maxi-mum power from the source?

4.6.2. If the entire circuit is replaced by its Thevenin equivalent exceptfor the load, as shown below, the power delivered to the load resistor RLis

p = i2RL where i =Vth

Rth +RL

RTh a

RLVTh

b

i

The derivative of p with respect to RL is given by

dp

dRL= 2i

di

dRLRL + i

2 = 2Vth

Rth +RL

(− Vth

(Rth +RL)2

)+

(Vth

Rth +RL

)2=

(Vth

Rth +RL

)2(− 2RLRth +RL

+ 1

).

We then set this derivative equal to zero and get

RL = RTH .


4.6.3. Maximum power transfer occurs when the load resistance RLequals the Thevenin resistance RTh. The corresponding maximum powertransferred to the load RL equals to

pmax =

(Vth

Rth +Rth

)2Rth =

V 2th4Rth

.

Example 4.6.4. Connect a load resistor RL across the circuit in Exam-ple 4.4.4. Assume that R1 = R2 = 14Ω, Vs = 56V, and Is = 2A. Find thevalue of RL for maximum power transfer and the corresponding maximumpower.

Example 4.6.5. Find the value of RL for maximum power transfer inthe circuit below. Find the corresponding maximum power.

12 Ω

6 Ω

2 A

a

RL

2 Ω

12 V

b

3 Ω

ECS203 2015 1 u3ECS203 2015 2 u6ECS203 2015 3 u3ECS203 2015 4 u5

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