Chapter 2

CHAPTER 2 Basic Laws

To determine the values of electrical

variables such as current, voltage and

power in a given circuit requires

understanding some fundamental laws for

examples; Ohm’s law and Kirchhoff’s laws

to analyze the circuit. In addition, some

techniques must be used together with those fundamental laws.

2.2 Ohm’s law

Materials in general have a characteristic behavior of resisting the flow of electric charge which is known as resistance (R). The resistance of any material depends on cross – sectional area A and its length ll

AMaterial with resistivity

R = Al (Ohm)

– resistivity of materials (ohm – meters)Good conductors low resistivity

insulators high resistivitySeeTable2.1

The circuit element used to model the current -resisting behavior of a material is the resistor

Resistance of the resistor

Ohm’s law states that the voltage V across a resistor is directly proportional to the current i flowing through the resistor

iRviv

Note : AVΩ /11

If current flows from a higher potential to a lower potential

iRv

If current flows from a lower potential to a higher potential

iRv

perfect conductor

short - circuit

R0

open - circuit

resistorfixed

variable

Wire wound

composition large resistance

Linear resistor obey ohm’s law

non - linear resistor does not obey

The reciprocal of resistance R, known as conductance and denoted by G

G = IR = i

vmho

Or Siemens (S)

G

iGvpGvi

22,

Example 2.2 In the circuit shown in Fig.2.8, calculate the

current i , the conductance G, and the power p.

30V k50

+

-

v

i

Solution: The Voltage across the resistor is the same as the

source voltage (30 V) because the resistor and the voltage

source are connected to the same pair of terminals. Hence,

the current is

mAxR

vi 6

105

303

mSxR

G 2.0105

113

mWxvip 180)106(30 3

mWxxxRip 180105)106( 3232

mWxxGvp 180102.0)30( 322

The conductance is

tsin20

tmAx

t

R

vi

sin4

105

sin203

tmWvip 2sin80

Example 2.3 A voltage source of

Hence,

V is connected across a 5-kΩ resistor. Find the current through theresistor and the power dissipated.Solution:

2.3 Nodes, Branches and Loops

A branch represents a single element such as a voltage source or a resistor

A Node is the point of connection between two or more branches

A Loop is any closed path in the circuit formed by starting at a node, passing through a set of nodes, and returning to starting node without passing through any node more than once

Two or more elements are in series if they exclusively share a single node and consequent by carry the same current

Two or more elements are in parallel if they exclusively connected to the same two nodes and consequent by have the same voltage across them

Example 2.4 Determine the number of branches and node in the circuit show in Fig.2.12. Identify which elements are in series and which are in parallel.Solution: Since there are four elements in the circuit, the circuit has four branches: 10 V, 5 Ω, 6 Ω, and 2 A. The circuit has three nodes as identified in Fig. 2.13. The 5 Ωresistor is in series with the 10-V voltage source because the same current would flow in both. The 6-Ω resistor is in parallel with the 2-A current source because both are connected to the same nodes 2 and 3.

DC 2A6

5

DC 2A6

51 2

3

2.4 Kirchhoff’s laws

Kirchhoff’s current law (KCL) states that the algebraic sum of currents entering a node is zero

01

N

nni

0)()( 54321 iiiii

52431 iiiii

entering leaving

A simple application of KCL is combining current sources in parallel. The combined or equivalent current source can be found by applying KCL to node a

IT = I1-I2+I3

IS = I1-I2+I3

ITa

b

I1 I2 I3

ITa

b

Kirchhoff’s voltage law (KVL) states that the algebraic sum of all voltages around a loop is zero

-v1+v2+v3-v4+v5 = 0

v2+v3+v5 = v1+v4

When voltage sources are connected in series, KVL can be applied to obtain the total voltage

Example 2.5 For the circuit in Fig.2.21 (a), find voltage v1 and v2.

20V

+ -

2

3+

-

1v

2v 20V

+ -

2

3+

-

1v

2vi

Solution: To find v1 and v2, we apply Ohm’s law and Kirchhoff’s voltage law. Assume that current i flow through the loop as shown in Fig. 2.21(b). From Ohm’s law,

v1 =2i , v2=-3i (2.5.1)

Applying KVL around the loop gives

-20+v1-v2=0 (2.5.2)

Substituting Eq. (2.5.1) into Eq. (2.5.2), we obtain

-20+2i+3i=0 or 5i=20

i =4 A

Substituting i in Eq. (2.5.1) finally gives

v1= 8 V , v2=-12 V

Example 2.6 Determine v0 and i in the circuit shown in Fig 2.23(a)+ -

4

12V 4V

+

6

ov -

ov2i+ -

4

12V 4V

+

6

ov -

ov2i

i

Solution: We apply KVL around the loop as shown in

Fig.2.23(b). The result is

-12+4i+2v0-4+6i = 0 (2.6.1)

Applying Ohm’s law to the 6-Ω resistor gives

v0 = -6i (2.6.2)

Substituting Eq.(2.6.2) into Eq.(2.6.1) yields

-16+10i-12i = 0

i = -8 A

and v0 = 48 V.

Example 2.8 Find the current and voltage in the circuit show in Fig. 2.27(a)

30V+

-

+ -

8

1v

32v+

-6

3v

a1i

2i

3i

20V+

-

+ -

8

1v

32v+

-6

3v

a1i

2i

3i

Loop1 Loop2

Solution: We apply Ohm’s law and Kirchhoff’s law. By Ohm’s law,

v1 = 8i1, v2=3i2, v3 = 6i3 (2.8.1)

Since the voltage and current of each resistor are related by Ohm’s law as show, we are really looking for three thing: (v1,

v2, v3) or (i1, i2, i3). At node a, KCL gives

i1- i2- i3 = 0 (2.8.2)

Applying KVL to loop as in Fig. 2.27(b),

-30 + v1 + v2 = 0We express this in terms of i1 and i2 as in Fig. (2.81) to obtain

-30 + 8i1 + 3i2 = 0

8

i330i 21

(2.8.3)

Applying KVL to loop 2,

-v2 + v3 = 0

v3 = v2

(2.8.4)

as expected since the two resistors are in parallel. We express v1 and v2 in term of i1 and i2 as in Eq. (2.8.1).Eq. (2.8.4) becomes

6 i3 = 3 i2

2

23

ii

Substituting Eqs. (2.8.3) and (2.8.5) into (2.8.2) gives

028

330 22

2 i

ii

or i2 = 2 A. From the value of i2 , we now use Eqs. (2.8.1) to (2.8.5) to obtain

i1 = 3 A, i3 = 1 A, v1 = 24 V, v2 = 6 V, v3 = 6 V.

2.5 Series Resistors and Voltage division

v1 = iR1 v2 = iR2

-v+v1+v2 = 0

v = v1+v2 = i (R1+R2)

v = i Req

Req = R1 + R2. ..

For N resistors in series then,

Req = R1 + R2+…RN = RnN

n=1

To determine the voltage across each resistor

v1viR1

i (R1+R2)= v1 R1+R2

R1 v=,v2 = R1+R2

R2 v

Vn = RnR1+R2+…+Rn

V

2.6 Parallel resistors and current divisionv = i1R1 = i2R2

i1 = R1v i2 = R2

v,at node a ; i = i1+ i2

1 Req

= R11

R21+

R1

R1R2

R2+=1

Req

i R1v

R2v+ v

R11

R21+ == v

Req=

R1+ R2

R1 R2=Req. . .Note : R1 = R2 then Req = R1 /2 ส ำหรบ R 2 ตวตอ ขนำน

For N resistors 1 Re

q

= R1

1 + R2

1RN

1+ … +

If R = R1 = R2 = … = RN Req = RN

It is more often to use conductance rather than resistance when dealing with resistors in parallel.

Geq = G1 + G2 + G3 + … + GN

Given the total current i enter node a, how do use obtain current i1 and i2

v = i Req = i R1R2R1+R2

i1= R1

vR1

= R1R2R1+R2

i iR2R1+R2

=

=i2 iR1R1+R2

Current divider

Note : larger current flow through the smaller resistance

Example 2.9 Find Req for the circuit shown in Fig. 2.34.4 1

2

368

5

eqR

Solution: To get Req , we combine resistors in series and in parallel. The 6-Ω and 3-Ω resistors are in parallel, so equivalent resistance is

236

363//6

x

The 1-Ω and 5-Ω resistors are in series; hence their equivalent resistance is

1Ω+5Ω = 6Ω

Thus the circuit in Fig. 2.34 is reduced to that in Fig. 2.35(a). In Fig. 2.35(a), we notice that the two 2-Ω resistors are in series, so the equivalent resistance is

2Ω + 2Ω = 4Ω

This 4-Ω resistor is now in parallel with the 6-Ω resistor in Fig. 2.35(a); their equivalent resistance is

4.264

646//4

x

The circuit in Fig. 2.35(a) is now replaced with that in Fig. 2.35(b). In Fig. 2.35(b), the three resistors are in series. Hence, the equivalent resistance for the circuit is

4.1484.24R eq

4

2

2

8

6eqR

4

8

4.2eqR

Example 2.10 Calculate the equivalent resistance Rab in the circuit in Fig. 2.37.

10

3eqR

1

4

1

5

a

b

c d

b b

12

6

Solution: The 3-Ω and 6-Ω resistors are in parallel because they are connected to the same two nodes c and b. Their combined resistance is

263

636//3

x

Similarly, the 12-Ω and 4- Ω resistors are in parallel since they are connected to the same two nodes d and b. Hence

3412

4124//12

x

Also the 1-Ω and 5-Ω resistors are in series; hence, their equivalent resistance is

651

With these three combinations, we can replace the circuit in Fig. 2.37 with that in Fig. 2.38(a). In Fig.2.38(a),3-Ω in parallel with 6-Ω gives 2-Ω, as calculated in Eq. (2.10.1). This 2-Ω

equivalent resistance is now in series with the 1-Ω resistance to give a combined resistance of 1Ω + 2Ω =3Ω . Thus, we replace the circuit in Fig. 2.38(a) with that in Fig. 2.38(b). In Fig. 2.38(b), we combine the 2-Ω and 3-Ω resistors in parallel to get

2.132

323//2

x

This 1.2-Ω resistor is in series with the 10-Ω resistor, so that

Rab = 10 + 1.2 = 11.2 Ω

10

2eqR

1

3 6

a

b

c d

b b b

10

2eqR

3

a

b

c

b b

Example 2.12 Find iO and vO in the circuit shown in Fig. 2.42(a). Calculate the power dissipated in the 3-Ω resistor.

Solution: The 6-Ω and 3-Ω resistors are parallel, so their combined resistance is

236

363//6

x

Thus our circuit reduces to that shown in Fig. 2.42(b). Notice that vO is not affected by the combination of the resistors because the resistors are in parallel and therefore have the same voltage vO. From Fig. 2.42(b), we can obtain vO in two ways. One way is to apply Ohm’s law to get

Ai 224

12

and hence, vO = 2i = 2x2 = 4 V. Another way is to apply voltage division, since the 12 V in Fig. 2.42(b) is divided between the 4-Ωand 2-Ω resistors. Hence,

VVvO 4)12(42

2

Similarly, iO can be obtained in two way. One approach is to apply Ohm’s law to the 3-Ω resistor in Fig. 2.42(a) now that we know vO; thus,

43 OO iv

AiO3

4

Another approach is to apply current division to the circuit in Fig. 2.42(a) now that we know I, by writing

AAiiO3

4)2(

3

2

36

6

The power dissipated in the 3-Ω resistor is

Wivp OOO 333.53

44

12V

4

6+

-3

ov

ai oi

b

12V

4

2+

-ov

ai

b

Example 2.13 For the circuit shown in Fig. 2.44(a), determine: (a) the voltage vO, (b) the power supplied by the current source, (c) the power absorbed by each resistor.

k6

k9 k12

+

-ov30mA k9 k18

+

-ov30mA

oi

1i

2i

Solution: (a) The 6-kΩ and 12-kΩ resistors are in series so that their combined value is 6 + 12 = 18 kΩ . Thus the circuit in Fig. 2.44(a) reduces to that shown in Fig. 2.44(b). We now apply the current division technique to find i1 and i2.

mAmAi 20)30(000,18000,9

000,181

mAmAi 10)30(000,18000,9

000,92

Notice that the voltage across the 9-kΩ and 18-kΩ

resistors are the same, and vO = 9,000i1 = 18,000i2 = 180 V, as expected.

(b) Power supplied by the source isWmWivp OOO 4.5)30(180

(c) Power absorbed by the 12-kΩ resistor isWxRiRiiivp 2.1)000,12()1010()( 232

222

Power absorbed by the 6-kΩ resistor isWxRip 6.0)000,6()1010( 232

2

Power absorbed by the 9-kΩ resistor is

WR

vp O 6.3

000,9

)180( 22

WmWivp O 6.3)20(1801

Notices that the power supplied (5.4W) equals the power absorbed (1.2 + 0.6 + 3.6) = 5.4 W). This is one way of checking results.

2.7 Wye – Delta transformations

When the resistors are neither in parallel nor in series. For example, the bridge circuit, this circuit can be simplified by using three – terminal equivalent networks, wye (y) and delta ( ) as will be shown in Ex.2.15.

Delta to Wye conversion Wye to delta

R1 = Ra = R1R2 + R2R3 + R3R1

R1

Rc = R1R2 + R2R3 + R3R1

R3

Rb = R1R2 + R2R3 + R3R1

R2

Y and balancewhen

R1 = R2 = R3 = Ry

Ra = Rb = Rc = R

Ry = R or R = 3Ry3

Rb Rc

Ra+Rb+Rc

Rc Ra

Ra+Rb+Rc

Ra Rb

Ra+Rb+Rc

R2 =

R3 =

Example 2.14 Convert the network in Fig. 2.50(a) to an

equivalent Y network.

25

a b

c

10 15aRbR

cRa b

c

5

1R5.7

3

2R

3R

Solution: Using Eqs. (2.49) to (2.51), we obtain

5050

250

151025

10251

x

RRR

RRR

cba

cb

5.750

15252

x

RRR

RRR

cba

ac

350

10153

x

RRR

RRR

cba

ba

The equivalent Y network is shown in Fig. 2.50(b).

Example 2.15 Obtain the equivalent resistance Rab for the circuit in Fig. 2.52 and use it to find current i.

a

c

5.12

15

10

20

120V

b

30n

a

b

5

i

Solution: In this circuit, there are two Y network and one

network. Transforming just one of these will simplify the circuit.

If we convert the Y network comprising the 5-Ω, and 20-Ω

resistors, we may select

101R 202R 53R

Thus from Eqs. (2.53) to (2.55) we have

10

1055202010

1

133221 xxx

R

RRRRRRRa

3510

350

5.1720

350

2

133221

R

RRRRRRRb

705

350

3

133221

R

RRRRRRRc

pairs of resistors in parallel, we obtain

213070

30x7030//70

2917.75.175.12

5.175.125.17//5.12

x

5.103515

351535//15

x

so that the equivalent circuit is shown in Fig. 2.53(b). Hence, we find

632.921292.17

21292.1721//)5.10292.7(

xRab

AR

vi

ab

s 458.12632.9

120

a

c

5.12

15

5.17

b

30

a

b

35

70

5.10

21

b

292.7

a

2.8 Application

Lighting systems, such as in a house, often consist of N lamps connected either in parallel or in series

2.8.1 Lighting systems

Assuming that all the lamps are identical and v0 is the power – line voltage, the voltage across each lamp is v0 for parallel connection and v0/N for series connection. The series connection is easy to manufacture but is seldom used in practice for two reasons. First, it is less reliable. Second, it is harder to maintain.

Potentiometer is a three – terminal device that operates on the principle of voltage division. It is essentially an adjustable voltage divider. As a voltage regulator, it is used as a volume or level control on radios, TVs and other devices.

2.8.2 Design of DC Meters

Potentiometer controlling potential levels

Vout = Vbc = VinRbcRac

Where Rac = Rab+Rbc

Thus, vout decreases or increases as the sliding contact of the pot moves toward c or a, respectively.

Another application where resistors are used to control current flow is the analog dc meters, ammeter, voltmeter and ohmmeter. Each of these meter employs the d’Arsonval meter movement.

The movement consists of a movable iron – core coil mounted on a pivot between the poles of a permanent magnet. When current flows through the coil, it creates torque which causes the pointer to deflect. The amount of current through the coil determines the deflection of the pointer.

VoltmeterIt measures the voltage across a load and is connected in

parallel with the element. It consists of a d’Arsonval movement in series with a resistor whose resistance Rm is deliberately made very large to minimize the current drawn from the circuit.

m

fs

fs

n

nmfsfs

RI

VR

RRIV

AmmeterIt measures the current through the load and is connected in

series with it. It consists of a d’ Arsonaval movement in parallel with a resistor whose resistance Rm is deliberately made very small to minimize the voltage drop across it.

mmfs

mn

fsmn

nm

RII

IR

IRR

RI

Ohmmeter

It consists of d’Arsonval movement, a variable resistor and a battery

)1.....(....................mm

x

mxm

RRI

ER

IRRRE

The resistor R is selected such that the meter gives a full-scale deflection when Rx = 0fsm II

)2........(....................fsm IRRE

Substitute eq. (2) in (1)

mm

fsx RR

I

IR

1

Example 2.16 Three light bulbs are connected to a 9-V battery as shown in Fig. 2.56(a). Calculate: (a) the total current supplied by the battery, (b) the current through each bulb, (c) the resistance of each bulb.

9V

15W

10W

20W 9V

+

-

+

-

+

-1v

2v

3v

1R

2R

3R

I 1I

2I

Solution: (a) The total power supplied by battery is equal to the total power absorbed by the bulbs, that is,

p = 15 + 10 + 20 = 45 W

Since p = VI, then the total current supplied by the battery is

AV

pI 5

9

45

(b) The bulbs can be modeled as resistors as shown in Fig. 2.56(b). Since R1 (20-W bulb) is in parallel with the battery as

the series combination of R2 and R3,

V1 = V2 + V3 = 9 V

The current through R1 is

AV

pI 222.2

9

20

1

11

By KCL, the current through the series combination of R2 and R3 is AIII 778.2222.2512

(c) Since p = I2R,

05.4222.2

2022

1

11

I

pR

945.1777.2

1522

2

22

I

pR

297.1777.2

1022

3

3

3I

pR

Example 2.17 Following the voltmeter setup of Fig. 2.60, design a voltmeter for the following multiple ranges:

(a) 0-1 V (b) 0-5 V (c) 0-50 V (d) 0-100 V

Assume that the internal resistance Rm = 2 kΩ and the full scale current .100 AI fs

Solution: We apply Eq. (2.60) and assume that R1, R2, R3, and R4 correspond with ranges 0-1 V, 0-5 V, 0-50 V, and 0-100

V, respectively.

For range 0-1 V,

kx

R 82000000,10200010100

161

For range 0-5 V,

kx

R 482000000,50200010100

562

For range 0-50 V,

kx

R 4982000000,500200010100

5063

For range 0-100 V,

kx

R 9982000000,000,1200010100

10064

2.9 Summary1. A resistor is a passive element in which the voltage

v across it is proportional to the current i through it

iRv ;R is the resistance of the resistor

Open circuit is a resistor with infinite resistance(R =∞)2. Short circuit is a resistor with zero resistance(R = 0)

3. The conductor G of a resistor is the reciprocal of its resistance

RG

1

4. A branch is a single two-terminal element in an electric circuitA node is the point of connection between two or more branchesA loop is a closed path in a circuit

5. KCL states that the sum of the currents entering a node equals the sum of currents leaving the node

6. KVL states that the voltage around a closed path algebraically sum to zero

Elements are in parallel if they have the same voltage across them

8. When two resistors are in series

7. Elements are in series if the same current flows through them

21 RRReq 21

21

GG

GGGeq

9. The voltage division principle for two resistors in series is

vRR

Rvv

RR

Rv

21

22

21

11 ,

10. When two resistors R1 and R2 are in parallel

2121

21 , GGGRR

RRR eqeq

11. The current division principle for two resistors in parallel is

iRR

Rii

RR

Ri

21

12

21

21 ,

12. Delta – to – Wye conversion

cba

ba

cba

ac

cba

cb

RRR

RRR

RRR

RRR

RRR

RRR

321 ,,

13. Wye – to – Delta conversion

3

133221

2

133221

1

133221

R

RRRRRRR

R

RRRRRRR

R

RRRRRRR

c

b

a