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Electric CircuitsElectric CircuitsElectric CircuitsElectric CircuitsByBy

Prof. Dr.Prof. Dr. Mousa AbdMousa Abd--AllahAllahDr. Mohammed EisaDr. Mohammed Eisa

ContentsCH1: Sinusoids and PhasorsCH2: Sinusoidal steady-State AnalysisCH3: AC Power AnalysisCH4: Three Phase CircuitsCH5: Transient Response of CircuitsCH6: Fourier Analysis and Circuit

ApplicationsCH7: Two-Port Circuits

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ReferencesMahmood Nahvi, Joseph A. Edminister, “Electric

Circuits”, Schaum’s Outlines, Mc Graw Hill. ISBN: 9780071633727, 2011.

James W. Nilsson, Susan A. Riedel, “Electric Circuits”, Prentice Hall, New Jersey, ISBN:9780137050512-0137050518, 2011.

Jhon Bird, “Electrical Theory and Technology”, Elsevier, Oxford, ISBN:9780080549798-0080549799, 2007.

Chapter IChapter IChapter IChapter ISinuSoidS, PhaSorS and SinuSoidS, PhaSorS and

ResonanceResonance

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sinusoids

Definitions

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Ex.1: Find the amplitude, phase, period, and frequency of the sinusoid: v(t)=12cos(50t+10o)

Solution:The amplitude is: Vm = 12 VThe phase is : = 10o

The angular frequency is = 50 rad/sThe period is: T =(2/) = (2/50) =0.1257 sThe frequency is f =(1/T) = (1/0.1257) = 7.958 Hz

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Ex.2 : Calculate the phase angle between v1 = -10 cos(t+50o) & v2 = 12 sin(t-10o)state which sinusoid is leading

Solution

Phasor diagram

)Im()sin()( )( tjmm eVtVtv

)Im()Im()( tjjtjm VeeeVtv

mj

mj

m VeVeVV )Im(

Phasors

mVV mII

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Ex. 3 : Transform these sinusoids to phasors:(a) v = - 4 sin(30t+50o)(b) i = 6 cos(50t-40o)

Solution(a) v = -4 sin(30t+50o) = 4 cos (30t+50+90o)

= 4 cos (30t + 140o)

The phasor form of v is V = 4140o

(b) i = 6 cos(50t - 40o)The phasor I = 6 -40o

Ex.4: Given i1(t)=4cos(t+30o) andi2(t)=5sin(t-20o), find their sum.

Solution: i1(t)= 4 cos(t+30o) I1 = 430o

i2(t)= 5 sin(t-20o) = 5 cos(t -20o-90o) = 5 cos(t-110o)I2 = 5 -110o

I = I1 + I2 = 430o + 5-110o = 3.464+j2-1.71-j4.698 = 1.754 –j2.698 = 3.218-56.97o

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Phasor relationship for circuit elements

(a) Resistori = Im cos(t + )

v = i R = RImcos(t + )I = Im v = R Im

(b) Inductori = Im cos(t + )v = L(di/dt) = -LIm sin(t + )v = LIm cos(t + +90o)I = Im and V = Vm+90o

The voltage lead the current by 90o → XL= j ω L

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(b) capacitor

)sin()cos( tCVtVdtdci mm

)90cos( tCVi m

mVV 90 mII

The voltage lag the current by 90o → XC= - j 1/ωC

v(t) = Vm cos(ωt+) i = C dv/dt

G conductance, B susceptance

Impedance and Admittance

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Series Circuit

Parallel Circuit

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Δ-Y-Conversion

T

rms dttvT

V0

2 )(1

T

av dttvT

V0

)(1

2m

rmsV

V

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Ex.5: consider the circuit,(a) Calculate the sinusoidal voltages v1 and v2

using phasors and voltage divider rule(b) Sketch the phasor diagram showing E, V1 and

V2.(c) Sketch the sinusoidal waveforms of e, v1 and v2.

Vjj

jV oo 69.334.78)071.70()4030()8040(

80401

Vjj

jV oo 87.829.43)071.70()4030()8040(

40302

Solution:(a) The phasor form of the voltage source is determined as

e = 100 sin(t) E = 70.710o V

and

(b) Phasor diagram

(c) v1(t) = Vm sin(t+) = 2x78.4 sin(t-33.69)= 111 sin(t-33.69)

v2(t) = Vm sin(t+) = 2x43.9 sin(t+82.87)= 62 sin(t+82.87)

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Ex. 6: Refer to the circuit;(a) Find the total impedance, ZT.(b)Determine the supply current, IT.(c) Calculate I1, I2, using current divider rule.(d)Verify Kirchhoff’s current law at node a.

Solution

201

11

11

2011.

jjZa

T kZ o

T 020

AAxxZ

VIb oo

TT 0250010250

0102005. 63

AxI

AxI

AxIc

oo

oo

oo

9050000250901020

9050000250901

020

02500250020020.

3

2

1

d. I1 + I2 + I3 = 2500 + 5000-90 + 500090 = 2500o

IT = I1 + I2 + I3 Kirchhoff’s current law is verified.

Ex. 7: Find the current I in the circuit.

Solution:

8.06.110

8168424

)42(4 jjjjjjZ an

2.310

)8(4 jjZ bn

2.36.110

)42(8 jjZ bn

Z1 = Zan+12 = 13.6 + j0.8 , Z2 = Zbn – j3 = j0.2 and Z3 = Zcn +j6+8 = 9.6+j2.8

16.1336.9

)8.26.9(2.08.06.1332

321 j

jjjj

ZZZZZZT

o

T jZVIcurrentsourceThe 2.4666.3

16.13050

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At balance V1=V2

21

21 ZZ

ZVV s

x

xs ZZ

ZVV

32

3321

2

ZZZ

ZZZ x

21

3 ZZZ

Z x

AC Bridges

Series Resonance

)1(C

LjRZ

At resonance Im(z) = 0

01

CL

CL

oo

1

sradLCo /1

HzLC

f o2

1

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22 )1(C

LR

VI

The highest power is at ωo

RVRIP o

22)(

III 707.02

RVPP

2

21 21)()(

RZ 2

RC

LR 2)1(2

The half power frequencies are at:

ω1 & ω2 can be get by setting

RC

LR 2)1(2

LCLR

LR 1

22

2

1

LCLR

LR 1

22

2

2

21 o

By solving the above equation for ω

Bandwidth “It is defined as the half power bandwidth”B = ω2 – ω1 = (R / L)

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Quality Factor Q“Measure the sharpness of resonance “

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Example: For the circuit shown:(a) Find the resonant frequency and the half power frequencies(b) Calculate the quality factor and bandwidth(c) Determine the amplitude of the current at ωo, ω1 and ω2

Solution

Parallel Resonance

At resonance Im(Y)=0

At resonance:(1)The parallel LC combination acts like open circuit.

i.e. the entire current flows through R(2) The inductor and capacitor currents can be much

more than the source current

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Example: For the circuit shown:(a) Calculate ωo, Q and B(b) Find ω1 and ω2(c) Determine the power dissipated at ωo, ω1 and ω2

Solution

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WxxRIP o3232 105.12)8000()1025.1(

WxxR

VP 3

22

1025.680002

102