Chap3_bekg1113_part1

CHAPTER 3: ALTERNATING CURRENT (AC) CIRCUITS PART I 1

INTRODUCTION • This chapter will focus on circuit analysis with varying source voltage or current(sinusoidally). • A sinusoid is a signal that has the form of the sine or cosine function. • A sinusoidal current is usually referred to as alternating current (ac).

2

INTRODUCTION contd.

• AC is an electrical current whose magnitude and direction vary sinusoidally with time. • Such a current reverses at regular time intervals and has alternately positive and negative values.

3

INTRODUCTION contd.

• The circuits analysis is considering the timevarying voltage source or current source. • Circuits driven by sinusoidal current or voltage sources are called ac circuits. • A sinusoid can be express in either sine or cosine form.

4

INTRODUCTION contd.

GENERATING AC VOLTAGE • One way to generate an ac voltage is to rotate a coil of wire at constant angular velocity in a uniform magnetic field. • The magnitude of the resulting voltage is proportional to the rate at which flux lines are cut.

5

INTRODUCTION contd.

6

WAVEFORM TERMS & DEFINITIONS • The definitions for all terms are stated as below, – Period – the time taken to complete a cycle, T (s) – Peak value – the maximum instantaneous value measured from its zero value, Vp @ Vm (V) – Peak-to-peak value – the maximum variation between the maximum positive instantaneous value and the maximum negative value, Vp-p (V)

7

WAVEFORM TERMS & DEFINITIONS contd.

8

SINUSOIDS • Consider the expression of a sinusoidal voltage

v (t )  Vm sin t where

Vm = the amplitude of the sinusoid

 = the angular frequency in radians/s t = the argument of the sinusoid

9

SINUSOIDS contd.

• The sinusoid repeats itself every T seconds, thus T is called the period of the sinusoid or the time taken to complete one cycle. (s) 10

SINUSOIDS contd.

• The number of cycles per second is called frequency, f. (Hz)

1 f  T • Angular frequency, ω. (rad/s)

  2 f • An important value of the sinusoidal function is its RMS (root-mean-square) value.

VRMS 

Vm 2

 Vdc 11

SINUSOIDS contd.

• Note: Radian measure – – – –

ω is usually expressed in radian/s 2π radians = 360° to convert from degrees to radians, multiply by π/180°. to convert from radians to degrees, multiply by 180°/π.

• From the general expression of the sinusoidal voltage, we can find the value of voltage at any given instant of time.

12

SINUSOIDS contd.

• If the waveform does not pass through zero at t=0, it has a phase shift. • For a waveform shifted left,

v (t )  Vm sin  t    • For waveform shifted right,

v (t )  Vm sin  t    where

 = phase angle of the sinusoid function

13

SINUSOIDS contd.

14

SINUSOIDS contd.

Example: 2. Find the amplitude, phase, period and frequency of the sinusoid

v (t )  12sin  50t  10  Solution: Amplitude, Vm Phase,  Angular frequency, ω

= 12V = 10˚ = 50rad/s

2 2   0.1257s thus the period, T =  50 The frequency, f

1  7.958Hz = T

15

SINUSOIDS contd.

•

A sinusoidal voltage is given by the expression V = 300 cos (120πt + 30°). a) b) c) d)

What What What What

is is is is

the the the the

frequency in Hz? period of the voltage in miliseconds? magnitude of V at t = 2.778ms? RMS value of V?

Solution: a) Given ω = 120π = 2πf, thus f = 60Hz b) T = 1/f = 16.67ms c) V = V = 300 cos (120πx2.778m + 30°) = 300 cos (60° + 30°) = 0V d) Vrms = 300/√2 = 212.13V 16

SINUSOIDS contd.

• Consider the following:

v1(t )  Vm sin t

v 2 (t )  Vm sin  t   

17

SINUSOIDS contd.

• The v2 is occurred first in time. • Thus it can be said that v2 leads v1 by lags v2 by  . • If • If

 or v1

 ≠ 0 we can say v and v are out of phase. 1 2 

= 0 we can say v1 and v2 are in phase.

• v1 and v2 scan be compared in this manner because they operate at the same frequency (do not need to have the same amplitude).

18

SINUSOIDS contd.

• Transformation between cosine and sine form

sin A  cos( A  90) cos A  sin( A  90)

Converting from negative to positive magnitude

 sin A  sin( A  180)  cos A  cos( A  180) where

A  t   19

SINUSOIDS contd.

Example: • For the following sinusoidal voltage, find the value v at t = 0s and t = 0.5s. v = 6 cos (100t + 60˚) Solution: at t = 0s v = 6 cos (0+60˚) = 3V Note: both ωt and



at t = 0.5s v = 6 cos (50 rads +60˚) = 4.26V

must be in same unit before adding them up. 20

SINUSOIDS contd.

1.

Calculate the phase angle between v1 = -10 cos (ωt + 50°) v2 = 12 sin (ωt - 10°) State which sinusoid is leading. Solution: In order to compare v1 and v2, we must express them in the same form (either in cosine or sine function) with positive magnitude. Note: the value of  must be between 0° to ±180° v1 = -10 cos (ωt + 50°) = 10 cos (ωt + 50° - 180°) = 10 cos (ωt - 130°) 21

SINUSOIDS contd. and v2 = 12 sin (ωt - 10°) = 12 cos (ωt - 10° - 90°) = 12 sin (ωt - 100°)

the equation v2 can be written in the following form v2 = 12 sin (ωt - 130° + 30°) ‘+30°’ in the above expression means v2 leads v1 by 30°

22

PHASORS •

Sinusoid can be express in terms of phasors, which are more convenient to work with than sine and cosine functions.

•

A phasors is a complex number that can be represents the amplitude and the phase of a sinusoid.

•

Recall: Complex number can be written in: a) Rectangular form:

z = x + jy

23

PHASORS contd. where j  1 ; x is the real part of z and y is the imaginary part of z. •

Polar form:

z  r  •

Exponential form:

z  re

j

where r is the magnitude of z and  is the phase of z.

24

PHASORS contd.

• The relationship between the rectangular form and the polar form is shown below:

25

PHASORS contd.

• Form Figure 9.6

r  x y 2

or

x  r cos 

2

y ;   tan x 1

;

y  r sin 

Thus, z may be written as

z  x  jy  r   r cos   j  r sin   Note: addition and subtraction of complex number better perform in rectangular form. Multiplication and division in polar form. 26

PHASORS contd.

• Basic properties of complex number: Addition:

z1 + z2 = (x1 + x2) + j(y1 + y2)

subtraction:

z1 - z2 = (x1 - x2) + j(y1 - y2)

multiplication:

z1z2  r1r 2   1  2 

division:

z1 r1    1  2  z2 r 2

reciprocal:

1 1      z r 27

PHASORS contd.

Square root: Complex conjugate:

note:

z  r    2 z *  x  jy  r      re  j

1  j j j  1

28

PHASORS contd.

• As we know, the sinusoidal voltage can be represented in sine or cosine function. • First, consider the cosine function as in:

v (t )  Vm cos  t    • This expression is in time domain. • In phasor method, we no longer consider in time domain instead in phasor domain (also known as frequency domain). 29

PHASORS contd.

• The cosine function will be represented in phasor and complex number such as:

v (t )  Vm cos  t     Vm 

 Vm cos   j  Vm sin   • For example: Transform the sinusoid: v(t) = 12 cos (377t - 60˚) thus,

V  12  60

or

6  j10.39 30

PHASORS contd.

• The phasor representation carries only the amplitude and phase angle information. • The frequency term is dropped since we know that the frequency of the sinusoidal response is the same as the source. • The cosine expression is also dropped since we know that the response and source are both sinusoidal.

31

PHASORS contd.

• Sinusoid-phasor transformation:

Vm cos  t    Vm sin  t   

Im cos  t    Im sin  t   



Vm 



Vm     90 



Im 



Im     90 

32

PHASORS contd.

• The phasor can be represented by a phasor diagram as shown below.

• Once a sinusoidal voltage or current is represented in its phasor form, simple arithmetic operation can be done. 33

PHASORS contd.

• After performing the arithmetic operations, the phasor form can then be inverse back to the time domain form.

Vm 

 Vm cos  t   

the ω of the response is the same as the source.

34

PHASORS contd.

Example: Given y1 = 20 cos (100t - 30°) and y2 = 40 cos (100t + 60°). Express y1 + y2 as a single cosine function. Solution: In phasor form

y1  20  30

&

y 2  4060

 y1  y 2  20  30  4060  17.31  j 10  20  j 34.64  37.32  j 24.64  44.7233.4 Thus, y1 +y2 = 44.72 cos (100t + 33.4°)

35

CIRCUIT ELEMENTS IN PHASOR DOMAIN • Circuit analysis is much simpler if it is done in phasor domain. • In order to perform the phasor domain analysis, we need to transform all circuit elements to its phasor equivalent. • Transform from time domain to frequency domain. • For sinusoidal sources, phasor transform will make the source eligible to enter the phasor domain.

36

CIRCUIT ELEMENTS IN PHASOR DOMAIN contd.

IMPEDANCE • To analyze the circuit in ac domain, first we have to convert all the resistance, inductance and capacitance into impedance, Z or admittance, Y. • Impedance, Z is the ratio of the phasor voltage V to the phasor current, I.

V Z I

or

V  Vm V

&

V  ZI

where

I  Im I 37

CIRCUIT ELEMENTS IN PHASOR DOMAIN contd.

• Z is a complex number and measured in ohms(Ω). • As a complex quantity, Z can be express in rectangular form, Z = R + jX where R = resistance (real part of Z) X = reactance (imaginary part of Z) • Reactance can be positive or negative. If: – X is positive – the impedance is inductive – X is negative – the impedance is capacitive 38

CIRCUIT ELEMENTS IN PHASOR DOMAIN contd.

• Example: – Impedance Z = R + jX is said to be inductance or lagging since current lags voltage. – Impedance Z = R – jX is said to be capacitive or leading since current leads voltage.

• The impedance, Z can also be expressed in polar form,

Z  Z 

where Z 

and

2

R X

R  Z cos 

2

X R

;

  tan

;

X  Z sin 

1

39

CIRCUIT ELEMENTS IN PHASOR DOMAIN contd.

ADMITTANCE • It is sometimes convenient to work with the reciprocal of impedance, known as admittance. • Admittance, Y is the ratio of the phasor current through it to the phasor voltage across it. • Y is measured in siemens (S).

1 I Y  Z V 40

CIRCUIT ELEMENTS IN PHASOR DOMAIN contd.

• Y can also be written in rectangular form, Y = G + jB where G = conductance (real part of Y) B = susceptance (imaginary part of Y) • Or

1 Y Z 1 G  jB  R  jX 41

CIRCUIT ELEMENTS IN PHASOR DOMAIN contd.

IMPEDANCE FOR RESISTOR • In phasor domain, impedance Z=R • Voltage-current relationship is given by V = IZ

42

CIRCUIT ELEMENTS IN PHASOR DOMAIN contd.

• In phasor diagram,

• From the figure it shows that voltage and current are in phase.

43

CIRCUIT ELEMENTS IN PHASOR DOMAIN contd.

IMPEDANCE FOR INDUCTOR • In phasor domain, the impedance ZL = jXL = jωL = j2πfL where L is the value of an inductor. • Voltage-current relationship is given by V = IZ = I(jωL)

44

CIRCUIT ELEMENTS IN PHASOR DOMAIN contd.

• In phasor diagram,

• From the figure it shows that voltage leads current by 90° or current lags voltage by 90°. 45

CIRCUIT ELEMENTS IN PHASOR DOMAIN contd. IMPEDANCE FOR CAPACITOR • In phasor domain, the impedance ZC = = = =

-jXC -j/(ωC) 1/(jωC) 1/(j2πfC)

where C is the value of an capacitor. • Voltage-current relationship is given by V = IZ = I/(jωC) 46

CIRCUIT ELEMENTS IN PHASOR DOMAIN contd.

• In phasor diagram,

• From the figure it shows that current leads voltage by 90° or voltage lags current by 90°. 47

CIRCUIT ELEMENTS IN PHASOR DOMAIN contd.

ELEMENT

RESISTANCE

REACTANCE

IMPEDANCE

ADMITTANCE

VOLTAGE

R

R

0

Z=R

Y = 1/R

V = RI

L

0

XL = ωL

ZL = jXc = jωL

Y = 1/ jωL

V = jωLI

C

0

XC = 1/ωC

ZC = -jXc = -j/ωC

Y = jωC

V= I/jωC

48

CIRCUIT ELEMENTS IN PHASOR DOMAIN contd.

EXAMPLE: •

A current in the 75mH inductor is 4 cos (40000t - 38°)mA. Calculate a) b) c) d)

The The The The

inductive reactance impedance of the inductor phasor voltage steady-state expression of v(t)

Solution: • Inductance reactance, XL = ωL = 40000(75mH) = 3000Ω •

Impedance ZL = jωL = j40000(75mH) = j3000Ω

49

CIRCUIT ELEMENTS IN PHASOR DOMAIN contd.

a) Phasor voltage, V = IZ  (4 x103   38)( j3000)

 1252

g) Steady-state expression, v(t) = 12 cos (40000t + 52°)V

50

CIRCUIT ELEMENTS IN PHASOR DOMAIN contd.

2)

Find v(t) and i(t).

51

CIRCUIT ELEMENTS IN PHASOR DOMAIN contd.

From the voltage source 10 cos 4t, ω = 4,

VS  100 The impedance is

1 1 Z 5 5  5  j2.5 j C j 4(0.1) Hence the current VS 100 100 I    Z 5  j2.5 5.5902  26.5651  1.788826.5651 A 52

CIRCUIT ELEMENTS IN PHASOR DOMAIN contd.

The voltage across the capacitor is V  IZC 

I 1.788826.5651   4.4720  63.4349V jC j 4(0.1)

In time domain, i(t) = 1.7888 cos (4t + 26.5651°)A v(t) = 4.4720 cos (4t – 63.4349°)V

53

KIRCHHOFF’S LAWS IN THE FREQUENCY DOMAIN • In phasor domain: – Recall: KVL – states that the algebraic sum of phasor voltages around a loop is zero. KCL – states that the algebraic sum of phasor currents at a node is zero.

• This principles used in dc analysis, are also applicable in the phasor domain. • The difference is simply the voltages, currents and resistance/inductance/capacitance are converted to phasor and impedance. 54

KIRCHHOFF’S LAWS IN THE FREQUENCY DOMAIN contd.

Example: Four branches terminate at a common node. The reference direction of each branch current (i1, i2, i3 and i4) is toward the node. If i1= 100 cos(ωt + 25°), i2 = 100 cos(ωt + 145°), i3 = 100 cos(ωt - 95°), i4 = 100 cos (ωt - 95°). Find i4. Solution: By using the phasor concept, I1 = 100 25°, I2 = 100 145°, I3 = 100 -95°. Applying KCL, I1 + I2 + I3 + I4 = 0 100 25° + 100 145° + 100 -95° + I4 = 0

55

KIRCHHOFF’S LAWS IN THE FREQUENCY DOMAIN contd.

90.63+j42.26 – 81.92+j57.36 – 8.72-j99.62 + I4=0 I4 = 0

56