AC Circuits 1004121

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AC CircuitsLecture Notes

EEE 165

Basic Electrical Technology

Course Teacher: Mr. Hafiz Imtiaz

BANGLADESH UNIVERSITY OF ENGINEERING AND TECHNOLOGY

Submitted by:

Saad Abd Ar Rafie

1004121

L-1/T-2, Section B

Department of Civil Engineering

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AC Circuits Lecture Notes

1004121 Page 1

LECTURE12 Date: 08-10-2011

Alternating Current (AC)

A current that reverses at regular time intervals and has alternately positive and

negative values is an alternating current. An alternating current is usually referred

to as sinusoidal current.

A sinusoid is a signal that has the form of the sine or cosine function. Thus, circuits

driven by sinusoidal current or voltage sources are calledac circuits.

Let us consider the alternating current ,

i (t) = Imsin (t)

where,

Im = the amplitude of the current

= the angular frequency in radians/s

t= the argument of the sinusoid

The magnitude of current will be 0 (zero) when its amplitude is also zero.

Alternating current will be transformed into direct current when= 0.

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It is evident that the sinusoid repeats itself every Tseconds ; thus Tis called

theperiodof the sinusoid. From the figure included above, we observe that

T = 2

T = i(t) repeats itself every Tseconds is shown by replacing t by t+T . So, we get

i (t+T) = Imsin (t+T) =Imsin (t + )

= Imsin (t + 2) = Imsin (t) = i (t)

ihas the same value at (t+T)as it does at tand so i(t)is said to beperiodic.

Cycle : The portion of a waveform contained in one period of time.

Frequency : The number of cycles that occur in 1 s. The frequency of the

waveform of Fig. 13.5(a) is 1 cycle per second, and for Fig. 13.5(b), 12

cycles per second. If a waveform of similar shape had a period of 0.5 s [Fig.

13.5(c)], the frequency would be 2 cycles per second.

Thus we get,

f =

.

So, finally we get the relation = 2f, which is the angular frequency.

A more general expression for sinusoid,

i (t) = Imsin (t + )

where,

(t + )is the argument and is thephase.

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Sinusoids are easily expressed in terms of phasors, which are more convenient to

work with than sine or cosine functions.

A phasor is a complex number that represents the amplitude and phase of a

sinusoid. A complex number, can be written in the rectangular form as := +

Where, z=, is the real part and is the imaginary part.Polar form: z= r

Exponential form: z= Given and, we can get r and as

r = + = t n

On the other hand we can obtain

and

as

= rs = r sin

So, z can be written as

=+= r= r (s + sin Eulers Identity:

= ssi n

Where,

s=s i n =

Im axis

Re axis

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Given a sinusoid for alternating voltage,

= s + = = =

where,

= = For alternating current,

= s + =

=

where,

= = = = s + = {} *

= {}

Thus, time domain representation may be transformed into phasor domain

representation.

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Instantaneous expression for reactance:

= s +

Part of this, phasor expression: = Instantaneous Phasor

s + s + Impedance: Inductance

= = sin= s+

= = =

Inductive resistance, = Note: Generally, the instantaneous expression is converted into the cosine

function to obtain the phasor domain representation.

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Impedance: Capacitance

= s =

= = * sin= s+ =

=For AC circuits,

=

= =2 2

=

Capacitative reactance,

=

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The RL Branch:

= + =

Impedance, = + The RC Branch:

= s +

= 2 Impedance, =

=

= 2 =

=

2

=

2

= 2 + = +

= s

= s +

=

2 =

=

=

sin=

s

=

2

= =

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The RLC Branch:

= s = s+ =

s

= 2 + = So,

Impedance,

= +

= 2 = 2 =

2

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Calculation of Power:

= + So, || = + 2

= tn

Instantaneous Power,

=

= ss= *s2+s

The angle difference between voltage and current is the angle of impedance

=

= =

= ||

= s = s || =

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To get the average, well divide the whole thing by 2 instead of T and integrate:

*s2+s

= s + s2

Here,

s2 =

s

= ,sin-

= sin+sin =

Instantaneous Power, = *s2+s

Average Power = s

= |||| sSo, the value for average power can be obtained by the multiplication of the

phasor amplitude by the modulus of voltage, current and cos.

Let, 2 = 2 = 0 2 2

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Stored Power for an Inductor:

= ||

= 2 sin2

= ,s2-

=

s s

= 2=

=

=

+

=

=

=

s

s

+

= 2 { sin2}

= sin= s +

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Stored Power for a Capacitor:

= =

ss+ =

2sin2

= || =

2 sin2

= ,s2- = 2

=

=

=

= s

= * sin=

s +

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Phasor Diagram:

In a series circuit the angle of the current is zero. So, it is taken as the reference

quantity. All the angles are considered with respect to the reference quantity.

VR=IR, so angle ofVRwill be zero

VL=I XL XL=XL= XL So, the angle of VLwill be +90with respect to I.

In the whole circuit

VL+ VR= VS

VL

VRI

VLVR+ + --

VS

= += tn

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Phasor Diagram:

At first a reference point has to be selected, which is parallel to the X axis. All the

angles are then calculated with respect to the X axis. Here, the current is selected

as the reference quantity and set in a direction long the X axis.

VS

R

L

VR= Voltage across the resistor

VL= voltage across the inductor

IXL

I VR

VL

=

+

= tn

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=

=

tn

=So, is at an angle of -90with respect to IR.

VR+ -

VC

+

-

R I

VC

VS

VR

Phasor diagram

VS

VC

VL+ - + -

VR1

VR2

R1

R2

XL

+-XC

-

+

-

VL

VC

VR2

VR1

Phasor Diagram

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Series Resonance:

|| = +

= . /In a series circuit the magnitude of Iis maximum at a certain frequency. This

frequency is called the resonance frequency.

Here, Ishould be differentiated with respect to frequency.

=

(

+ . /)=

= = = At resonance point,

XL= XC

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Wye-Delta Connection:

Balanced 3-phased Circuits:

Wye

Z1

Z2Z3

ZBC

a

ZA

a

b bcc

n

Van

Vbn

Vcn

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The voltage between a to b, b to c or c to a is called line to line voltage. The

voltage from the phase to neutral is said to be the line to neutral voltage.

Delta

Line to line voltage =

Zbc

aa

b bcc

Van

Vbn

Vcn

Vbc

Vca

Vab

b 2

c 24

Reference quantity

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AC Circuits 1004121