Module 1AC Circuits: Basic

Principles

Engr. Gerard AngSchool of EECE

Types of Electrical Current

• Direct Current (DC). It is electric current which flows in one direction only.

• Alternating Current (AC). It is electric current that reverses direction periodically usually many times per second.

Generation of Alternating Current and Voltages

• Alternating voltage may be generated by:Rotating a coil in a magnetic field Rotating a magnetic field within a stationary coil

• Altering the direction of the magnetic field

Generation of Alternating Current and Voltages

Faraday’s Laws of Electromagnetic

Induction First Law. An emf is induced in a coil whenever the flux linking the coil changes with time.

Second Law. The magnitude of the induced emf in an N-turn coil is equal to the time rate of change of the magnetic flux through it

φde N

dt=

Where:e = induced emf in voltsN = number of turns of the

coildφ/dt = rate of change of

magnetic flux in Webers per sec

Induced EMF

It is emf resulting from the motion of a conductor through a magnetic field, or from a change in the magnetic field that threads a conductor.

e B v=

Where:e = dynamically induced

emf in volts

B = flux density of uniform magnetic field in Tesla

l = length of the inductor in m

v = velocity of the conductor in m/sec

Fleming’s Left Hand Rule

Thumb = direction of forceForefinger = direction of field.Middle finger = direction of current

Fleming’s left hand rule is used to determine the direction of the force acting on a conductor. With your left hand, stretch out the thumb, forefinger and middle finger so that these are at right angles with each other.

Importance of AC

It can be generated at comparatively high voltage and can be raised or lowered by means of a transformer.

Transmission of power over long distances is much more economical with alternating than direct current.

It can be built in large unit of high speed, unlike in dc due to commutation problems.

Induction motor (ac motor) are more efficient than dc motor at constant speed work and less in first cost because ac motor does not have commutator.

AC Waveform Terminologies

2ππ0

NegativePeak (-Em)

PositivePeak (Em)

CycleWavelength

Period

+

-

Sinusoidal Wave

1. Waveform – it is the shape of the curve obtained by plotting the

instantaneous values of voltages or currents as the ordinate against time as the abscissa.2. Cycle – it is a complete set of

positive and negative value alternation of sinusoidal wave.3. Alternation – It is one half cycle of a complete set of positive and negative values.

1 Revolution = 360 Electrical deg.

= 180P Mechanical deg.

Where: P = number of poles (even)

AC Waveform Terminologies

4. Period of the wave (T) – It is the time taken in seconds by an alternating quantity to complete one cycle.

5. Frequency of the wave (f) – It is the number of cycles produced per second or Hertz (Hz) by an alternating quantity.

Where: n = shaft speed rotation in rev per min (rpm)

f = frequency in Hertz

6. Wavelength (λ) – it is the length of one complete wave or cycle or the distance traveled by the wave form in one cycle.

𝐟=𝐩𝐧𝟏𝟐𝟎

𝐟=𝐩𝐧𝟏𝟐𝟎

𝐓=𝟏𝐟

𝐓=𝟏𝐟

AC Waveform Terminologies7. Instantaneous value of a sinusoidal wave – it is the magnitude of the

wave at any instant.

8. Maximum or Peak value of a sinusoidal wave – it is the maximum value (positive or negative) attained by an alternating voltage or current.

9. Peak-to-peak value of a sinusoidal wave – it is the value from the positive peak or negative peak or vice versa. It is always twice the peak value.

θ

10. Phase or Phase Angle (θ) – it is the fractional part of a period or cycle though which the quantity has advanced or

delayed from selected origin.

Note:If the waveform starts before the y-axis, it will have a positive phase angle.If the waveform starts after the y-axis, it will have a negative phase angle.

AC Waveform Terminologies

11. Phase Difference – it is the difference between the phases of two or more alternating quantity of the same frequency which do not reached their

maximum or zero value simultaneously.

θlag’g

θlead’g

θpd

Where:θpd = phase differenceθleading = phase of leading quantityθlagging = phase of lagging quantity

Note: Ifθpd = (+) denotes “leading” phase difference = (-) denotes “lagging” phase difference

Sample ProblemsFind the phase difference of the following waveforms:

a. i = 10sin(377t + 25°)

v = 200cos(377t – 20°)

b. i1 = 5cos(377t – 20°)

i2 = 10cos(377t – 30°)

c. i1 = -10sin(314t – 30°)

i2 = 40cos(314t – 10°)

d. v1 = 20cos377t

v2 = 50sin(314t + 20°)

AC Waveform Terminologies

12. Average value or mean value of a sinusoidal wave – it is defined as that steady quantity which transfers across any circuit the same charge as is transferred by that alternating quantity during the same time. It is also the arithmetical average of all the values of an alternating quantity over one cycle.

13. Root-mean-square (RMS) value or effective value of a sinusoidal wave – It is defined as that steady current which when flowing through a given resistance for a given time produces the same amount of heat as produced by the alternating current when flowing through the same resistance for the same time.

𝐕𝐚𝐯𝐞=𝟏𝐓∫𝟎

𝐓

𝐯 (𝐭 )𝐝𝐭𝐕𝐚𝐯𝐞=𝟏𝐓∫𝟎

𝐓

𝐯 (𝐭 )𝐝𝐭

𝐕 𝐫𝐦𝐬=√ 𝟏𝐓∫𝟎

𝐓

[𝐯 (𝐭 )]𝟐𝐝𝐭𝐕 𝐫𝐦𝐬=√ 𝟏𝐓∫𝟎

𝐓

[𝐯 (𝐭 )]𝟐𝐝𝐭

AC Waveform Terminologies

14. Form Factor – It is the ratio of the RMS value or effective value to the average value of an alternating quantity.

15. Peak factor or Crest factor or Amplitude Factor – It is the ratio of the maximum value to the RMS value or effective value of an alternating quantity.

𝐅𝐨𝐫𝐦𝐅𝐚𝐜𝐭𝐨𝐫=𝐑𝐌𝐒𝐕𝐚𝐥𝐮𝐞𝐀𝐯𝐞𝐫𝐚𝐠𝐞𝐕𝐚𝐥𝐮𝐞

𝐅𝐨𝐫𝐦𝐅𝐚𝐜𝐭𝐨𝐫=𝐑𝐌𝐒𝐕𝐚𝐥𝐮𝐞𝐀𝐯𝐞𝐫𝐚𝐠𝐞𝐕𝐚𝐥𝐮𝐞

𝐏𝐞𝐚𝐤 𝐅𝐚𝐜𝐭𝐨𝐫=𝐌𝐚𝐱𝐢𝐦𝐮𝐦𝐕𝐚𝐥𝐮𝐞𝐑𝐌𝐒𝐕𝐚𝐥𝐮

𝐏𝐞𝐚𝐤 𝐅𝐚𝐜𝐭𝐨𝐫=𝐌𝐚𝐱𝐢𝐦𝐮𝐦𝐕𝐚𝐥𝐮𝐞𝐑𝐌𝐒𝐕𝐚𝐥𝐮

Average and RMS Value and Form and Peak Factor for Various Waveforms

Type of Waveform

Wave ShapeRMS Value

Average Value

Form Factor

Peak Factor

Sine Wave 1.11 1.41

Half-Wave Rectified Sine Wave

1.57 2.0

Full-Wave Rectified Sine Wave

1.11 1.41

Average and RMS Value and Form and Peak Factor for Various Waveforms

Type of Waveform

Wave ShapeRMS Value

Average Value

Form Factor

Peak Factor

Rectangular Wave Vm Vm 1.0 1.0

Triangular Wave 1.16 1.73

Sample Problems1. Compute for the average and effective values of thesquare voltage wave shown below.

2. Calculate the RMS value of the function shown below if it is given that for 0 < t < 0.1, v = 10(1 – e-100t) and for 0.1 < t < 0.2, v = 10e-50(t – 0.1).

0 0.1 0.2 0.40.3t

20 V

v

0 0.1 0.2 0.40.3 t (seconds)

10 V

v

Sample Problems

3. Find the average and effective values of the saw-tooth waveform shown.

4. The waveform of an output current is as shown in the figure. It consists of a portion of the positive half cycle of a sine wave between the angle θ and 180°. Determine the effective value for θ = 30°.

10 V

v

v

1 320

Sample Problems5. Calculate the r.m.s. and average value of the voltage wave shown in the figure below.

4

1 2 3

2

-4

0

-2

Equations of AlternatingCurrent and Voltage

Any sinusoidal quantity can be expressed as

𝐞 ( 𝐭 )=𝐄𝐦𝐬𝐢𝐧 (𝛚𝐭 ±𝛉)𝐞 ( 𝐭 )=𝐄𝐦𝐬𝐢𝐧 (𝛚𝐭 ±𝛉)Where:e(t) = instantaneous value of voltagei(t) = instantaneous value of currentEm = maximum value of voltage Im = maximum value of currentt = time in secondsθ = angle of rotation or phase angle in

degreesN = number of turns of the coilBm = maximum flux densityA = area of the coilω = angular velocity in rad per sec

ω = 2πf

ii

𝐞 ( 𝐭 )=𝛚𝐍𝛟𝐦=𝛚𝐍𝐁𝐦𝐀𝐞 ( 𝐭 )=𝛚𝐍𝛟𝐦=𝛚𝐍𝐁𝐦𝐀

Sample Problems

1. The maximum values of the alternating voltage and current are 400 V and 20 A respectively in a circuit connected to a 50 Hz supply and these quantities are sinusoidal. The instantaneous values of the voltage

and current are 283 V and 10 A respectively at t = 0 both increasing positively. Write down the expression for current and voltage at time t.

2. An alternating current of frequency 60 Hz has a maximum value of 120 A. Write down the equation for the instantaneous value. Reckoning time from the instant the current is zero and is becoming positive, find (a) the instantaneous value after 1/360 second and (b) the time taken to reach 96 A for the first time.

Sample Problems3. An alternating current of frequency 50 Hz has a positive maximum value of 100 A. Calculate (a) its value after 1/600 second after the instant the current is zero and its value decreasing there afterwards (b) How many seconds after the instant the current is zero (increasing thereafter wards) will the current attain the value of 86.6 A? 

4. An alternating current varying sinusoidally with a frequency of 50 Hz has an RMS value of 20 A. Write down the equation for the instantaneous value and find this value (a) 0.0025 second (b) 0.0125 second after passing through a positive maximum value. At what time, measured from a positive maximum value, will the instantaneous current be 14.14 A?

Harmonics

Harmonics or Non-Sinusoidal or Distorted or Complex waveform - these are alternating waveforms which deviate to a greater or lesser degree. Complex waveforms are produced due to superposition of sinusoidal waves are different frequencies. Such waves occur in speech, music, TV, rectifier outputs and many applications of electronics.

Harmonics• Types of Harmonics

a. Even Harmonics - these are waves having frequencies of 2f, 4f, 6f, etc. or 2w, 4w, 6w.

b. Odd Harmonics - these are waves having frequencies of 3f, 5f, 7f, etc. or 3w. 5w, 7w.

• General Equation of a Complex Wave

The general equation of a complex wave is given as:

𝐞=𝐄𝟏𝐦 𝐬𝐢𝐧(𝛚𝐭+𝛟𝟏)+𝐄𝟐𝐦𝐬𝐢𝐧(𝟐𝛚𝐭+𝛟𝟐)+⋯+𝐄𝐧𝐦 𝐬𝐢𝐧(𝐧𝛚𝐭+𝛟𝐧)𝐞=𝐄𝟏𝐦 𝐬𝐢𝐧(𝛚𝐭+𝛟𝟏)+𝐄𝟐𝐦𝐬𝐢𝐧(𝟐𝛚𝐭+𝛟𝟐)+⋯+𝐄𝐧𝐦 𝐬𝐢𝐧(𝐧𝛚𝐭+𝛟𝐧)

Where: E1m sin (ωt + φ1) = fundamentalE2m sin (ωt + φ2) = second harmonicEnm sin (ωt + φn) = nth harmonic

 

Harmonics

• RMS Value of a Complex Wave

𝑬 𝒓𝒎𝒔=√𝑬𝒅𝒄𝟐+

𝑬𝟏𝒎𝟐

𝟐+𝑬𝟐𝒎

𝟐

𝟐+⋯+

𝑬𝒏𝒎𝟐

𝟐𝑬 𝒓𝒎𝒔=√𝑬𝒅𝒄

𝟐+𝑬𝟏𝒎

𝟐

𝟐+𝑬𝟐𝒎

𝟐

𝟐+⋯+

𝑬𝒏𝒎𝟐

𝟐

Where: Edc = dc component of the harmonic

Similarly,

𝑰 𝒓𝒎𝒔=√ 𝑰𝒅𝒄𝟐+ 𝑰𝟏𝒎𝟐𝟐 +𝑰𝟐𝒎

𝟐

𝟐+⋯+

𝑰𝒏𝒎𝟐

𝟐𝑰 𝒓𝒎𝒔=√ 𝑰𝒅𝒄𝟐+ 𝑰𝟏𝒎𝟐𝟐 +

𝑰𝟐𝒎𝟐

𝟐+⋯+

𝑰𝒏𝒎𝟐

𝟐

Harmonics

• Power Supplied by a Complex Wave

The total average power supplied by a complex wave is the sum of the average power supplied by each harmonic component acting independently.

𝑷=𝑬𝟏𝒎 𝑰𝟏𝒎𝟐

𝐜𝐨𝐬 (𝜶𝟏− 𝜷𝟏 )+𝑬𝟐𝒎 𝑰𝟐𝒎𝟐

𝐜𝐨𝐬 (𝜶𝟐− 𝜷𝟐 )+⋯+¿𝑬𝒏𝒎 𝑰𝒏𝒎𝟐

𝐜𝐨𝐬 (𝜶𝒏− 𝜷𝒏 )¿𝑷=𝑬𝟏𝒎 𝑰𝟏𝒎𝟐

𝐜𝐨𝐬 (𝜶𝟏− 𝜷𝟏 )+𝑬𝟐𝒎 𝑰𝟐𝒎𝟐

𝐜𝐨𝐬 (𝜶𝟐− 𝜷𝟐 )+⋯+¿𝑬𝒏𝒎 𝑰𝒏𝒎𝟐

𝐜𝐨𝐬 (𝜶𝒏− 𝜷𝒏 )¿

Sample Problems

A complex voltage is given by e = 60 sin ωt + 24 sin (3ωt + π/6) + 12 sin (5ωt + π/3) is applied across a certain circuit the resulting current is given by i = 0.6 sin (ωt - 2π/10) + 0.12 sin (3ωt - 2π/24) + 0.1 sin (5ωt - 3π/4).

Find:

(a) rms value of current and voltage

(b) total power supplied