Chapter 1 et201

CHAPTER 1

A L T E R N A T IN G V O L T A G E S A N D

C U R R E N T( W e e k 2 & 3 )

ET201: ELECTRICAL CIRCUITP O L IT E K N IK S U L T A N A B D U L H A L IM M U ’ A D Z A M S H A H

1.1) UNDERSTAND ALTERNATING CURRENT (AC)

1.1.1 Differentiate between AC and DC

• Direct Current (DC), which is electricity flowing in a constant direction, and/or possessing a voltage with constant polarity.

• Electricity made by a battery.

• DC circuit and waveform:


• AC is short for alternating current.

• This means that the direction of current flowing in a circuit is constantly being reversed back and forth.

• This is done with any type of AC current/voltage source.

• AC circuit and waveform:


1.1.2 AC is used in preference to DC

AC is commonly used because: - More voltage can be generated than DC - AC voltage can be increased and decreased with the help of static machine called transformers. - AC transmission and distribution is more economical as line material can be saved by transmitting power at higher voltage - AC can be converted to DC easily and required but DC cannot be converted to DC easily and will not be economical.

1.1.3 List the sources of alternating current AC

Source of AC: i) Rotating electrical machine (AC generator) - Rotating conductor - Rotating flux magnetic ii) Electronic Oscillator Circuit (electrical signal generator)


1.2 ) GENERATION OF ALTERNATING CURRENT1.2.1 Faraday’s and Lenz’s Law involved in generating a.c current

FARADAY LAW• The amount of voltage induced in a coil is directly proportional to the rate of

change of the magnetic field with respect to the coil.

• The amount of voltage induced in a coil is directly proportional to the number of turns of wire in the coil (N).


LENZ’S LAW

• Lenz’s Law is used to find the direction of induced e.m.f and hence current in a conductor or coil.

• Lenz’s Law is stated as follows:

The direction of the induced current is such as to oppose the change causing it.


1.2.2 AC waveforms produced by a simple alternating current generator (one loop in 2-pole magnet)


• Coil voltage versus angular position.


Sinusoidal voltages are produced by ac generators and electronic oscillators.

N S

M o t i o n o f c o n d u c t o r C o n d u c t o r

B

C

D

A A

B

C

D

A B

B

C

D

A C

B

C

D

A D

When a conductor rotates in a constant magnetic field, a sinusoidal wave is generated.

When the conductor is moving parallel with the lines of flux, no voltage is induced.

When the loop is moving perpendicular to the lines of flux, the maximum voltage is induced.

B

C

D

A


1.2.3 Develop an equation of a sinusoidal waveform, e = Em sin (ωt + θ) • From emf formula:

e = - d(Nφ) / dt volt

= - N dφ / dt

= - N d(φm cos ωt) /dt

= -Nφm ω (-sin ωt )

• So, formula become:

e = Nφm ω sin ωt

= Nφm ω sin θ (1)

• When the coil has turned through 900

i.e. when θ = 900, then sin θ = 1, hence e has maximum values, say Em .

• Therefore, from Eq. we get

Em = Nφm ω (2)


• Substituting eq (1) and (2), we get

e = Em sin θ = Em sin ωt

• Similarly, the equation of induced alternating current is

i = Im sin ωt.

• Since ω = 2πf , where f is the frequency of rotation of the coil, the above equations of the voltage and current can be written as

e = Em sin 2πf t and i = Im sin 2πf t.

• Instantaneous value : e = Em sin ωt

• Xne = Em sin (ωt + θ) where,

Em = maximum coil voltage (peak voltage)ωt = instantaneous angular position of the coil

ω is angular velocity = 2πfθ = phase shift (angle of lag or lead)

1.3) UNDERSTANDING A SINUSOIDAL VOLTAGE AND CURRENT VALUES


ET201: ELECTRICAL CIRCUITP O L IT E K N IK S U L T A N A B D U L H A L IM M U ’ A D Z A M S H A H1.3.1 Define frequency, period, peak value or amplitude and their relations

Cycle

Period (T)

Frequency (f)

Peak value (Vp) / Amplitude

The time interval for one complete cycle of a periodic waveform.

A complete consist of ½ cycle positive and ½ cycle negative.

A cycle within one second. Unit used is hertz (Hz)f = 1/T

T = 1/f

Value of voltage or current at the positive or negative maximum (peaks) with respect to zero.

Peak-to-peak value

rms value

Average value

Voltage or current from the positive peak to the negative peak.

Vpp = 2Vp or Ipp = 2Ip

Term rms stands for root mean square. RMS value referred to as the effective value and actually a measure of the heating effect of sine wave.

Vrms = 0.707Vp or Irms = 0.707Ip

Defined over a half cycle rather than over a full cycle.

Vavg = 0.637Vp or Iavg = 0.637Ip


1.3.2 Determine the various voltage and current values of a sine wave

Instantaneous value The value at different points along the curve.

Peak Factor Peak value over RMS valueForm Factor = Peak value RMS value

= Vp . 0.707Vp = 1.414


Form Factor RMS over average value.Form Factor = RMS value Average value

= 0.707Vp 0.637Vp = 1.11

1.3.3 Calculate mean value, rms value and peak factor for a given waveform.


0 V

1 0 V

- 1 0 V

1 5 V

- 1 5 V

- 2 0 V

t ( s )µ0 2 5 3 7 . 5 5 0 . 0

2 0 V

The peak-to-peak voltage is 40 V.

The rms voltage is VPP

Vrms

14.1 V


0 V

1 0 V

- 1 0 V

1 5 V

- 1 5 V

- 2 0 V

t ( s )µ0 2 5 3 7 . 5 5 0 . 0

2 0 V

The average value for the sinusoidal voltage is

12.7 V.

Vavg


Calculate Vp, Vpp, Vrms, Vavg, frequency and peak factor.

Vpp = 60V

Vrms = 21.21V

Vave = 19.11V

f = 125 Hz

Vp = 30V

Peak factor = 1.414

1.4) UNDERSTAND ANGULAR MEASUREMENT OF A SINE WAVE

1.4.1) Show how to measure a sine wave in terms of angles• Sinusoidal voltage can be produced electromagnetically by

rotating electromechanical machines. As the rotor of the ac generator goes through a full 3600 of rotation, the resulting output is one full cycle of a sine wave


1.4.2 Define radian

• A radian (rad) is the angle formed when the distance along the circumference of a circle is equal to the radius of the circle.

• One radian is equivalent to 57.300.


1.4.3 Convert radians to degree

Because there are 2π radians in one complete revolution and 360o in a revolution, the conversion between radians and degrees is easy to write. To find the number of radians, given the number of degrees:

Rad = degree X π180

To find the number of degrees, given the radians:

Degree = rad X 180π


1.4.4 Determine the phase angle of a sine wave

wherePhase angle

( )θω ±= tEe m sin

θ =


The phase angle of a sine wave is an angular measurement that specifies the position of that sine wave relative to a reference.

Phase difference refers to the angular displacement between different waveforms of the same frequency.

e = Em sin ωt

i = im sin ωt

e = Em sin ωt

i = im sin (ωt + θ)

e = Em sin ωt

i = im sin (ωt – θ)



Example of a wave that lags the referenceV

olt

ag

e (

V)

2 7 0 ° 3 6 0 °0 ° 9 0 ° 1 8 0 °

4 0

4 5 ° 1 3 5 ° 2 2 5 ° 3 1 5 °0

A n g le (° )

3 0

2 0

1 0

- 2 0

- 3 0

- 4 0

4 0 5 °

P e a k v o lt a g e

R e f e re n c e

Notice that a lagging sine wave is below the axis at 0o

v = 30 V sin (θ − 45o)

…and the equation has a negative phase shift


Vo

lta

ge

(V

)

2 7 0 ° 3 6 0 °0 ° 9 0 ° 1 8 0 °

4 0

4 5 ° 1 3 5 ° 2 2 5 ° 3 1 5 °0

A n g le (° )

3 0

2 0

1 0

- 2 0

- 3 0

- 4 0

P e a k vo lt a g e

R e f e re n c e

- 4 5 °- 1 0

Notice that a leading sine wave is above the axis at 0o

v = 30 V sin (θ + 45o)

…and the equation has a positive phase shift

Example of a wave that leads the reference

1.5) UNDERSTAND A PHASOR TO REPRESENT A SINE WAVE

1.5.1 Define phasor

The sine wave can be represented as the projection of a vector rotating at a constant rate. This rotating vector is called a phasor.


Phasors allow ac calculations to use basic trigonometry.

h y p o t e n u s e

θr i g h t a n g l e

o p p o s i t e s i d e

a d j a c e n t s i d e h y p o t e n u s eo p p o s i t e s i d e

s i n = θ

1.5.2 Explain how phasors are related to the sine wave

formula


00 9 0

9 0

1 8 01 8 0 3 6 0

1.5.3 Draw a phasor diagram


The position of a phasor at any instant can be expressed as a positive angle, measured counterclockwise from 0° or as a negative angle equal to θ − 360°.

positive angle of θ

negative angle of θ − 360°

phasor

v = = 1 9 . 2 V V p s i nV p

9 0 °

5 0 °0 °= 5 0 °

V p

V p

= 2 5 V

A plot of the example in the previous slide (peak at 25 V) is shown. The instantaneous voltage at 50o is 19.2 V as previously calculated.


1.5.4 Identify angular velocity


When a phasor rotates through 360° or 2π radians, one complete cycle is traced out.The velocity of rotation is called the angular velocity (ω).

ω = 2πf

(Note that this angular velocity is expressed in radians per second.)

• Sometimes voltages and currents are expressed in terms of cos ωt rather than sin ωt. For sines or cosines with an angle, the following formulas apply:

To illustrate, consider cos(ωt + 30°). From Equation above,

cos(ωt + 30°) = sin(ωt + 30° + 90°) = sin(ωt + 120°)


• Figure below illustrates this relationship graphically. The red phasor in as was shown


• Figure (a) generates cos ωt• Figure (b) generates sin ωt, and the green phasor generates a waveform that leads it by 120°

1.6) UNDERSTAND THE BASIC CIRCUIT LAWS OF RESISTIVE AC CIRCUITS

1.6.1 Apply Ohm's Law to resistive circuits with AC sources

• In general, Ohm's law for resistive AC circuits is given by:

Where,

V = voltage

I = current

R = resistance

V = IR

• When an AC circuit consists of a voltage source and a resistor, the current is in phase with the voltage, meaning that each quantity rises and declines in step.



This expression is also accurate for the maximum values of the potential and current. Where;

iRVR

Vi

ti

tR

Vi

iRt

iR

tV

mm

m

m

==∴

=

=

===

or

1sin when maximum become

sin

sinV therefore

sin

m

ω

ω

ων

ων

maxmaxor

RpeaktopeakV

peaktopeakI

R

VI

R

VavgavgI

−−=−−

==

tVm ων sin=

R


A series circuit consists of two resistors (R1 = 5 Ω and R2 = 15 Ω) and an alternating voltage source of 120 volts. What is Iavg?

Find the current and voltage drop at all the resistors in the circuit shown below:

1.6.2 Apply Kirchhoff's Voltage Law and Current Law to

resistive circuits with AC sources

Kirchhoff’s Law can be divided into 2:

• Kirchhoff’s Current LawAt any point in an electrical circuit where change density is not changing in time, the sum of current flowing towards that point is equal to the sum of currents flowing away from that point.

or

• Kirchhoff’s Voltage Law

The algebraic sum of various potential drops across an electrical circuit is equal to the electromotive force acting on the circuit

Vs = V1 + V2 + V3


Calculate value current that flows to each branch use Kirchhoff’s Law

Kirchhoff Current Law


Determine Vx

Kirchhoff Voltage Law

1.6.3 Determine power in resistive AC circuits

• This power, measured in watts, is the power associated with the total resistance in the circuit

• To calculate true power, the voltage and current associated with the resistance must be used.

1.7) USE AN OSCILLOSCOPE TO MEASURE WAVEFORM

1.7.1 Identify common oscilloscope controls

Oscilloscopes are commonly used to observe the exact wave shape of an electrical signal.

Type of electronic test instrument that allows observation of constantly varying signal voltages

http://en.wikipedia.org/wiki/Waveform

http://en.wikipedia.org/wiki/Electronic_test_instrument

http://en.wikipedia.org/wiki/Voltage

1.7.2 Use an oscilloscope to measure the amplitude,

period and frequency of a waveform

Chapter 1 et201

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