electric and electronic lecture presentation - chapter04.ppt

7/29/2019 Electric and Electronic Lecture Presentation - Chapter04.ppt

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AC NETWORK ANALYSIS

AC VS DCENERGY STORAGE CIRCUIT ELEMENTS

RESISTORSLINEAR ELEMENTS, DISSIPATE ENERGY

CAPACITORS AND INDUCTORS

IDEAL CAPACITORA DEVICE THAT CAN STORE ENERGY INTHE FORM OF CHARGE SEPARATION WHEN VOLTAGE IS

APPLIED


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Figure 4.2,

4.3

Structure of parallel-plate capacitor

DOES NOT ALLOW DC CURRENT DUE TOINSULATION BETWEEN TWO PLATES

IN A C, CHARGE SEPARATION ISPROPORTIONAL TO APPLIED VOLTAGE

Q = CV or q(t) = Cv(t)

VOLTAGE-CURRENT RELATIONSHIP

i(t) = C dv(t)/dt

vc(t) = 1/C ic dt + V0(how does this relationship come up?reading assignment)


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Combining capacitors in a circuit

Rule of thumbC in parallel add. C

in series combined according to R inparallel

(Prove this)

(Prove this)


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Exdetermining current for a C given voltage

Plot q(t) and i(t)?

V(t) V

t (s)

10

0 2 4 5

+

+

i(t)

C = 1 F


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Energy stored in C

Energy is the integral of power ,

recall : p = v.i

p(t) = v(t). C dv/dt

pc(t) = C.v

c(t) dv

c/dt

So, Wc(t) = Pc(t) dt

= Cvc2(t) or vc(t)q(t) (prove this)


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Figure 4.10

Inductance and practical inductors

IDEAL INDUCTORresistance is

zero , so for DC current, acts as ashort cct

If v(t) is applied, i(t) will result

vL(t) = L diL/dt

iL(t) = 1/L vL dt + I0


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Figure 4.12

Combining inductors in a circuit

(Prove these)


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Energy stored in L

Energy is the integral of power ,

recall : p = v.i

p(t) = L diL/dt . iL(t)

pL

(t) = L . iL

(t) diL

/dt

So, WL(t) = PL(t) dt

= LiL2(t) (prove this)


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TIME DEPENDENT SIGNAL SOURCES

Periodic signal waveformsx(t) = x(t + nT) ; n=1,2,3,

Sinusoidal waveformsgeneralized sinusoid; x(t) = A Cos(t + )


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Figure 4.17,

4.18

Sinusoidal waveforms

4-4

x1(t) = A Cos (t)

x2(t) = A Cos (t + )f = natural freq = 1/T cycles/s or Hz

= radian freq = 2f rad/s

Can represent a sine wave in terms ofcosine wave by introducing a phase

shift of/2 rad

: A Sin(t) = A Cos (t - /2)

Q : why are we interested in sinusoid?

x1(t)

x2(t)


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Figure 4.22

AVERAGE AND RMS VALUES

-TO MEASURE THE STRENGTH OF TIME VARYING SIGNALS :

1. THE AVERAGE (OR DC) VALUE

2. THE ROOT MEAN SQUARE (RMS) VALUETAKES INTO ACCOUNT THEFLUCTUATION OF THE SIGNAL ABOUT ITS AVERAGE VALUE

DEFINE THE AVERAGE VALUE OF x(t):

Ex: Compute the average value of x(t) = 10 Cos(100t), w = 100 rad/s

< x(t) > = 1/TT

0

x(t) dt - ave value

T period of integration

< x(t)> = 1/T 10 cos 100t dt

= 0

T

0


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Ex2: v(t) = 155.6 Sin (377t + /6), express v(t) in Cosine form, w = 377 rad/s

V(t) = 155.6 cos (377t + /6 /2)

= 155.6 cos (377t /3) V

Ex3: compute the average value of v(t)

V(t) V

t (ms)

3

0 5 10

< V(t) > = 1/T V(t) dt

= 1.5 V

T

0


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A useful measure of the strength of AC signals is rms value defined as:

Xrms =

We use Vrms or V and Irms or I to refer to rms values of a voltage and current

Ex 1: find rms value of i(t) = I Cos(t)

Irms = I/ 2 = 0.707 (I)

For a sinusoid signal the rms value is equal to 0.707 times the peak value, independent of its

amplitude and frequency.

1/T x (t) dtT

0

Only for

sinnusoidal signals


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Phasors (frequency domain)

A notation to represent sinusoid signals as complex numbers and eliminate the need to

solve differential equation.

We can write sinusoid signals in terms of complex numbers based on Eulers Identity

Where from Eulers identity: e j = Cos + j Sin

Can also write in terms of polar form (magnitude and angle:

A e j = A

A Cos(t + ) = Re (A e j(t + ) )

Summary: any sinusoid in time domain v(t) = A Cos(t + )can be represented in terms of frequency domain (phasor)

form V(jw) = A e j= A


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ex

+

+

V1(t)

V2(t)+

Vs(t)

V1(t) = 15 cos (377t + /4) V

V2(t) = 15 cos (377t + /12) V Find eqVs(t)

Write in phasor forms


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Figure

4.29

4-7

The impedance

In AC circuits, 3 elements, R, L, C will be described as impedance (complex resistance)


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Figure4.33

4-8

Impedances ofR, L, and C in the complex plane

Fi 4 37


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Figure 4.37

4-9

Figure 4.37


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Figure4.41

4-10

AC circuit analysis methods

1. Identify sinusoidal source(s) and note the freq

2. Convert source(s) to phasor form

3. Represent each cct element with its impedance4. Solve using appropriate method

5. Convert phasor form to time domain equivalent


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AC Thevenin equivalent circuits

ex

RLR+

Rs

C

Vs(t)

b

aRs = R1 = RL = 50

C = 0.1 F

L = 10 mH

Vs(t) = 10 cos 1000t V

Find the Thevenin eq. cct seen by RL

1)

Vs(jw) +

Zs

ZL

ZC

ZR ZL


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Vs(jw)+

Zs

ZL

ZC ZR

ZT

ZT = ((Zs + ZL//Zc))//ZR

= 25.37 5.6

= 25.25 + j 2.48

Zs + ZL//Zc ZR

ZT

Vs(jw)

= 100

+

Zs ZL//Zc

ZR

+

Voc

2) Find VT


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Voc = Vs ZR

Zs + ZL//Zc + ZR

= (10 0)(50)50 + j 10.01 + 50

= 4.985.72 V

VT = Voc = 4.985.72 V

I = Vs(jw)

Zs + ZL//Zc + ZR

Voc = I ZR

+

ZT

= 25.375.6 V

VT = 4.985.7 V


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R1 = 1 M

C1 = 1x 10 F9R1 C1

ZR1 ZC1

Find equivalent Impedance

ZR1 = R1

ZC1 = 1/jwC1

Zeq = ZR1//ZC1

= R1 (1/jwC1)

R1 + (1/jwC1)

= R1

1 + jwC1R1

= 1 x 10

1 + j (377)(1x10 )(1x10 )

= 9.346 x 10 20.66 = R x j X

6

69

5


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Ex 2:

R

L

R = 0.136

L = 0.098 H

ZR

ZL

Zeq

Zeq = ZR+ ZL

= R + jwL


25/29

AC Power

When a cct is excited by sinusoidal source, all voltage and currents are alsosinusoidal with same freq

The most general form for voltage and current delivered to a load v(t) = V Cos (tv) and i(t) = I Cos (ti) Instantaneous power is given by the product of voltage and current, so

p(t) = v(t).i(t) = VI Cos(t).Cos(t), where = viusing trigo identity; p(t) = VI Cos + VI Cos(2 t - )

Power is equal to sum of an average component and a sinusoidal component

Pave = VI Cos - Average Power

Or = V2/ZCos

Note: to eliminate the factor, in AC power analysis, normally use rms values

Vrms= V/2 ; Irms= I/ 2

So Pave = Vrms . Irms Cos


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Power Factor

The ave power is dependent on the cosine of the angle of impedance

To recognise its importance, the term Cos() is referred to as the power factor (pf)

Pf can range from 0 to 1. for purely resistive load pf=1, for purely inductive or

capacitive pf=0

Pave = Vrms . Irms Cos ; pf = Cos = Pave / (Vrms . Irms)

Complex Power

From expression of instantaneous power

P(t) = VI/2 cos + VI/2 cos (2wt - )

= Vrms/|Z| [ cos + cos cos (2wt) + sinsin(2wt)]

= Irms |Z| [cos + coscos(2wt) + sinsin(2wt)]

= Irms |Z| cos[1 + cos (2wt)] + Irms |Z| sinsin(2wt)


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This suggest p(t) consists of 3 comps:

1) Average comp.ave power (Pave) (also called real power)

Pave = Irms R ; R = ReZ2) A time varying comp with zero ave power due to fluctuation and resistive

component ; PR(t) = Irms Rcos(2wt)

3) A time varying component with zero ave power due to power fluctuation in

reactive component

Px(t) = Irms X sin 2wt = Q sin 2wt

where X = Im Z and Q is called the reactive power

Pave = Preal = Irms R (W)

Q = Irms X (VAR)

SIMPLIFIED FORMULA FOR COMPLEX POWER; S = VRMSIRMS*

S = P + jQ


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Ex:

+

Vs

C

Rc

Rs

+

VL

Load

Vs = 1100

Rs = 2

RL

= 5C = 2000 F

f = 60 Hz

1) Calculate real/average and reactive power for the

load

2) Calculate pf = cos

Rules for impedance and admittance reduction


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Figure4.45

4 12

Rules for impedance and admittance reduction

Figure 4.45

electric and electronic lecture presentation - chapter04.ppt

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