Ee1 chapter11 ac_circuits

Apr 14, 2023

Lecturer Name [email protected]

Contact Number

IT2001PAEngineering Essentials (1/2)

Chapter 11 – Alternating Current Circuits


IT2001PA Engineering Essentials (1/2) 2

Lesson Objectives

Upon completion of this topic, you should be able to: Explain and calculate the fundamentals of alternating

current waveform.


IT2001PA Engineering Essentials (1/2)

Specific Objectives

Students should be able to : Describe the basic construction of a simple alternator. Describe the characteristics of an alternating quantity as

fluctuating over a given period of time. Define the following terms with reference to an alternating

quantity : Frequency Average value Instantaneous value Maximum value


IT2001PA Engineering Essentials (1/2)

Specific Objectives

Students should be able to : State that the frequency of the generated emf is

proportional to the speed and number of poles of the alternator.

Explain the term Root Mean Square (RMS) value with reference to an alternating current.

State the following equations for a sinusoidal waveform :

RMS value = 0.707 x Maximum value

Average value = 0.637 x Maximum value



Alternating Waveform

Alternating Current (ac)

current that is continuously reversing

direction, alternately flowing in one

direction and then in the other.

can be similarly described.

The designation ac is normally applied to

both current and voltage.

Alternating Voltage



Alternating Waveform

Vertical axis : current (I) or voltage (V)

t

V+

V-

t

I+

I-

Horizontal axis : time (t)



Why not Direct Current (DC)?

Problems of using DC as power distribution system: suffers from rapid power loss

in the wires due to their resistance, which dissipates energy as heat

DC power stations had useful ranges of about two kilometers

Once generated, DC power cannot be modified

t

V+

V-



Power Distribution System

1)To prevent risk of sparks and short circuits in the generator itself, power needs to be generated with low voltages.

V low I is high

V=RI

2)To avoid serious heating effects, present at high current, it needs to be transmitted at the lowest current practical.

I low V is high

3)For the safety of the user, the voltage needs to be low in the home.

V low I is high

DC is not suitable



Advantages of A.C.

--- readily available from generators and power supply sockets in homes and workshops.--- can be stepped up or down by the use of a transformer.--- for transmission purpose.



Alternator

An alternator is a machine that converts mechanical energy to electrical energy.A simple alternator basically consists of two parts:

a) coils or windingsb) magnetic poles



Generation of Alternating CurrentAn alternating quantity may be generated by

a) rotating a coil in a magnetic field

b) rotating a magnetic field within a stationary coil

Carbon Brushes 1, 2

Slip Rings a, b



Generation of Alternating Current

x

xx x

x

How does AC generate a sine waveform?








x

x




Generation of Alternating Current AC generator consists of a magnet and a loop of

wire which rotates in the magnetic field of the magnet.

As the wire rotates in the magnetic field, the changing strength of the magnetic field through the wire produces a force which drives the electric charges around the wire.

The force initially generates an electric current in one direction along the wire. Then as the loop rotates through 180 degrees the force reverses to give an electric current in the opposite direction along the wire. Every time the loop rotates through 180 degrees the direction of the force and therefore the current changes.

The changing direction of the force after every 180 degrees of rotation gives the alternating current.

AC generator also has slip rings which make sure that the ends of the wire are always connected to the same side of the electric circuit. This makes sure that the direction of the current changes every half revolution of the wire.



Generation of Alternating emf

The magnitude of the emf is given by,

e = Em sin



Sine Wave

A common type of ac.

Also referred assinusoidal waveformsinusoid

Symbol for a sine wave voltage source :

OR



What is an Alternating Quantity?

An alternating quantity is one which acts in alternate directions and whose magnitude undergoes a definite cycle of changes in definite interval of time.



Polarity

When the voltage changes polarity at its zero crossing,the current correspondingly changes direction.

V+

t

V-

VS

+

-R

-

+VS R

Positive Alternation

Negative Alternation

I

I

t

V+

V-



Amplitude

The maximum value of the sine wave.

t

V+

V-

Positive Maximum (peak)

Negative Maximum (peak)



What is Cycle of a Waveform?

One complete waveform is known as one cycle.Each cycle consists of two half-cycles.During 1st half-cycle, the quantity acts in one direction andduring the second half-cycle, in the opposite direction.

Each cycle :

consists of 2 alternations.

consists of 2 peaks.

reaches maximum amplitude 2 times.



Cycle

A sine wave repeats itself in identical cycles.

t

V+

V- 1st cycle 2nd cycle 3rd cycle



Period

The time required for a given sine wave to complete one full cycle.

symbol - T

unit - second (s)

Time taken is the same for each cycle;

thus fixed value for a given sine wave.



Period Measurement

From one zero crossing to the corresponding zero crossing in the next cycle

t

V+

V- period period period

positive zero crossing to positive zero crossing.



Period Measurement

From one peak to the corresponding peak in the next cycle

t

V+

V- period period

positive peak to positive peak.



Period Measurement

From any point to the corresponding point in the next cycle.

t

V+

V-

period

period



Example 11-1

What is the period of the sine wave?

V

t(s)2 4 60

1 cycle takes 2s to complete.Therefore the period is 2s.



Example 11-2

What is the period of the sine wave?

5 cycles takes 12s to complete

V

t(s)120

T =time taken

no. of cycles

1 cycle takes (12/5)s to complete

1 cycle takes 2.4s to completeTherefore the period is 2.4s.



Frequency

The number of cycles a sine wave can complete in 1 second.

symbol : f

unit - Hertz (Hz) Examples :

160Hz - 160 complete cycles in 1s.

50Hz - 50 complete cycles in 1s.



Frequency vs Period

Reciprocal relationship :

More cycles in 1s Lesser cycles in 1s

V

t0

1s

TV

t0

1s

T

1f

T =

- higher frequency. - lower frequency.- shorter period. - longer period.

f = 1T



Example 11-3

Which sine wave has the higher frequency? Determine the period and the frequency of both waveforms.

V

0 1s

Waveform A

t t

V

0 1s

Waveform B



Example 11-3

continue …V

t0 t

V

01s 1s

Waveform A Waveform B

Waveform B complete more cycles in 1s.Therefore Waveform B has the higher frequency.



Example 11-3

continue …

V

t0 1s

Waveform A

3 cycles in 1s -- frequency is 3Hz

3 cycles take 1s-- 1 cycles takes (1s/3) = 0.333s-- period is 0.333s

OR

period (T) = 1/f = 1/3Hz = 0.333s

frequency (f) = 1/T = 1/0.333s = 3Hz



Example 11-3

continue … 5 cycles in 1s -- frequency is 5Hz

t

V

0 1s

Waveform B 5 cycles take 1s-- 1 cycles takes (1s/5) = 0.2s-- period is 0.2s

period (T) = 1/f = 1/5Hz = 0.2s

OR

frequency (f) = 1/T = 1/0.2s = 5Hz



Example 11-4

The period of a sine wave is 10ms. What is the frequency?

period (T) = 10ms

frequency (f) = 1/T

= 1/10ms

= 100Hz



Example 11-5

The frequency of a sine wave is 60Hz. What is the period?

frequency (f) = 60Hz

period (T) = 1/f

= 1/60Hz

= 16.7ms



Voltage Sources

If an ac voltage is applied to a circuit, an ac current flows.The voltage and current will have the same frequency.

t

V+

V-

t

I+

I-

I



Sine Wave Angles

The horizontal axis can be replaced by angular measurement (degrees).

angle

V+

V-

0 o

½ cycleone cycle

90 o

peak270 o

peak360o

zero180o

zero



Sine Wave Values

Ways to express and measure the value of a sine wave :

1) instantaneous value.2) peak value.3) peak-to-peak value.4) root-mean-square value.5) average value.



Instantaneous Value

The voltage or current value of a waveform at a given instant in time.

Symbol :voltage - vcurrent - i



Instantaneous Value

Different at different points of time.

t

V+

V-

t

I+

I-

5.53

8.5

5

-6

t1 t2

t3

-4.5

t1 t2

t3



Instantaneous Value Different at different points of angle.

degree

V+

V- v = Vmax sin

degree

I+

I- i = Imax sin

Vp

v

Ip

i

v = Vmax sin 2ft i = Imax sin 2ft



Example 11-6

A current sine wave has a maximum value of 200A. How much is the current at the instant of 30 degrees of the cycle?

i = Imax sin

i = 200 sin 30 o

= 200 x 0.5 = 100A



Peak Value

The voltage or current value of a waveform at its maximum positive or negative points.

Symbol :

voltage - Vp or Vmax

current - Ip or Imax



Peak Value

The peaks are equal for a sine wave and is characterized by a single peak value.

t

V+

V-

Positive Maximum (peak)

Negative Maximum (peak)



Peak-to-Peak Value

The voltage or current value of a waveform measured from its minimum to its maximum points.

Symbol :

voltage - Vpp

current - Ipp



Peak-to-Peak Value

Vpp = 2Vmax

t

V+

V-

Vp

Vp

Vpp

Vpp = 2Vmax

12 Vpp = Vmax 1

2 Vmax = Vpp

Vmax = 0.5Vpp



Root Mean Square Value

Equal to the dc voltage that produces the same amount of heat in a resistance as does the AC (sine wave) voltage.

I

+

-Vdc R

sameamount of

radiated heat.

I

+

-VS R

radiatedheat

Vmax = 10 V

Vrms = 7.07V = Vdc




Also known as the effective value.

Symbol :voltage - Vrms

current - Irms

The voltage and current values given are usually as rms unless otherwise stated.




Vrms = Vmax

12

t

V+

V-

Vp

VrmsVrms = 0.707Vmax

2 Vrms = Vmax

Vmax = 2 Vrms

Vmax = 1.414 Vrms



Average Value

The average of a sine wave over half-cycle.

The average value taken over a complete cycle is always zero.

Symbol :voltage - Vavg

current - Iavg



Average Value

Vavg = 0.637 Vmax

t

V+

V-

Vp

Vavg

10.637 Vmax = Vavg

Vmax = 1.57 Vavg

10.637 Vavg = Vmax



Form & Peak Factor for Sine Wave

Form factor =

= 1.11

Avg value 0.707 X Max value0.637 X Max value=

Peak factor =

= 1.414

Rms value Max value

Max value0.707 X Max value=

Rms value



Example 11-7

Determine Vmax when :

Vpp = 3V ; Vrms = 5V ; Vavg = 4V.

Vmax = 0.5 Vpp

= 0.5 (3) = 1.5V

Vmax = 1.414 Vrms

= 1.414 (5) = 7.07V

Vmax = 1.57 Vavg

= 1.57 (4) = 6.28V



Example 11-8

Determine Vrms when

Vp = 20V ; Vpp = 10V ; Vavg = 30V.

Vrms= 0.707 Vmax

= 0.707 (20) =14.14V

Vmax= 0.5 Vpp

= 0.5 (10) = 5V

Vmax = 1.57 Vavg

= 1.57 (30) = 47.1V

Vrms = 0.707 Vmax

= 0.707 (5) = 3.535V

Vrms= 0.707 Vmax

= 0.707 (47.1) = 33.33V



Example 11-9

Determine Vavg when Vmax = 37V ; Vpp = 28V ;

Vrms = 46V.

Vavg= 0.637 Vmax

= 0.637 (37) = 23.569V

Vmax= 0.5 Vpp

= 0.5 (28) = 14V

Vmax=1.414 Vrms

= 1.414 (46) = 65. V

Vavg= 0.637 Vmax

= 0.637 (65.) = 41.5V

Vavg= 0.637 Vp

= 0.637 (14) = 8.918V



Sine Wave Values

Vmax = 0.5Vpp

Vmax = 1.414 Vrms

Vmax = 1.57 Vavg

Vpp = 2Vmax

Vrms = 0.707Vmax

Vavg = 0.637 Vmax

10.637 = 1.57

10.707 = 1.414

Note:



Summary



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Ee1 chapter11 ac_circuits

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