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THEVENIN EQUIVALENT CIRCUIT

Thevenin equivalent circuit consist of anindependent voltage source, VTh in serieswith a resistor RTh.

Thevenin equivalent circuit

ThV

ThR

a

b

Thevenin voltage, VTh = Open circuit voltage inthe original circuit.

Thevenin resistance, RTh is the ratio of open-circuit voltage to the short-circuit current.

sc

ThTh

Th

Thsc

i

VR

R

Vi

Example

V25

5

a

b

20

4

A3

1V

abV

Step 1

Node-voltage equation for open-circuit

ThVVV

VV

32

03205

25

1

11

Step 2

Short-circuit condition at terminal a-b.

V25

5 a

b

20

4

A3

2V scI

Node-voltage equation for short-circuit

VV

VVV

16

04

3205

25

2

222

84

32

sc

ThTh

I

VR

AI sc 44

16

Short-circuit current

Thevenin resistance

Thevenin equivalent circuit

V32

8

a

b

NORTON EQUIVALENT CIRCUIT

A Norton equivalent circuit consists of anindependent current source in parallel with theNorton equivalent resistance.

Can be derive from a Thevenin equivalentcircuit simply by making a sourcetransformation.

Norton current, IN = the short-circuit current atthe terminal of interest.

Norton resistance, RN = Thevenin resistance,RTh

Example

V25

5

a

b

20

4

A3

Step 1Source transformation.

A5 5

a

b

20

4

A3

Step 2Parallel sources and parallel resistors combined.

a

b

4

A8 4

Step 3Source transformation, series resistorscombined, producing the Thevenin equivalentcircuit.

V32

8

a

b

Step 4Source transformation, producing the Nortonequivalent circuit.

a

b

A4 8

Example

Step 1

Deactivated all sources except voltage source

V0 is calculated using voltage divider

Vk

kV 54

1020

Step 2

Deactivated all sources except current source

V0 is calculated by using current divider

VkmV

mAmk

ki

2)2)(1(

1)2(4

2

0

0

V0 =2+5=7V.

Step 3

Sum all the resulting voltages:

Question For Thevenin

Open-circuit voltage, Voc

Node-voltage equation for Voc

VV

V

VV

VV

oc

oc

ococ

ococ

10

202

0424

0222

24

Thevenin resistance, RTh

5422THR

Thevenin equivalent circuit

VV 88.6)10(16

110

Question For Norton

Open-circuit current, Isc

AIsc 64

123

Norton resistance, RN

RN = 4Ω

Norton equivalent circuit

VV 18)3(612460

PLT 121

CHAPTER 2

AC Circuit Analysis

• Sinusoids’ features

• Phasors

• Phasor relationships for circuit elements

• Impedance and admittance

• Kirchhoff’s laws in the frequency domain

• Impedance combinations

SINUSOIDAL STEADY-STATE ANALYSIS SINUSOIDAL AND PHASOR

• Basic Approach

• Nodal Analysis

• Mesh Analysis

• Thevenin Equivalent Circuits

• A sinusoid is a signal that has the form ofthe sine or cosine function.

• A general expression for the sinusoid,

whereVm = the amplitude of the sinusoidω = the angular frequency in radians/sФ = the phase

Sinusoids

)sin()( tVtv m

A periodic function is one that satisfiesv(t) = v(t + nT), for all t and for all integers n.

2T Hz

Tf

1

f 2

• Only two sinusoidal values with the samefrequency can be compared by theiramplitude and phase difference.

• If phase difference is zero, they are inphase; if phase difference is not zero, theyare out of phase.

Example 1

Given a sinusoid, , calculate itsamplitude, phase, angular frequency, period,and frequency.

SolutionAmplitude = 5, phase = –60o, angular frequency = 4 rad/s, Period = 0.5 s, frequency = 2 Hz.

)604sin(5 ot

Example 2

Find the phase angle between and ,

does i1 lead or lag i2?

)25377sin(41

oti )40377cos(52

oti

Solution

Since sin(ωt+90o) = cos ωt

therefore, i1 leads i2 155o.

)50377sin(5)9040377sin(52

ooo tti

)205377sin(4)25180377sin(4)25377sin(41

oooo ttti

PHASORS

• A phasor is a complex number that represents the amplitude and phase of a sinusoid.

• It can be represented in one of the following three forms:

rzjrez

)sin(cos jrjyxza.Rectangular

b.Polar

c.Exponential

22 yxr

x

y1tanwhere

Example 3

• Evaluate the following complex numbers:

a.

b.

Solution

a. –15.5 + j13.67

b. 8.293 + j2.2

]605j4)1j2)([(5 o

oo

3010j43

403j510

)()( 212121 yyjxxzz )()( 212121 yyjxxzz

212121 rrzz

21

2

1

2

1 r

r

z

z

rz

11

2 rz

jrerjyxz

sincos je j

Mathematic operation of complex number

• Addition

• Subtraction

• Multiplication

• Division

• Reciprocal

• Square root

• Complex conjugate

• Euler’s identity

• Transform a sinusoid to and from the time domain to the phasor domain:

(time domain) (phasor domain)

)cos()( tVtv m mVV

• Amplitude and phase difference are two principal concerns in the study of voltage and current sinusoids.

• Phasor will be defined from the cosine function in all our proceeding study. If a voltage or current expression is in the form of a sine, it will be changed to a cosine by subtracting from the phase.

Example 4

Transform the following sinusoids to phasors:

i = 6cos(50t – 40o) A

v = –4sin(30t + 50o) V

Solution

a. I A

b. Since –sin(A) = cos(A+90o);

v(t) = 4cos(30t+50o+90o) = 4cos(30t+140o)V

Transform to phasor => V V

406

1404

Example 5

Transform the sinusoids corresponding to phasors:

a.

b.

V 3010 V

A j12) j(5 I

Solution

a) v(t) = 10cos(t + 210o) V

b) Since

i(t) = 13cos(t + 22.62o) A

22.62 13 )12

5( tan 512 j512 122

I

The differences between v(t) and V:

• v(t) is instantaneous or time-domainrepresentation. V is the frequency orphasor-domain representation.

• v(t) is time dependent, V is not.

• v(t) is always real with no complexterm, V is generally complex.

Note: Phasor analysis applies only when

frequency is constant; when it is

applied to two or more sinusoid

signals only if they have the same

frequency.

Relationship between differential, integral

operation in phasor listed as follow:

)(tv

dt

dvVj

vdtj

V

VV

Example 6

Use phasor approach, determine thecurrent i(t) in a circuit described bythe integro-differential equation.

Answer: i(t) = 4.642 cos(2t + 143.2o) A

)752cos(50384 tdt

diidti

Phasors Relationship for circuit elements

Resistor Inductor Capacitor

Summary of voltage-current relationship

Element Time domain Frequency

domain

R

L

C

Riv RIV

dt

diLv LIjV

dt

dvCi Cj

IV

Example 7

If voltage v(t) = 6 cos(100t – 30o) is applied to

a 50 μF capacitor, calculate the current, i(t),

through the capacitor.

Answer: i(t) = 30 cos(100t + 60o) mA

IMPEDANCE AND ADMITTANCE

• The impedance, Z of a circuit is the ratio ofthe phasor voltage V to the phasor currentI, measured in ohms Ω.

• where R = Re, Z is the resistance andX = Im, Z is the reactance. Positive X isfor L and negative X is for C.

• The admittance, Y is the reciprocal ofimpedance, measured in siemens (S).

jXRI

VZ

V

I

ZY

1

RY

1

LjY

1

CjY

Impedances and admittances of passive elements

Element Impedance Admittance

R

L

C

RZ

LjZ

CjZ

1

LjZ

CjZ

1

Z

Z

;

0;0

0;

;0

Z

Z

Example 8

Refer to Figure below, determine v(t) and i(t).

Answers: v(t) = 2.236sin(10t + 63.43o) V

i(t) = 1.118sin(10t - 26.57o) A

)10sin(5 tvs

Kirchhoff’s Law in the Frequency Domain

• Both KVL and KCL are hold in the phasor

domain or more commonly called frequency

domain.

• Moreover, the variables to be handled are

phasors, which are complex numbers.

• All the mathematical operations involved

are now in complex domain.

IMPEDANCE COMBINATIONS

• The following principles usedfor DC circuit analysis all applyto AC circuit.

• For example:– voltage division

– current division

– circuit reduction

– impedance equivalence

Example 9

Determine the input impedance of the circuit in

figure below at ω =10 rad/s.

Answer: Zin = 32.38 – j73.76

• Basic Approach

• Nodal Analysis

• Mesh Analysis

• Thevenin Equivalent Circuits

BASIC APPROACH

1. Transform the circuit to the phasor or

frequency domain.

2. Solve the problem using circuit techniques

(nodal analysis, mesh analysis, etc.).

3. Transform the resulting phasor to the time

domain.

Time to Freq Solve

variables in Freq

Freq to Time

Steps to Analyze AC Circuits:

NODAL ANALYSIS

Exercise 1

Calculate V1 and V2 in the circuit shown in

figure below .

Answer

V1 = 19.3669.67 V V2 = 3.376165.7 V

MESH ANALYSIS

Exercise 2

Find Io in the following figure using mesh

analysis.

Answer : Io = 1.19465.44 A

THEVENIN EQUIVALENT CIRCUITS

Thevenin transform

Answer : Zth =12.4 – j3.2

VTH = 18.97-51.57 V

Find the Thevenin equivalent at terminalsa–b of the circuit below.

Assignment 1.3