lecture 1 basics of electric circuits

Lectures on Electrical

and Instrument

Engineering

Focused, To-The-Point and Practical!

Personal IntroductionEngr. Arslan Ahmed Amin is a professional Electrical and

Instrumentation Engineer serving Pakistan’s pioneer Oil and Gas

Organization, Pakistan Petroleum Limited. He obtained his Bachelor's

degree in Electrical Engineering from the prestigious University of

Engineering and Technology, Lahore in 2010 and started his

professional career with Pakistan Petroleum Limited. He has served

this organization for more than 5 years and achieved lots of

accomplishments in the development of the systems of newly installed

210 MMSCFD gas compression facility. He actively contributed his

services in commissioning, testing, maintenance and upgradation of

the E&I systems. He completed his Master’s in Business Administration

(M.B.A.) in 2014 from Virtual University of Pakistan, Lahore through

distance learning and afterwards obtained Masters (M.Sc.) in Electrical

Engineering from University of Engineering and Technology, Lahore in

M.Sc. Electrical Engineering (University of Engineering

and Technology, Lahore)

M.B.A. Online (Virtual University of Pakistan, Lahore)

B.Sc. Electrical Engineering with Honors (University of

Engineering and Technology, Lahore)

Education

05 years’ experience in industrial process controls and

electrical power systems domain with Pakistan

Petroleum Limited (PPL) in CMMS (SAP) environment.

Experience of commissioning, testing and maintenance

of latest systems regarding Power Generation, Field

Instrumentation, Distributed Control System, Safety

Instrumented System, Gas Turbines, PLCs, Analyzers and

Utility packages.

Experience

‘Circuits and Electronics’ from Massachusetts Instituteof Technology (MIT) USA.

‘Project Management’ from Virtual University ofPakistan (VU).

‘Production and Operations Management’ from VirtualUniversity of Pakistan (VU).

‘Conflict Management’ from Virtual University ofPakistan (VU).

‘Crisis Management’ from Virtual University of Pakistan(VU).

Professional Courses

QMS 9001, ISO 14001 EMS and OHSAS 18001, ERP System, Cost ofQuality, Productivity Improvement Techniques, Process SafetyManagement, Hazard Identification and Risk Assessment, HAZOP,SIL systems, Occupational health and Safety, Permit to worksystem, Safety Modules (Complete), Communication Skills, TeamWork Skills, Decision Making Skills (Organized by PPL)

SAP System (R3P version) Maintenance Work Orders Processing,Contracts Management, Spares and Material.

‘Instrumentation and Controls Fundamentals’ from OMS Institute ofManagement and Technology, Lahore.

Installation, calibration and maintenance of Fire and Gas detectorsby Det-tronics.

Generation, Transmission and Distribution at WAPDAEngineering Academy Faisalabad.

Professional Trainings

Among Top 10 students in the session of 240 students in B.Sc.Electrical Engineering.

Received Dean’s Honor Role award in consecutive fivesemesters for excellent academic performance in B.Sc.Electrical Engineering.

Overall Topped in F.Sc. in Board of Intermediate andSecondary Education, Faisalabad 2006.

Winner of Quaid-e-Azam Scholarship.

Gold medal winner in District Science Quiz CompetitionFaisalabad.

Represented as ‘Talent of Pakistan Youth’ in China in 2007 byMinistry of Youth, Pakistan.

Academic Achievements

Lets Start!

Lecture-1

Basics of Electric Circuits

Electric Circuit

An electric circuit is an interconnection of

electrical elements.

Systems of UNITS

Quantity Basic Unit Symbol

Length meter m

Mass kilogram kg

Time second s

Electric current ampere A

Thermodynamic

Temperature

kelvin K

Luminous

intensity

candela cd

The SI Prefixes

Charge and Current

Electric Current

Current is the flow of

electricity, much like

the flow of water in a

pipe. It is measured in

Amperage as opposed

to gallons per minute

of water.

Conductors

Free Electrons (e)

Easily Directed

Usually metals

Copper

Aluminum

Platinum

Semi-Conductors

Dielectrics

4 Valence Electrons

Polarize with Some

Electron Flow due to

Electrical Fields

Insulators

No Free Electrons

No Current Flow with

Why Does Current Flow?

A voltage source provides the energy (or work) required to produce a current

Volts = joules/Coulomb = dW/dQ

A source takes charged particles (usually electrons) and raises their potential

so they flow out of one terminal into and through a transducer (light bulb or

motor) on their way back to the source’s other terminal

Voltage

Voltage is a measure of the potential energy that causes a current to flow

through a transducer in a circuit

Voltage is always measured as a difference with respect to an arbitrary

common point called ground

Voltage is also known as electromotive force or EMF outside engineering

Voltage (Volts - V or E)

Voltage is the electrical

pressure in the system,

much like water pressure.

Electrical pressure is

measured in Volts as

opposed to Pounds per

Square Inch. (ie: 110V like

water from a tap, 4160 like

a fire hose)

Voltage

Resistance (Ohms - R or Ω)

Resistance is simply the

restriction of current

flow in a circuit.

Smaller wire

(conductors) and poor

conductors have higher

resistance.

Resistance

Many Collisions = Heat!

Fewer Collisions = Less Heat!

Ohm’s Law

Current, Voltage, and Resistance relate as follow:

I = E / R

A Circuit Current flows from the higher voltage

terminal of the source into the higher

voltage terminal of the transducer before

returning to the source

SourceVoltage

+ Transducer -Voltage

The source expendsenergy & the transducerconverts it into something useful

Passive Devices

A passive transducer device functions only when energized by a source in a

circuit

Passive devices can be modeled by a resistance

Passive devices always draw current so that the highest voltage is present on

the terminal where the current enters the passive device

+ V > 0 -

Notice that the voltage ismeasured across the device Current is measured

through the device

Active Devices

Sources expend energy and are considered active devices

Their current normally flows out of their highest voltage terminal

Sometimes, when there are multiple sources in a circuit, one overpowers

another, forcing the other to behave in a passive manner

The rate at which energy is transferred from an active source or used by a

passive device

P in watts = dW/dt = joules/second

P= V∙I = dW/dQ ∙ dQ/dt = volts ∙ amps = watts

W = ∫ P ∙ dt – so the energy (work in joules) is equal to the area under the

power in watts plotted against time in seconds

The power consumed or created is just the Voltage

multiplied by the Current

P = V x I

If 3 amps flowing through a component

generate 12 volts across the component

the power is 3 x 12 = 36 watts

Some power calculations

Current

Voltage

2 Amps 5 Volts

9 Amps 36 Watts

10 Watts

4 Volts

Conservation of Power

Power is conserved in a circuit - ∑ P = 0

We associate a positive number for power as power absorbed or used by a

passive device

A negative power is associated with an active device delivering power

If I=1 amp

V=5 volts

Then passive

P=+5 watts

(absorbed)

If I= -1 amp

V=5 volts

Then active

P= -5 watts

(delivered)

If I= -1 amp

V= -5 volts

Then passive

P=+5 watts

(absorbed)

Example

A battery is 11 volts and as it is charged, it increases to 12 volts, by a current

that starts at 2 amps and slowly drops to 0 amps in 10 hours (36000 seconds)

The power is found by multiplying the current and voltage together at each

instant in time

In this case, the battery (a source) is acting like a passive device (absorbing

energy)

Energy

The energy is the area under the power curve

Area of triangle = .5 ∙ base ∙ height

W=area= .5 ∙ 36000 sec. ∙ 22 watts = 396000 J.

W=area= .5 ∙ 10 hr. ∙ .022 Kw. = 110 Kw.∙hr

So 1 Kw.∙hr = 3600 J.

Since 1 Kw.∙hr costs about $0.10, the battery costs $11.00 to charge

AC and DC Current

•DC Current has a constant value

•AC Current has a value that changes sinusoidally

Notice that AC current

changes in value and

direction

No net charge is

transferred

AC v DC

• DC can be produced chemically or mechanically; AC must be produced mechanically

• DC can be easily stored; AC cannot

• AC is easier, and thus cheaper, to produce

• AC can easily be transformed to other voltages

• AC can be transmitted more economically

Polarity

• Some components (like a bulb) can be connected either way round – they will still work

Polarity

• Some components (like a diode) can be connected either way round – they work one way but not the other

Passive Sign Convention

PSC: Example I

PSC: Example II

PSC: Example III

Circuit Elements

Ideal Independent Source: provides a specified

voltage or current that is completely independent of

other circuit variables

Ideal Independent Voltage Source:

Circuit Elements

Ideal independent current source

Circuit Elements

Ideal dependent voltage source

Ideal dependent current source

Automotive circuits

Equivalent electrical circuit

Itail Istart Ifan Ilocks Idash

Electrical vehicle battery

DC-AC converter(electric drive)

12 V12 V12 V12 V12 V

AC motor

Vbatt1 Vbatt2 Vbattn

Various representations of an

electrical system

HeadlightCar

battery

(a) Conceptualrepresentation

Power flow

(b) Symbolic (circuit)representation

(c) Physicalrepresentation

Volt-ampere characteristic of

a tungsten light bulb

–0.5

–0.4

–0.3

–0.2

0–20–30–40–50–60 –10 5040302010 60

–0.1

i (amps)

v (volts)

Variablevoltagesource

Currentmeter

The resistance element

i-v characteristicCircuit symbolPhysical resistorswith resistance R.

Typical materials arecarbon, metal film.

Resistor color code

b 4 b 3 b 2 b 1

Color bands

blackbrownredorangeyellowgreen

012345

bluevioletgraywhitesilvergold

6 7 8 9

Resistor value = ( b 1 b 2 ) 10b3;b4 = % tolerance in actual value

The current

1.5 V+_

v+ –

v– +

i flows through each of the four series elements. Thus, by KVL,

1.5 = v1+ v2+ v3

N series resistors are equivalent to a single resistor equal to the sum of the individual resistances.

Parallel circuits

KCL applied at this node

The voltage v appears across each parallel

element; by KCL, iS = i1 + i 2+ i 3

N resistors in parallel are equivalent to a single equivalent resistor with resistance equal to the inverse of the sum of

the inverse resistances.

RN REQR1 R2 R3 Rn

i1 i2 i3

iS R1 R2 R3

Wheatstone bridge circuits

v S a+_

v bva b

R3R1+_

v bva b

A force-measuring instrument

Beam cross section

R2 , R3 bondedto bottom surface

Practical voltage source

Practical

voltagesource

iS =vS

rS + R L

lim iS =vS

rSRL 0

r S i S max

The maximum (short circuit) current which can be supplied by a practical voltage source is

iS max = vS

Practical current source

R Li S

v Sr S

A model for practical current sources consists of an ideal source in parallel with an internal resistance.

v Sr S

Maximum output voltage for practical current source with open-circuit load:

vS max = iS rS

Measurement of current

A seriescircuit

Symbol forideal ammeter

Circuit for the measurementof the current i

Measurement of voltage

A seriescircuit

Idealvoltmeter

Circuit for the measurementof the voltage v2

Models for practical ammeter

and voltmeter

Practicalammeter

Practicalvoltmeter

Measurement of power

Internal wattmeter connections

Measurement of the powerdissipated in the resistor R2:

P2 = v2 i

Definition of a branch

Practicalammeter

Idealresistor

A battery

A branch

Branchvoltage

Branchcurrent

Examples of circuit branches

Definition of a node

Examples of nodes in practical circuits

Node a

Node b

Node c Node a

Node b

Definition of a loop

Loop 1 Loop 2

Loop 3

1-loop circuit 3-loop circuit(How many nodes in

this circuit?)

Note how two different loopsin the same circuit may in-clude some of the same ele-ments or branches.

iSR1 R2

Magnetics

NorthSouth

Magnetic Flux

Magnet

Current Flow in Conductor

Current Flowing in a Conductor

Generated Field Around

Conductor

Magnetic Field With Coil

North Magnetic Pole

South Magnetic Pole

Interaction with Medium

NorthSouth

Magnetic Flux

MagnetMetal NS

Electrical Properties

Frequency

Inductance (L)

Mutual

Inductive Reactance (XL)

Capacitance (C)

Capacitive Reactance (XC)

Phase Angle/Power Factor

Impedance (Z)

Frequency

0 90 180 270 360

Inductance

Stores electromagnetic energy

in its magnetic field

)0()(1

I lags V

Mutual Inductance

When 2 coils in close

proximity, a changing

current in one coil will

induce a voltage in a

second coil

0 90 180 270 360

N1 = 5 Turns

100 Volts

N2 = 5 Turns

100 Volts

Inductive Reactance XL

Inductive Reactance is

the AC Resistance of a

Presented as a

resistance in Ohms

Frequency and

Inductance Dependant

fLX L 2

Capacitance

Stores energy in an electric field

Dielectric between 2 plates

The charged condition is maintained until a discharge path is present

Causes current to lead voltage

Capacitive Reactance XC

Phase Angle / Power Factor

In a coil or motor,

current lags behind

voltage

This is represented as

an angle or a fraction

of ‘unity’

Adding C can improve

0 90 180 270 360

Impedance Z

1fLX L 2

Resistance

Complex

Resistance

22 )( CL XXRZ

Summary

Atomic Structure and Electron Movement

Conductors, Semi-Conductors, Insulators

Basic Electricity: Current, Voltage and Resistance

Electrical and Magnetic Fields

Alternating Current Electricity: L, C, XL, XC, Z

Generation & Distribution

Generator

10.6 KV

GT220 KV

Step down

transformer

Distribution

Power plant Transmission

systemDistribution system

• AC generators (“alternators”) generate

electricity

• Electricity generated at 9-13 KV

• Power generated from 67.5 to 1000 MW

• Power stations: generating transformers

(GTs) to increase voltage to 132-400 KV

• Substations: step-down transformers to

reduce voltage before distribution

Benefits of high voltage transmission

• Less voltage drop: good voltage regulation

• Less power loss: high transmission

efficiency

• Smaller conductor: lower costs

Single phase AC circuit:

• Two wires connected

to electricity source

• Direction of current

changes many times

per second

Phase of Electricity

3-phases of an electric system

Three phase systems:

• 3 lines with electricity from 3 circuits

• One neutral line

• 3 waveforms offset in time: 50-60 cycles/second

Star connection

Phase of Electricity

Delta connection

Review of Phasors

Goal of phasor analysis is to simplify the

analysis of constant frequency ac systems

v(t) = Vmax cos(wt + qv)

i(t) = Imax cos(wt + qI)

Root Mean Square (RMS) voltage of sinusoid

v t dtT

Phasor Representation

Euler's Identity: e cos sin

Phasor notation is developed by rewriting

using Euler's identity

( ) 2 cos( )

( ) 2 Re V

v t V t

v t V e

Then drop the constant terms

( ) Re 2

V cos sin

I cos sin

V V e V

v t Ve e

Advantages of Phasor Analysis

Resistor ( ) ( )

( )Inductor ( )

1 1Capacitor ( ) (0)

Z = Impedance

R = Resistance

X = Reactance

XZ = =arctan( )

v t Ri t V RI

di tv t L V j LI

i t dt v V Ij C

R jX Z

Device Time Analysis Phasor

RL Circuit Example

( ) 2 100cos( 30 )

4 3 5 36.9

100 30

5 36.9

20 6.9 Amps

i(t) 20 2 cos( 6.9 )

Complex Power

max max

( ) ( ) ( )

v(t) = cos( )

(t) = cos( )

1cos cos [cos( ) cos( )]

1( ) [cos( )

cos(2 )]

p t v t i t

p t V I

max max

1( ) [cos( ) cos(2 )]

1cos( )

cos( )

V I V I

p t V I t

P p t dtT

q q w q q

Power Factor

Average

Power Triangle Inductive Load, lagging Power

Factor.

Power Triangle Capacitive load, Leading

Power Factor

[cos( ) sin( )

P = Real Power (W, kW, MW)

Q = Reactive Power (var, kvar, Mvar)

S = Complex power (VA, kVA, MVA)

V I V IS V I j

q q q q

Power Factor (pf) = Cosø

If current leads voltage then pf is leading

If current lags voltage then pf is lagging

Relationships between real, reactive and complex power

Example: A load draws 100 kW with a leading pf of 0.85.What are (power factor angle), Q and S?

-cos 0.85 31.8

100117.6

117.6sin( 31.8 ) 62.0 kVarQ

Conservation of Power

At every node (bus) in the system

Sum of real power into node must equal zero

Sum of reactive power into node must equal zero

This is a direct consequence of Kirchoff’s

current law, which states that the total

current into each node must equal zero.

Conservation of power follows since S = VI*

Conversation of Power

Example

Earlier we found

I = 20-6.9 amps

100 30 20 6.9 2000 36.9 VA

36.9 pf = 0.8 lagging

S 4 20 6.9 20 6.9

S 3 20 6.9 20 6.9

1200var

Power Consumption in

Devices

2Resistor Resistor

2Inductor Inductor L

Capacitor Capacitor C

Capacitor

Resistors only consume real power

Inductors only consume reactive power

Capacitors only generate reactive power

jI X X

j C Cw w

Capacitor

Example

40000 0400 0 Amps

40000 0 (5 40) 400 0

42000 16000 44.9 20.8 kV

S 44.9 20.8 400 0

17.98 20.8 MVA 16.8 6.4 MVA

First solve

basic circuit

Example

Now add additional

reactive power load

and resolve

70.7 0.7 lagging

564 45 Amps

59.7 13.6 kV

S 33.7 58.6 MVA 17.6 28.8 MVA

LoadZ pf

Balanced 3 Phase () Systems

A balanced 3 phase () system has

three voltage sources with equal magnitude, but with an angle shift of 120

equal loads on each phase

equal impedance on the lines connecting the generators to the loads

Bulk power systems are almost exclusively 3

Single phase is used primarily only in low voltage, low powersettings, such as residential and some commercial

Balanced 3 -- No Neutral Current

* * * *

(1 0 1 1

n a b c

an an bn bn cn cn an an

I I I I

S V I V I V I V I

Advantages of 3 Power

Can transmit more power for same amount of

wire (twice as much as single phase)

Torque produced by 3 machines is constant

Three phase machines use less material for

same power rating

Three phase machines start more easily than

single phase machines

Three Phase - Wye Connection

There are two ways to connect

3 systems

Wye (Y)

Delta ()

Wye Connection Voltages

Wye Connection Line Voltages

VabVca

(1 1 120

ab an bn

V V V V

Line to line

voltages are

also balanced

Wye Connection

Define voltage/current across/through device

to be phase voltage/current

Define voltage/current across/through lines to

be line voltage/current

3 1 30 3

Line Phase Phase

Line Phase

Phase Phase

V V V e

Delta Connection

IabIbc

For the Delta

phase voltages equal

line voltages

For currents

Phase Phase

Three Phase Transmission Line

Wye Connection Line Voltages

VabVca

(1 1 120

ab an bn

V V V V

Line to line

voltages are

also balanced

Wye Connection Line Voltage

Define voltage/current across/through device to be phase

voltage/current

Define voltage/current across/through lines to be line

voltage/current

3 1 30 3

Line Phase Phase

Line Phase

Phase Phase

V V V e

Delta Connection

IabIbc

For the Delta

phase voltages equal

line voltages

For currents

a ab ca

b bc ab

a ca bc

Phase Phase

Three Phase Example

Assume a -connected load is supplied from a 3 13.8 kV

(L-L) source with Z = 1020W

13.8 0

13.8 0138 20

138 140 138 0

kVI amps

I amps I amps

138 20 138 0

239 50 amps

239 170 amps 239 0 amps

3 3 13.8 0 kV 138 amps

5.7 MVA

5.37 1.95 MVA

pf cos 20 lagging

a ab ca

Delta-Wye Transformation

To simplify analysis of balanced 3 systems:

1) Δ-connected loads can be replaced by 1

Y-connected loads with Z3

2) Δ-connected sources can be replaced by

Y-connected sources with V3 30

Per Phase Analysis

Per phase analysis allows analysis of balanced 3 systems

with the same effort as for a single phase system

Balanced 3 Theorem: For a balanced 3 system with

All loads and sources Y connected

No mutual Inductance between phases

Per Phase Analysis

All neutrals are at the same potential

All phases are COMPLETELY decoupled

All system values are the same sequence assources. The sequence order we’ve been using(phase b lags phase a and phase c lags phase a) isknown as “positive” sequence; later in the coursewe’ll discuss negative and zero sequence systems.

Per Phase Analysis Procedure

To do per phase analysis

1. Convert all load/sources to equivalent Y’s

2. Solve phase “a” independent of the other phases

3. Total system power S = 3 Va Ia*

4. If desired, phase “b” and “c” values can be

determined by inspection (i.e., ±120° degree phase

shifts)

5. If necessary, go back to original circuit to

determine line-line values or internal values.

Per Phase Example

Assume a 3, Y-connected generator with Van = 10volts supplies a -connected load with Z = -j through a transmission line with impedance of j0.1 per phase. The load is also connected to a -connected generator with Va”b” = 10 through a second transmission line which also has an impedance of j0.1per phase.

1. The load voltage Va’b’

2. The total power supplied by each

generator, SY andS

Per Phase Example

First convert the delta load and source to equivalent

Y values and draw just the "a" phase circuit

Per Phase Example

' ' 'a a a

To solve the circuit, write the KCL equation at a'

1(V 1 0)( 10 ) V (3 ) (V j

Per Phase Example

' ' 'a a a

' 'a b

' 'c ab

To solve the circuit, write the KCL equation at a'

1(V 1 0)( 10 ) V (3 ) (V j

10(10 60 ) V (10 3 10 )

V 0.9 volts V 0.9 volts

V 0.9 volts V 1.56

j j j j

Per Phase Example

S 3 5.1 3.5 VA0.1

3 5.1 4.7 VA0.1

a aa a a

a agen a

V VV I V j

V VS V j

Delta-Star Transformation

Start-Delta Transformation

THANK YOU!

For Contact

Email: arslan_engineer61@yahoo.com

LinkedIn: https://pk.linkedin.com/pub/arslan-ahmed-amin-p-e-b-sc-ee-m-sc-ee-m-b-a/24/853/68

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