ELE130 Electrical Engineering 1

Week 5Module 3

AC (Alternating Current) Circuits

Administration

Quiz - next week in lecture material from weeks 4, 5 & 6

Laboratory Test postponed from next week to the week following

Easter break (from week 6 to week 7) will include work from laboratory 1, 2 & 3 laboratory no. 4 will be done in week 6 instead of

week 7

Help Desk

Why use ac?

In power applications easier to generate (no brushes or rectifiers) easier to transform voltages (reduced losses) easier to distribute (easier switching) easy utilization (brushless motors, switching)

In communications ac required to reproduce audio signals, to transmit over

distance (radio, microwave, etc.)

DC circuits are special case of ac circuits (when frequency is zero)

Parameters of an ac wave

v a va= v a( t) = V a( 0 -p k ) s in ( t+) = V m s in (t+)

t

Vm

-Vm

T

vs = Vm sin(t + )

Vm : amplitude of the wave

: radian frequency = 2f, f in Hz, in rad/sec

t : time in seconds t : (in radians)

: relative angle (in degrees or radians, but don’t MIX them up!)

Generation of an ac wave

t0o

45o

90o

135o

180o

225o270o

315o

360o 0 o 4 5 o 9 0 o 1 3 5 o

1 8 0 o

2 2 5 o 2 7 0 o 3 1 5 o 3 6 0 o

Note

f (cycles / sec) and (radians/sec) 1 cycle = 2 radians = 360o

angle can be specified in degrees or radians 90° = /2 radians

generally, frequencies will be the same (if not big hassles!)

generally waves will have different amplitudes is relative to reference. Of more importance

however is relative phase, or phase difference

Consider two waves

Phase difference (1- 2) (if frequency different, then relative phase changes

with time)

Leading and lagging waves using time line, the wave whose peak is occurs first

is leading

Phase ‘wraps’ around i.e. 20° = 380 ° usually work in range -180 ° to +180 °

Lagging and Leading Waveformsv

va

vb

a b

= a - b

Say a = + 30o an d b = -3 0o

th erefore = + 30o - (-30o) = 6 0o

Th erefore vb is lagging va by 6 0o

va is lead in g vref by a (i.e. by +3 0o)

vb is lagging vref by b (i.e. by -3 0o)

vref All vref, va an d vb areof that same freq uency

Example - what is the phase difference?

x1(t) = X1 sin(t - 30°) & x2(t) = X2 cos (t - 40°)

convert sin to cos : sin(x) = cos (x - 90o)

x1(t) = X1 cos ( t - 30o - 90o) = X1 cos (t - 120°)

Phase difference is |-120° - (-40°)|= 80°

x2(t) leads x1(t)

Inductor response to an ac source

L vL

iL

~vs(t) = V m sin (t)+

-

The current iL is:

as vL = vs:

by Ohm’s Law: |ZL|= L current lags voltage by 90o

therefore angle of ZL = +90o

multiplying by j ZL = jL

dtvL

idtdiLv LL

LL

1

dttVL

dtvL

i msL )sin(11

)90sin()cos( otLmVt

LmV

)90sin( om tLV

Capacitor response to an ac source

The current iC is:

by Ohm’s Law: ZC

current leads voltage by 90o

angle of ZC = -90o

multiplying by -j

C vC

iC

~vs(t) = V m sin (t)+

-

)cos( tCVdtsdvC

dtdv

Ci mC

C

C1

CjCjZC

11

Voltage - Current Relationships

element phase v/ i ratio

R i in phase i proportional v/R

C i leads v by 90o i proportional vC

L i lags v by 90o i proportional v/L

Example - (solving circuits with an ac source)

RL circuit

solve in dc. - steady state + transient response (revision)

solve in time domain

solve in frequency domain (need complex numbers) real (time domain ) circuits can be solved by adding an

imaginary part, do the calculations in the complex domain, and then simply take the magnitude part of the complex answer as the solution to the real circuit!

Revision of DC Transients -RC

+_

R

Vs C vC

iC

Vs is given Time constant:

iC may change instantaneously

vC can not change instantaneously also:

Which implies that capacitor “appears” like an open circuit to DC

teIIIi FinitFC

)(

RC

teVVVv FinitFC

)(

dtdv

Ci CC

Revision of DC Transients-RL

+_

R

Vs L vL

iL Vs is given Time constant

iL can not change instantaneously

vL may change instantaneously also:

Which implies that an inductor “appears” like a short circuit to DC

RL

teFIinitI

FI

Li

)(

teFVinitV

FV

Lv

)(

dtdiLv L

L

Time Domain Analysis

Assume: i(t) = Imcos(t - )

i(t) = ?

vs= Vmcos(t)

Solution

KVL : vs = vR + v

= i(t) R + L di/dt Vmcos(t) = RImcos(t - ) + L d(Imcos(t - ))/dt

= RImcos(t - ) - LIm sin(t - )

= RIm [ cos(t)cos + sin(t)sin ] - LIm [sin(t) cos - sin cos(t)]

= cos(t)[RIm cos + LIm sin ] + sin(t) [RIm sin - LIm cos ]

cos (t) & sin (t) are othoganal functions ( ie no multiples of cos (t) will ever equate to sin (t) )

Hence, the 2 terms may be considered separately Vm = RIm cos + LIm sin ………………... (1)

0 = RIm sin - LIm cos …………………(2)

Solution (cont.)

From (2) sin / cos = LIm / RIm tan = L / R = tan -1 L/R Followes that sin = L / R2 + (L)2 & cos = R / R2 + (L)2

From (1) Vm = RIm cos + LIm sin

= RIm R / R2 + (L)2 + LIm L / R2 + (L)2

= Im [R2 + (L)2 / R2 + (L)2 ]

= Im [R2 + (L)2 ]

Im = Vm / R2 + (L)2

i(t) = Vm / R2 + (L)2 cos (t - tan -1 L/R )

Complex number revision Electrical Engineers already use i therefore use ‘j’

What is j2 ? if multiplication by j is a rotation by 90°, then j2 must be a

rotation by 180° j2 = -1 j = -1 (which cannot really exist so must be

imaginary)

Complex numbers can be expressed in cartesian co-ordinates (addition & subtraction) or polar co-ordinates (multiplication & division)

Euler’s Identity : z=a+jb = M[costjsin t] = Mej t

where M = magnitude

Phasor notation X = Xm

where is assumed or stated as a side note

Phasor notation - omits the time depend part and allows us to write the magnitude and phase only

Polar: X = Xm

Cartesian: X = XR + jXI

XR = Xmcos

XI = Xm sin

Xm = (X2R + X2

I)

= tan -1 (XI / XR)

XR

jXI Xm

Complex Impedance's

the ratio of voltage to current across and element is known as IMPEDANCE

Impedance is frequency complex, containing both real and imaginary components

Symbol is Z (which is complex) : Z = R + jX

R & X are both real R is always 0 or positive X is reactive component, can be positive or negative

Impedance

ZR = R

Inductance gives a positive reactance ZL = j L

Capacitance gives negative reactance

CjCjZC

11

Steps in solve AC Problems

1. Confirm sources are of the same frequency

2. Convert to Phasor notation

3. Treat as a DC problem (with complex numbers)

4. Convert Back to Time domain

Example (cont. in phasor notation)

Total impedance : ZT = R + jL = R2 + (L)2 tan -1 (L/R)

Ohms Law : Vs = IT . ZT

IT = Vm0° / R2 + (L)2 tan -1 (L/R)

= [Vm / R2 + (L)2 ] 0° - tan -1 (L/R)

= Vm / R2 + (L)2 cos ( t - tan -1 (L/R))

i(t) = ?vs=

Vmcos(t)Vm0°

jL