Course: EE 1103

Subject: Basic Electrical Engineering

Topic: Phasor Algebra – Cont’d

Edited and Presented by-

Dr. Mohiuddin Ahmad Dept. of EEE, KUET, Khulna-9203,

Bangladesh

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Contents of the Topic

Phasor Relationships For Circuit Elements

Impedance expressed in polar form

Impedance expressed in Cartesian form

Addition and subtraction of voltages and currents

Power calculations employing complex forms

Reactive volt-ampere calculations employing complex

forms

Conjugate method of calculating real and reactive

power

Mathematical problems

Reference Books

References

1. Alternating current circuits – Kerchner & Corcoran

2. Fundamentals of Electric Circuit – Alexander & Sadiku

3. Electrical Circuit analysis - Boylestad

Impedance expressed in polar forms

Impedance in series – R, L, C branches

000 jRRR Z

CC

C

RC

jXRR

XXR

R

C

CR

122

1

2

2

tan

1

tan1

Z

LLL jXXL 00 9090Z

CCC jXXC

00 90901

Z

LL

LRL jXRR

XXR

R

LLR

122122 tantan)(

Z

Impedance expressed in polar forms

RLC branch

Impedance

Abbreviated term:

here

+θ represents the angle of lead of the voltage w.r.t. current or

a lag of the current with respect to voltage

CL

RLC

jXjXRC

jLjR

CjLjR

R

CL

CLR

1

1

tan1 1

2

2Z

ZZ Z

Impedance expressed in polar forms

Phasor voltage

Impedance

Resulting current,

vV V

zv

z

v

Z

V

Z

V

I

zZ Z

Impedance expressed in Cartesian forms

The Cartesian form of the impedance function of a

given RLC branch or circuit is, in general

R is the equivalent resistance of the branch or circuit

with respect to the terminals considered

(XL – XC) is the equivalent reactance of the branch or

circuit with respect to the terminals considered.

)( CL XXjR Z

Phasor Diagram

Phasor diagram of an RLC circuit

Phasor Diagram

Phasor diagram of an RLC circuit

Example – 1

A given RL branch have R=3.5Ω and L=0.092 henry.

Find the vector current which flows through the

branch if a 60cycle voltage is V=110∠300 volts, is

applied to the RL branch.

Solution:

0

122

122

25.848.34

5.3

092.0377tan)092.0377(5.3

tan)(

R

LLRRL

Z

0

0

9

25.5416.325.848.34

30110

RLZ

VI

Example – 1

Vector representation

Example – 2

An RLC series branch consists of R=12.9Ω and L=0.056 henry,

and C=78μF. (a) What is the vector or complex impedance of the

RLC branch? (b) If a 60-cycles vector current I=10∠300 amperes

flows through the branch, find the vector voltage V across the

terminals of series branch. Draw a vector diagram illustrating

the vector positions of V and I and the magnitude of phase

angle V with respect to I.

Solution: (a)

ohms 4524.18 9.12

341.21tan341.219.12

9.12

1078377

1056.0377

tan1078377

1056.03779.12

1

tan1

0122

61

2

6

2

1

2

2

R

CL

CLRRLC

Z

Example – 2

An RLC series branch consists of R=12.9Ω and L=0.056 henry, and C=78μF. (a) What is the

vector or complex impedance of the RLC branch? (b) If a 60-cycles vector current I=10∠300

amperes flows through the branch, find the vector voltage V across the terminals of series

branch. Draw a vector diagram illustrating the vector positions of V and I and the

magnitude of phase angle V with respect to I.

Solution: (b) In Rectangular form

(C) Vector diagram – Draw it by yourself

ohms 341.219.12 jjjXjXR CLRLC Z

volts154.182)4524.18)(3010( 000 RLCIZV

Example – 3

Example: The terminal of an a-c generator which has an internal resistance of 2 –ohm and an equivalent internal inductive reactance of 6-ohms are connected to an RLC series branch, the R of which is 10 ohms, the ωL is 20 ohms, and the 1/ωC of which is 40 ohms. If the magnitude of the internally generated emf is 500 volts, find the current that flows in the series circuit and the terminal voltage of the generator.

Solution:

Internal impedance of the generator

Total impedance of the series circuit

ohms 6.7132.6 ohms 62 0 jgZ

ohms 4.4944.18ohms 1412

40201062

0

j

jjjRLCgtotal ZZZ

Example – 3

Example: The terminal of an a-c generator which has an internal resistance of 2 –ohm and an equivalent internal inductive reactance of 6-ohms are connected to an RLC series branch, the R of which is 10 ohms, the ωL is 20 ohms, and the 1/ωC of which is 40 ohms. If the magnitude of the internally generated emf is 500 volts, find the current that flows in the series circuit and the terminal voltage of the generator.

Solution:

Generated emf

Current flows in the series circuit

Terminal voltage of the generator

volts05000500 0 jg E

amperes 4.491.274.4944.18

0500 0

0

0

total

gI

Z

E

volts14606j147588.3

volts)6.7132.6()4.491.27(0500

0

000

ggg IZEV

Example – 3

Draw the complete vector diagram Vector or Phasor diagram:

Example – 3

Calculate the total power generated and total

absorbed by the external RLC branch

Solution:

Total power generated

Total Branch power

watts8818)4.49cos(1.27500cos gE

Ig IEP

watts1.7344101.27 22 RIP

Additions and Subtraction of voltages and Currents

Complex expressions for voltages and currents

specify both the magnitudes and relative phase

positions of these quantities.

1. Voltage drops in series may be added to obtain the

combined voltage drop of the series elements

considered.

If the combined voltage drop and one component are

known, the remaining voltage drop may be determined

by subtracting the component in question from the

combined voltage drop.

Additions and Subtraction of voltages and Currents

2. Generated emf'e connected in additive or

subtractive aeries may be added or subtracted,

depending upon the relative polarities of the terminal

which are joined together to found the series

connection.

Series connections of generated emf’s will be

considered in more detail when poly-phase systems

are studied.

3. Two or more currents flowing away from a junction

may be added to find the current flowing toward the

junction, or vice versa.

Circuit Directions of voltages and currents

Average power absorbed by a branch

V is the magnitude of the voltage drop across the

branch or circuit.

I is the magnitude of the current flowing through the

branch or circuit in the same circuit direction as that

which has been taken for the + V direction.

is the angle of lag (or lead) of I with respect to V.

• In a normal dissipative type of branch or circuit, θ will not

be as great as ±900.

VIVIP cos

Circuit Directions of voltages and currents

Similarly, average power generated by a generating

device

E is the magnitude of the generated voltage.

I is the magnitude of the current flowing through the

branch or circuit in the same circuit direction as that

which has been taken for the + E direction.

is the angle of lag (or lead) of I with respect to E.

• In case, the generating device actually delivering power,

θ is less than 900.

EIEIP cos

Power Calculations Employing Complex Forms

Power Expression – absorbed power

VIcosVIP

ivvi

IVIV

IVIV

VIVI

VI

VI

VIP

iviv

iviv

iviv

iviv

iviv

iv

)sin)(sin()cos)(cos(

sinsincoscos

sinsincoscos

)sinsincos(cos

)sinsincos(cos

)cos(

i

v

i

v

Ii

Vv

Ii

Vv

sin

sin

cos

cos

Power Calculations Employing Complex Forms

In rectangular form

Absorbed power

Generated power

vjv

jVV vv

sincos V

i

v

i

v

Ii

Vv

Ii

Vv

sin

sin

cos

cos

iji

jII ii

sincos I

ivviP

ieeiP

Power Calculations Employing Complex Forms

Power Expression – Generated power

EIcosEIP

ieei

IEIE

IEIE

EIEI

EI

EI

EIP

iviv

iviv

iviv

iviv

iviv

iv

)sin)(sin()cos)(cos(

sinsincoscos

sinsincoscos

)sinsincos(cos

)sinsincos(cos

)cos(

i

v

i

v

Ii

Ee

Ii

Ee

sin

sin

cos

cos

Example – 4

Find the generated power, when the generated

voltage is E = 200+j40 volts and that the current

flowing in the positive circuit direction is I = 30 – j10

amperes.

Solution:

You can get the same results by applying

watts5600

4006000

)10)(40()30)(200(

ieeiP

EIcosEIP

Reactive Volt-ampere Calculations Employing Complex Forms

Volt-ampere calculations

VIsinVIPX

ivivP

IVIV

IVIV

VIVI

VI

VIP

X

iviv

iviv

iviv

iviv

ivX

)sin)(cos()cos)(sin(

sincoscossin

sincoscossin

)sincoscos(sin

)sin(

i

v

i

v

Ii

Vv

Ii

Vv

sin

sin

cos

cos

Reactive Volt-ampere Calculations Employing Complex Forms

Example – 5

If V= 200∠300 =173.2+j100 volts and I= 10∠600

=5+j8.66 amperes, find the real power, the reactive

volt-amperes, and the total voIt-amperes involved.

Solutions:

Real-power,

Reactive volt-ampere,

Volt-ampere,

watts1732866866

)66.8)(100()5)(2.173(

ivviP

vars1500500)66.8)(2.173()5)(100( ivivPX

amperes- volt2000)1000()1732( 22 S

Conjugate Method of Calculating Real and Reactive Power

A method of conjugates is sometime employed in

the determination of real power and reactive volt-

amperes.

It affords a convenient means of calculating these

quantities when both the voltage and current are

expressed in Cartesian form

iji Ivjv V iji *I

xjPP

ivivjivvi

ivivjijvvi

ijivjv

)(

))((*VI

Conjugate Method of Calculating Real and Reactive Power

Example – 6

If V= 173.2+j100 volts and I= 5+j8.66 amperes, find

the real power, the reactive volt-amperes, and the

total voIt-amperes by the method of conjugates.

Solutions:

Volt-ampere,

Real-power,

Reactive volt-ampere

Here, + sign indicates capacitive vars when the

conjugates of the voltage is used.

Volt-ampere,

100017325008661500866

)66.85)(100173(

jjj

jjPva

amperes- volt2000)1000()1732( 22 S

vars1000XP

watts1732P

Exercise Problems

Problems [Corcoran book] – Self study

Exercise Problems

Problems [Corcoran book] – Self study

Exercise Problems

Problems [Corcoran book] – Self study

Exercise Problems

Problems [Corcoran book] – Self study

Exercise Problems

Problems [Boylestad] – Self study

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