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 1

4-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. 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

2.2.9. 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 2.2.10. The primary signals within a circuit are its currentsand 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.

i5

i4

i3

i2

i1

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

2.3. KIRCHHOFF’S LAWS 23

Example 2.3.2. 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)

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.

v1

+ – v2 v3+ –

v4

v5 + –

24 2. BASIC LAWS

Example 2.3.3. 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

Example 2.3.4. Find v1 and v2 in the following circuit.

10 V

+ – v2

v1+ –

8 V

4 Ω

2 Ω

Example 2.3.5.

2.3. KIRCHHOFF’S LAWS 25

Example 2.3.6 (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

j1

1Rj

Req n

j1 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 et/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 q0et/RC (discharging a capacitor). (27-39)

During the discharging, the current is

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

q0

RC et/RC

i dqdt

R et/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.)

Fig. 27-19 Question 4.

+ –

R1 R2

a b c d e

x

+–

+– +–

(a) (b) (c)

Fig. 27-20 Question 5.

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. Series Resistors and Voltage Division

2.4.1. When two resistors R1 and R2 ohms are connected in series, theycan be replaced by an equivalent resistor Req where

Req = R1 +R2

.In particular, the two resistors in series shown in the following circuit

v

+ – v1 v2+ –

a

b

R1 R2i

can be replaced by an equivalent resistor Req where Req = R1+R2 as shownbelow.

v

+ – v

a

b

Reqi

The two circuits above are equivalent in the sense that they exhibit thesame voltage-current relationships at the terminals a-b.

Voltage Divider: If R1 and R2 are connected in series with a voltagesource v volts, the voltage drops across R1 and R2 are

v1 =R1

R1 +R2v and v2 =

R2

R1 +R2v

2.5. PARALLEL RESISTORS AND CURRENT DIVISION 27

Note: The source voltage v is divided among the resistors in direct propor-tion to their resistances.

2.4.2. In general, for N resistors whose values are R1, R2, . . . , RN ohmsconnected in series, they can be replaced by an equivalent resistor Req

where

Req = R1 +R2 + · · ·+RN =N∑j=1

Rj

If a circuit has N resistors in series with a voltage source v, the jthresistor Rj has a voltage drop of

vj =Rj

R1 +R2 + · · ·+RNv

2.5. Parallel Resistors and Current Division

When two resistors R1 and R2 ohms are connected in parallel, they canbe replaced by an equivalent resistor Req where

1

Req=

1

R1+

1

R2

or

Req = R1‖R2 =R1R2

R1 +R2

v

i1

R1

i

i2

R2

Node a

Node b

Current Divider: If R1 and R2 are connected in parallel with a currentsource i, the current passing through R1 and R2 are

i1 =R2

R1 +R2i and i2 =

R1

R1 +R2i

Note: The source current i is divided among the resistors in inverse pro-portion to their resistances.

28 2. BASIC LAWS

Example 2.5.1.

Example 2.5.2. 6‖3 =

Example 2.5.3. (a)‖(na) =

Example 2.5.4. (ma)‖(na) =

In general, for N resistors connected in parallel, the equivalent resistorReq = R1‖R2‖ · · · ‖RN is

1

Req=

1

R1+

1

R2+ · · ·+ 1

RN

Example 2.5.5. Find Req for the following circuit.

Req

4 Ω

8 Ω

1 Ω

5 Ω

2 Ω

3 Ω 6 Ω

Example 2.5.6. Find the equivalent resistance of the following circuit.

2.5. PARALLEL RESISTORS AND CURRENT DIVISION 29

12 V

i a

b

+

–

v0 3 Ω 6 Ω

4 Ω i0

Example 2.5.7. Find io, vo, po (power dissipated in the 3Ω resistor).

Example 2.5.8. 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

30 2. BASIC LAWS

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.