Chapter 6a.ppt

BEX 17003ELECTRICAL AND

ELECTRONICS TECHNOLGY

CHAPTER (6a)

Chapter 6Sinusoidal a.c. Circuit

Content

Sinusoids Complex numbers Phasors AC circuit analysis Power in ac circuit

Sinusoids Consider a sinusoidal wave in Figure

6.1. The sinusoidal signal can be expressed as

where Vm is the peak value (amplitude), ω is the angular frequency, θ is the phase angle and T is the period.

Sinusoidal signals are periodic, repeating the same pattern in interval T, which is

θωtcosVtv m

Figure 6.1

tvTtv

Sinusoids (cont.) The sine/cosine function completes 1 cycle in every 2 radian,

therefore

Frequency of sinusoidal signal is the number of completed cycle in one second, which is

The angular frequency is given as

Angular frequency has a unit of radian/second (rad/s).T 2

f 2

Tf 1

2πωT

Sinusoids (cont.) Phase relationship:

Assume there are two sinusoidal voltage signal as shown in Figure 6.2.

Since v2(t) reaches the peak value first, therefore it is said that v2(t) leads v1(t) by or v1(t) lags v2(t) by .

In this situation, both v1 and v2 are also said to be out of phase.

If = 0, then v1(t) and v2(t) are said to be in phase, in which they reach the minimum and maximum point at the same time.

Figure 6.2

Sinusoids (cont.) Sinusoids can be expressed either in sine or cosine function. Using the relationship shown below, we can transform a sine

function into cosine function and vice versa. The transformation can be done using this relationship:

ωtsin 90ωtcos

ωt cos90ωtsinωt cos180ωtcos

ωtsin 180ωtsin

Sinusoids (cont.) The root-mean-square (rms) value of a sinusoidal voltage

signal is also known as the effective value or DC-equivalent value . The rms periodic voltage v(t) and its average power for a

sinusoidal voltage across a resistor is given as

Similarly for a periodic current i(t) flows through a resistor, the rms value and its average power is given as

dttvT

VT

2

0

1rms R

VP

2rms

avg

dttiT

IT

2

0

1rms

RIP 2rmsavg

RMS In a circuit whose impedance consistsof a pure

resistance, the rms value of an ACwave is often called the effective value or DC-equivalent value.

For example, if an AC source of 100 volts rms is connected across a resistor, and the resulting current causes 50 watts of heat to be dissipated by the resistor, then 50 watts of heat will also be dissipated if a 100-volt DC source is connected to the resistor.

http://searchcio-midmarket.techtarget.com/definition/impedance

http://searchcio-midmarket.techtarget.com/definition/resistance

Sinusoids (cont.) Sinusoidal signal is usually described in its amplitude. The value

of signal amplitude can be used to find the rms value of the specified signal.

To find rms value from its amplitude

This formula is applies only to sinusoidal signal, which means that this formula cannot be used to square waves and triangular waves.

2mVV rms

Example on Vm

For a sine wave, the rms value is 0.707 times the peak value, or 0.354 times the peak-to-peak value. Household utility voltages are expressed in rms terms.? Aso-called “240-volt" AC circuit carries about 339 volts peak (pk), or 679 voltspeak-to-peak (pk-pk).

http://whatis.techtarget.com/definition/peak-pk

http://whatis.techtarget.com/definition/peak-to-peak-pk-pk

Complex numbers A complex number involves a real number and an imaginary

number. An imaginary number is represented as j where Examples of complex numbers are 2 + j4 and 5+j6 where 2 and

5 are real numbers whereas j4 and j6 are imaginary numbers. Complex numbers can be represented in complex plane, shown

in Figure 6.3, in which the real part is in horizontal coordinate and the imaginary part is in vertical coordinate.

1j

Figure 6.3

Complex numbers (cont.) Complex numbers in the form of x+jy are

in rectangular form. The conjugate of rectangular form is obtained by changing the sign of imaginary part.

For example, if Z = 3 - j4, therefore the complex conjugate of Z is Z*= 3 + j4.

Complex number can also be represented in polar form, as in Figure 6.4, by giving the magnitude and the angle between the arrow and the positive real axis.

Examples of complex number in polar form are

305Z1 4510Z 2and

Figure 6.4

Complex numbers (cont.) Complex numbers can be converted from polar to rectangular

form, or vice versa, using the following relationship:

rrectangula to polarθsinZy

θcosZx

polar to rrectangula

xy

tanθ

yxZ

1

22

The third form of complex number is known as exponential form. Exponential form of a complex number can be written as follows:

Examples of exponential complex number are

θ sinjAθ cos AθAAe jθ

j20j60 4e and10e

Using calculator for conversion

Casio fx-570MXConversion from

rectangular to polarShift (+) => Pol(Type 3, 4)RCL E and F.

Casio fx-570MXConversion from polar

to rectangularShift (-) => Rec(Type 5, 53.13)RCL E and F.

Complex numbers (cont.) Addition and subtraction of complex numbers are easier to be done in

rectangular form, and multiplication and division are better performed in polar form.

Given a pair of complex numbers as follows:

Addition: Subtraction: Multiplication:

Division:

Reciprocal:

Square root:

11111 θrjyxz 22222 θrjyxz 212121 yyjxxzz 212121 yyjxxzz

212121 θθrrzz

212

1

2

1 θθrr

zz

θrz

11

2θ/rz

Phasors For a sinusoidal voltage of the form

we define the phasor as Therefore, we define phasor of a sinusoid is a complex number

having a magnitude equals to the amplitude and the same phase angle as the sinusoid.

If the sinusoid is in the form of we have to convert into

Thus, the phasor are defined as

Similarly, for sinusoidal currents i1(t) and i2(t)

they can be expressed in phasor form

θωtsintv 2

111 θωtcosVtv

11 θV 1V

90222 θωtcosVtv

90222 θVV

222111 θωtsinIti and θωtcosIti

90222111 θI and θI II

Phasors (cont.) Adding sinusoids using phasors:

Step 1: Determine the phasor for each term. Step 2: Add the phasors using complex arithmetic. Step 3: Convert the sum to polar form. Step 4: Write the result as a time function.

Refer to the given example that illustrates adding sinusoids using phasors.

Phasors (Resistor) How to relate the phasor expression of sinusoidal voltage and

current with other circuit elements ? Assuming we have a current defined as

If this current is passing through a resistor, the voltage across the resistor isand the phasor form is where

θωtcosIi m

θωtcosRIiRv m

IV RθI mI

Phasors (Inductor) Assuming we have a current defined as

If this current is passing through an inductor L, the voltage across the inductor is

and the phasor form is

Since and , thus

θωtcosIi m

θωtsinωLIdtdiLv m

90θωtcosωLIv m

j90m

j90jθm

90θjm e θωLIeeωLIeωLIV

IθI m je j90 IV L jω

V = jωLI

Phasors (Capacitor) If the voltage across a capacitor is given as

the current through the capacitor is and the phasor form is

θωtcos Vv m

dtdvCi

C jωC jω IVVI

I = jωCV

Phasors (cont.) The current-voltage (I-V) relationship can be summarised as follows:

Element Time domain Frequency domain

Impedance Reactance

L

C

v = Ri

dtdiLv

dtdvCi

V = RI

V = jωLI

CjωIV

R R

jωL

j(-1/ωC)

–

ωL

-1/ωC

Phasors (cont.) To determine phase relationships from a phasor diagram,

consider the phasors to rotate counterclockwise. Then looking at a fixed point, if V1 arrives first followed by V2

after a rotation of θ , we say that V1 leads V2 by θ . Alternatively, we could say that V2 lags V1 by θ . (Usually, θ is

taken as the smaller angle between the two phasors.) Consider two voltages that can be described as

and the phasors are

From phasor diagram in Figure 6.6, V1 leads V2 or V2 lags V1.

4031 ωtcostv

2042 ωtcostv

4031V 204 2V

Figure 6.5

AC Circuit analysis Kirchhoff’s Laws in Phasor Form

We can apply KVL directly to phasors. The sum of the phasor voltages equals zero for any closed path.

Similarly, KCL in phasors, the sum of the phasor currents entering a node must equal the sum of the phasor currents leaving.

Circuit Analysis Using Phasors and Impedances 1. Replace the time descriptions of the voltage and current sources

with the corresponding phasors. (All of the sources must have the same frequency.)

2. Replace inductances by their complex impedances ZL = jωL. Replace capacitances by their complex impedances ZC = 1/(jωC). Resistances have impedances equal to their resistances.

3. Analyze the circuit using any of the techniques studied earlier in Chapter 3, performing the calculations with complex arithmetic.

AC Circuit analysis Series-parallel combination

The rules in combining series-parallel impedances is the same as resistors.

The only difference is that combining impedances involves complex numbers.

Node-voltage method In Chapter 3, you have seen node-voltage method applied in d.c.

circuit. In ac circuit, the same procedure applies but all the circuit

elements must be converted into frequency domain. Mesh-current method

Mesh-current method also has been introduced in d.c. analysis, and this method can be applied in ac circuit with all circuit elements expressed in frequency domain.

AC Circuit analysis (cont.) Source transformation

Figure 6.6 shows the source transformation in ac circuit. The same configuration can be seen in dc circuit, but now resistance (R) is replaced with impedance (Z).

Figure 6.6

AC Circuit analysis (cont.) Thevenin-Norton equivalent circuits

Figure 6.7 and Figure 6.8 illustrate the frequency domain version of Thevenin and Norton equivalent circuit respectively.

The technique in finding ZTh and VTh is the same in dc circuit but now it involves the complex numbers.

Figure 6.8

Figure 6.7

Power in ac circuit Power for Resistive load

Consider a network with pure resistive circuit.

Plot of these quantities is shown in Figure 6.9. Note that both current and voltage are in phase.

Since p is always positive, it is said that power transferred from source to load is the average power or real power.

ωtcosIVtitvtp

ωtcosItiωtcosVtv

mm

m

m

2

Figure 6.9

Power in ac circuit (cont.) Power for Inductive load

Consider a network with pure inductive circuit.

Plot of these quantities is shown in Figure 6.10. Note that current lags voltage by 90°.

When p is positive, power is delivered from source to load and when p is negative, power returns from inductor to source.

In inductive circuit, average power is zero, the only power exists is known as reactive power.

ωtsinIV

tp

ωtsinωtcosIVtitvtpωtsinI-ωtcosIti

ωtcosVtv

mm

mm

mm

m

22

90

Figure 6.10

Power in ac circuit (cont.) Power for Capacitive load

Consider a network with pure capacitive circuit.

Plot of these quantities is shown in Figure 6.11. Note that current leads voltage by 90°.

When p is positive, power is delivered from source to load and when p is negative, power returns from capacitor to source.

Similarly in capacitive circuit, only reactive power exists with average power is zero.

ωtsinIV

tp

ωtsinωtcosIVtitvtpωtsinIωtcosIti

ωtcosVtv

mm

mm

mm

m

22

90

Figure 6.11

Power in ac circuit (cont.) Consider an ac circuit in Figure 6.12 consists of a sinusoid

voltage source connected to resistor, inductor and a capacitor.

Now, we want to find the average power, reactive power and apparent power for this circuit.

Figure 6.12

Power in ac circuit (cont.) Average power

This power is total power transferred to resistive load and given as

where Vrms = Vm/√2 and Irms = Im/√2. Power factor

The term cos(θ) is called the power factor (PF) or power angle. θ is the difference between voltage angle and current angle.

Reactive power This power exists in inductor and capacitor. It is given as

Apparent power This power is defined as the product of effective voltage and effective

current.

cosrmsrmsIVP

cosPF iv

sinrmsrmsIVQ

rmsrmspower apparent IV

Power in ac circuit (cont.) The relationship between real power P, reactive power Q,

apparent power and power angle θ can be represented by power triangle in Figure 6.13.

Power triangle in Figure 6.13 (a) for inductive load with both θ and Q are positive and Figure 6.13 (b) for capacitive load with both θ and Q are negative.

Figure 6.13

Phasor Relationships for Circuit Elements

Example 6

The voltage is applied to a 0.1 H inductor. Find the current through the inductor.

)4560cos(12 0 tv

Solution

For the inductor, where and

Hence,

AxjLj

VI 00

00

4529064512

1.0604512

Converting to the time domain,

Atti )4560cos(2)( 0

LIjV srad /60 VV 04512

• The impedance Z, of the circuit is the ratio of the phasor voltage V to the phasor current I, measured in ohm.

6.8 IMPEDANCE and ADMITTANCE

CjLjR

IV,IV,IV

• The impedance Z is expressed as

IVZ

Z

V

I

+ -

• In the preceding section, we obtained the voltage-current relations for the three passive elements as

• These equations may be written in terms of the ratio of the phasor voltage to the phasor current as

CjLjR

1

IV,

IV,

IV

IMPEDANCE and ADMITTANCE cont..

• The impedance consist of the real component (resistance, R) and imaginary component (reactance, X).

• The total impedance connected in series is

nT ZZZZZ ......321

• The total impedance connected in parallel is

nT ZZZZZ1......1111

321


RZR

LL jXLjZ fLLX L 2

Cc jXCj

Z 1

fCCXC 2

11

• For resistor

• For capacitor

• For inductor


CjX

• The admittance Y, is the reciprocal of impedance, measured in siemens (S).• The admittance Y of the element is the ratio of the phasor current to the phasor

voltage.

• The admittance of the passive elements is expressed asVI

ZY

1

CjYLj

YR

Y

11

LjX1

jBGY • In complex quantity,

• Where,

jXRjBG

1


• The total admittance connected in series is

nT YYYYY ......321

• The total admittance connected in parallel is

nT YYYYY1......1111

321


Example 7

Find the total impedance for the following circuit.

8F300

Ati(t ) 37 7s in3 0

5.2)1067.6)(377(

67.61020

)10)(20(

95.210900377

900600300

43.36868

/377:

3

6

jjjXZ

mHmmmmL

jj

Cj

jXZ

FFFC

ZR

sradSolution

LL

parallel

cc

parallel

Rparallel

)68.031.3(

69.1138.369.11296.0

1

69.11296.006.029.034.04.029.0

95.21

5.21

43.311111

jS

Z

Sjjj

jjZZZZ

T

CLRT

impedance Total

8 6 F600F300 mH10 mH20

Ati(t) 377sin30

• The triangle impedance represent the relationship between resistance, reactance and impedance.

TRIANGLE IMPEDANCE

• Inductive Impedance

LjRjXRZ L

Where

00 90 ; 0 LXLjRR

Z

R

LjX L

22 LRZ

RL 1tan

TRIANGLE IMPEDANCE

• Capacitive Impedance

CjRjXRZ C

Where

00 90 ; 0

CXCjRR

2

2 1

CRZ

CR 1tan 1

Z

R

CXC

1

a) Series RL Circuit

.

6.9 RLC CIRCUITS ANALYSIS

The equivalent impedance

LjRjXRZ L

Where

RL

and

LRZ

1

22

tan

)(

From Ohm Law

IZV

LR VVLjIIRLjRI

)(

v(t)

i(t )

R

L

I R

VR

Lj00rmsV

V l

b) Parallel RL Circuit

.

RLC CIRCUITS ANALYSIS cont..


LjRLjRZ

)(

From Ohm Law

)( LjRLjRVI

RL II

RV

LjV

v(t)

i(t)

R L

I

R Lj00rmsV

Exercise 1Voltage v(t)= 200 sin (500t+60°) V supplied to the parallel circuit below.

Find the current i(t).

.


v ( t )

i ( t)

R L5Ω

20mH

Answer: i(t)= 44.74sin(500t+33.43°) A

c) Series RC Circuit

.



CjRjXRZ C

From Ohm LawIZV

CR VV

CjIRI

CjRI

v(t)

i(t)

R

C

I R

VR

Cj1

00rmsVVC

Where

RCand

CRZ

1tan1 1

22

d) Parallel RC Circuit

.



CjR

CRj

Z

1

From Ohm Law

jRjCVI

RC II

RVCVj

v(t)

i(t)

R C

I

R Cj1

00rmsV

Exercise 2Calculate the total impedance of a series RC circuit if R=27Ω, C=0.005 uF,

and the source frequency =1 kHz.


Answer: ZT =31.8 kΩ

e) RLC Circuit in Series

.


From Ohm Law

IZV

)( CLR

CL

VVjV

IXIXjIR

v(t)

i(t)

R

C

I R

VR

Cj1

00rmsVVC

L

LjL


CL jXjXRZ where

R

XXand

XXRZ

CL

CL

1

22

tan

f) RLC Circuit in Parallel

.



CL jXjXRZ1111

Therefore

v(t)

i(t)

R C

I

R cjX00rmsV

L Lj

))(()()()(( )

CLCL

CL

jXjXjXRjXRjXjXR

Z

CLCL

CL

XXjRXjRXXRX

RLC Circuit in Parallel.


From Ohm Law

IZV

CL

CLCL

XRXXXjRXjRXV

ZVI )(

RLC

LC

IjIjI

RV

XjV

XjV

Exercise 3Determine v0(t) in the circuit below.


v ( t )

i ( t ) 60Ω

10mF5Hv0+

-=20 cos (4t-15°)

Answer: v0(t)= 17.15 cos (4t + 15.96°) V

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