Electricity CPUs Ethics Refresher

Electrical; Computers; Ethics

Refresher Notes

Spring 2013

1FE Electrical General Notes

Electrical Circuits, Computers,

and Ethics

2Introduction I. Electromagnetics

Electrostatics/Magnetostatics II. DC Circuits

KCL, KVL Series/Parallel Equivalent Circuits Node and Loop Analysis Thevenin/Norton Theorems Transient Response Sampling Theorem Instrumentation

III. AC Circuits Average and RMS Values of Waveforms Impedance and Admittance Complex Power Resonance Analog-to-Digital Conversion Transformers Rotating Machines

IV. Computers Terminology Spreadsheets Structured Programming

V. Ethics and Business Practices

3I. Electromagnetics

Electrostatics/Magnetostatics

4Introduction

Equivalent Units Amperes A = W/V = N/Tm Charge C = J/V = Nm/V Capacitance F = C/V = C/J = C/Nm = S/ = J/V Inductance H = Vs/A = Tm/A Energy J = Nm = VC = C/F = (Kgm)/s Force N = J/m = VC/m = (Kgm)/s Flux Density T = Ns/Cm = N/Am = Wb/m Voltage V = W/a = C/F = J/C Power W = J/s = VA = V/ Flux Wb = Vs = HA = Tm

5Electrostatics

Charges

E F

+Q E = F/Q

Coulombs Force Law

+Q1 +Q2 F1 = -Q1E2 F2 = Q2E1

r

E1 E2

Page 193 from Supplied Reference Handbook

6Electrostatics

Permittivity Note: On the FE exam, assume the permittivity is 0 = 8.85 x 10 F/m unless another value is provided.

ar

QE 24

Law sCoulomb'

+Q

E1 F1 F2 E2

Q1 +Q2 F1 = (-Q1)(-E2)

F1 = Q1E2

F2 = (+Q2)(-E1)

F2 = -Q2E1


7Electrostatics

Electric Field Intensity

+Q Es

s

-Q Es

s

EL

L = Q/d = C/m

s = Q/A = C/m

(V/m or N/C)


8Electrostatics

Example A point charge of 0.001 C is placed 10 m from a sheet charge of 0.001 C/m, and a 10 m diameter sphere of charge 0.001 C is placed half-way in between on a straight line, all in a vacuum. What is the Qp force on the point charge? F = Fsheet + Fsphere = Qpoint (Esheet + Esphere)

F

5m 10m

QP

QS

S

ESH Esph

(-)

+ +

S = - 0.001 C/m

Qs = 0.001 C

QP = 0.001 C

NmCm

C

mF

CF

rQQ

rQ

QF spheresheetpospheresheetpo

42

2

12-

2int

2int

10 x 61.5)5(4

001.02

001.0

10 x 85.8

001.0

4242

9Electrostatics

Electric Flux Gauss Law

R = radius of the charge distribution r = radius of the sphere

dS Surface of Integration

Asphere = 4r = d


10

Electrostatics

The Concept of Work Work (W) done by an external agent in moving a charge Q in an electric field from a point p1 to a point p2:

For a uniform field, the work done by moving a charge Q a distance d parallel to the uniform field:

VQWQEddFW

Note that equation, W = - QV, is always true for all fields. (V may not be easy to compute.)

E Q d V

+

-

ES +Q +W

-W W = 0 +Q

+QS +Q

(Nm or Joules)


(Joules) 21

p

pldEQW

11

Electrostatics

Voltage Gradient points in the direction where the E-field increases or decreases most with voltage potential.

V = -W/Q

E Q

V

(V/m or N/C)

E

d

+ _

V


12

Electrostatics

Example A source at zero potential emits electrons (mass: ) at negligible velocity. An open grid at 18 V is located 0.003 m from the source. An electron has a charge of C. At what velocity will the electrons pass through the grid?

Kg 10 x 11.9 -31

-1910 x 6022.1

18V

0.003m

QP

0V

13

Electrostatics

Example/Contd.

2

21 T Handbook, FE of 59 Pg.in Equation Energy Kinetic mv

14

Electrostatics

Current Electric current i(t) through a surface is defined as the rate of charge transport through that surface or i(t) = dq(t)/dt, which is a function of time t, since q(t) denotes instantaneous charge. A constant current i(t) is written as I, and the vector current density in amperes/m is defined as J.

J (Amps/m)

(C/s) or Amps

(Current direction matters)

( = Amps/m)


15

Magnetostatics

Magnetism

(A/m)

(Wb/m or Teslas)

Right Hand Rule Generators

Left Hand Rule Motors


16

Magnetostatics

Example

1 m (L) 30

B

|B|Sin

I

17

Magnetostatics

Example What is the magnitude of the force on an electron moving at 0.1c in a uniform magnetic field of ?

T 10 x 0.5 -5

N 10 x 4.2)10 x m/s)(5.0 10 x C)(0.3 10 x 6022.1(

16-

5-819-

FTF

vBQF

B

v (-) electron velocity

magnetic field flux density

1 T = 1 Ns/Cm

18

Magnetostatics

Induced Voltage v = the induced voltage = the average flux (Weber) enclosed by each turn N = number of loops L = distance AdBA

s

I

B

L ds/dt

ds/dt = velocity

s = distance that conductor travels.

B

Area


19

Magnetostatics

Voltage The potential difference V between two points is the work per unit charge required to move the charge between points. For two parallel plates with potential difference V, separated by distance d, the strength of the E field between the plates is E = V/d directed from the + plate to the - plate.

+

-

20

DC Circuits

Circuit Analysis

21

DC Circuits

Kirchhoffs Laws (KCL, KVL)

The basic tools for solving DC circuit problems are Kirchhoffs laws: the current law and the voltage law, Ohms law, and the power relationship. The following configurations are typical:

RVRIVIP

IVRR

VIIRV

VVKVL

IIKCL

DROPSRISES

OUTIN

22

PowerResistive

LawsOhm'

LawssKirchhoff'

Voltage Drop (- V)

+

_ Voltage Drop (- V)

Voltage Rise (+V)

Voltage Rise (+V)

I1 I2

I3

I1 = I2 + I3


22

DC Circuits KCL This law is also called Kirchhoff's first rule or Kirchhoff's nodal rule.

The principle of conservation of electric charge implies that at any node (junction) in an electrical circuit, the sum of currents flowing into that node is equal to the sum of currents flowing out of that node or the algebraic sum of currents in a network of conductors meeting at a point is zero. (Assuming that current entering the junction is taken as positive and current leaving the junction is taken as negative). Recalling that current is a signed (positive or negative) quantity reflecting direction towards or away from a node, this principle can be stated as: n is the total number of branches with currents flowing towards or away from the node. The current entering any junction is equal to the current leaving that junction: i1 + i4 = i2 + i3 The law is based on the conservation of charge whereby the charge (measured in coulombs) is the product of the current (in amperes) and the time (which is measured in seconds).

23

DC Circuits (KCL)

Example Consider the following circuit with the following parameters:

V1 = 15V V2 = 7V R1 = 20 R2 = 5 R3 = 10

Find current through R3 using Kirchhoff's Current Law.

The circuit above shows voltages at nodes a, b, c and d. We use node a as common node (ground if you like). Thus, Va = 0V. From Node b we get:

Vb = V1 = 15V From Node d we get:

Vd = V2 = 7V

24

DC Circuits (KCL)

Example/Contd. Thus, we must solve Vc, in order to complete voltage definitions at all nodes. Vc will be found by applying KCL at node c and solving resulting equations follows: i1 = i2 + i3 Substituting values we get

VVV CC 857.1 Thus, A 65.0207

0 321

acacdbc V

RVV

RVV

RVV

25

DC Circuits (KCL)

Example/Contd. Thus, now we can calculate current through R3 as follows:

IR3

AI

I

RVI

R

R

CR

186.010857.1

3

3

33

26

DC Circuits

KVL This law is also called Kirchhoff's second law, or Kirchhoff's loop (or mesh) rule.

The principle of conservation of energy implies that the directed sum of the electrical potential

differences (voltage) around any closed circuit is zero.

Here, n is the total number of voltages measured. The voltages may also be complex:

The sum of all the voltages around the loop is equal to zero. v4 - v1 - v2 - v3 = 0

27

DC Circuits (KVL)

Example

Consider the following circuit with the following parameters: V1 = 15V V2 = 7V R1 = 20 R2 = 5 R3 = 10 Find current through R3 using Kirchhoff's Voltage Law. We can see that there are two closed paths (loops) where we can apply KVL in, Loop 1 and 2 as shown in the circuit above.

From Loop 1 we get:

V1 VR3 VR1 = 0 From Loop 2 we get:

V2 VR2 VR3 = 0

28

DC Circuits (KVL)

Example/Contd. The above results can further be simplified as follows: V1 (I1 I2) * R3 I1 * R1 = 0 (1) and

V2 I2 * R2 (I2 I1) * R3 = 0 (2)

By equating above (1) and (2) we can eliminate I2 and hence get the following: (3) We end up with the above three equations and now substitute the values given in the above equations and solve the variables. Only after we have arrived at a simplified equation then that we can substitute in values of Resistors, Voltages and Current.

32

3122 RR

RIVI

233213

323211 ))((

)(RRRRR

RVRRVI

29

DC Circuits (KVL)

Example/Contd. The positive sign for I2 tells us that current I2 flows in the same direction to our initial assumed direction. Thus, now we can calculate current through R3 as follows: The negative sign for IR3 only tells us that current IR3 flows in the same direction as I2.

AVI

xAVI

AVI

VVI

03.115

4.15)510(

)1084.0(7

84.0350295

)10()105(*)2010()10()7()105()15(

2

2

1

21

IR3

AI

AAI

III

R

R

R

19.0

03.184.0

3

3

213

30

DC Circuits

Series/Parallel Equivalent Circuits

Resistors in Series Resistors in Parallel

nRRRRRIf

RRIf

eq

2RR

321

eq21


31

DC Circuits

Example What is the resistance, as seen from the battery, of the parallel and series resistors in this circuit? (A) 4 (B) 5 (C) 6 (D) 8 Two 8 resistors in parallel are 4 . Two 10 resistors in parallel are 5 . Thus, we have two 10 resistors in parallel are 5 . Therefore, the correct answer is (B).

32

DC Circuits

Example

33

DC Circuits

Node Voltage Circuit Analysis: 1. Convert all current sources to voltage sources. 2. Choose one node as reference (usually ground). 3. Identify unknown voltages at other nodes compared to reference. 4. Write Kirchhoffs current equation for all unknown nodes except reference node. 5. Write all currents in terms of voltage drops. 6. Write all voltage drops in terms of the node voltages.

34

DC Circuits

Example Find the voltage potential at point A and the current i1.

A

VVVVi

VV

VVVVViii

A

A

AAA

35.72

3.35502

503.35

100

420

250

1

321

35

DC Circuits

Loop Current Circuit Analysis

1. Select one less than the total number of loops. 2. Write Kirchhoffs voltage equation for each loop. 3. Use the simultaneous equations to solve for the current you want.

36

DC Circuits

Example Find the current through the 0.5 resistor. The voltage sources around the left loop are equal to the voltage drops across the resistances. 20 V 19 V = 0.25 i1 + 0.4 (i1 i2) The same is true for the right loop. 19 V = 0.4 (i2 i1) + 0.5 i2 Solvetwo equations and two unknowns. 0.65 i1 0.4 i2 = 1 V 0.4 i1 + 0.9 i2 = 19 V i1 = 20 A i2 = 30 A The current through the 0.5 resistor is 30 A.

37

DC Circuits

Voltage Divider The voltage across a resistor R in a loop with equivalent resistance Req with a voltage source V is In the general case, the voltage on impedance Zi in a loop with equivalent impedance Zeq with a Voltage source v is NOTE: Each symbol is a complex number in the general case.

VRRVeq

R

VZZ

Veq

ii

38

DC Circuits

Example What is the voltage across the 6 resistor? (A) 5 V (B) 6 V (C) 8 V (D) 10 V Two 8 resistors in parallel equal 4 . The voltage across the 6 resistor is 6V. Therefore, the answer is (B).

VV 646

610

39

DC Circuits

Current Divider The current through a resistor R in parallel with another resistance Rparallel and a current into the node of I is: (Resistance R does not appear explicitly. The denominator is the sum of the resistances in parallel.) In the general case, the current through impedance Z connected to a node in parallel with equivalent Impedance Zparallel with a current i into the node is: (The denominator is the sum of the impedances in parallel.) NOTE: Each symbol is a complex number in the general case. Procedure: 1. Identify the component you want the current through. 2. Simplify the circuit. 3. Determine the current into the node that is connected to the component of interest. 4. Allocate current in proportion to the reciprocal of resistance.

IRR

RIPARALLEL

PARALLELR )(

IZZ

ZIPARALLEL

PARALLELZ )(

40

DC Circuits

Example What is the current through the 6 resistor? (A) 1/10 A (B) 1/3 A (C) 1/2 A (D) 1 A i = i1 + i2 Simplify the circuit. 3 in parallel with 6 = 2 2 in series with 4 = 6 Rparallel = 3 and Req = 3 + 6 = 9 Therefore, the answer is (B).

AVi 166

AAi31

63312

41

DC Circuits

Superposition Theorem The net current/voltage is the sum of the current/voltage caused by each current/voltage source. Procedure: 1. Short all voltage sources, and open all current sources, then turn on only one source at a time. 2. Simplify the circuit to get the current/voltage of interest. 3. Repeat until all sources have been used. 4. Add the results for the answer.

42

DC Circuits

Example Using the Superposition Theorem, determine the current through the center leg of the circuit. Short the 20 V source. Short the 50 V source. Ieq = 25 A + 5 A = 30 A

AVI 25250

AVI 5420

43

DC Circuits

Thevenin Equivalent Circuit


NTHSC

OCeq RRI

VR

44

DC Circuits

Example Find the Thevenin equivalent voltage and resistance of the circuit as seen by the 10 resistor. The Thevenin resistance is the same as the Norton resistance in the previous example, which is 1.33 with the 10 resistor open-circuited, apply the Kirchhoff voltage law around the loop and find VTH = 40V.

2 4

33.1424 x 2

THR

45

DC Circuits

Example/Contd. ( 50 V 20 V) = I(2 + 4 ) I = 5 A V = 50 V I(2 ) = 50 V (5 A)(2 ) = 40 V

46

DC Circuits

Norton Equivalent


47

DC Circuits

Example Find the Norton equivalent current and resistance of the circuit as seen by the 10 resistor. With the 10 resistor open circuited, and the voltage sources shorted, the circuit is 4.0 and 2.0 in parallel. With the 10 resistor shorted, the circuit looks just like the previous superposition example. IN = 30A

33.1424 x 2

NR 4 2

48

DC Circuits

Maximum Power Transfer DC Circuit Maximum power is transferred when the load resistance is the same as the Thevenin Equivalent circuit resistance, Rs = RL AC Circuit Maximum power is transferred when the load impedance is the complex conjugate of the Thevenin equivalent circuit impedance, Zs = x + jy and ZL = x jy = Zs*

49

DC Circuits

Example A load resistance is to be connected to the circuit below. (a) What value of load resistance will provide maximum power transfer to the load? (b) what is the value of that maximum power? (a) The maximum power theorem tells us that the load resistance should be, RL = RTH = 0.8

Ohms. (b) We can find the maximum power by finding the current through the load. Essentially the

11.2V source is across two 0.8 ohm resistors in series. Thus, the current is, I = 11.2V/1.6 ohm = 7.0 A. Thus, the power is WohmsAP 4.78)6.1()7( 2

50

DC Circuits

Capacitors

Q = CV

C = Q/V

V = Q/C

Pages 194, 195 from Supplied Reference Handbook

(Joules) 222

Cvenergy

v(t)][varying )()(22

C CCC

CC

vqC

q

tCvtq

51

DC Circuits

Example

52

DC Circuits

Example

53

DC Circuits

Inductors

2

2LLiEnergy

L = Inductance

N = # of loops

= Flux

I = Current

(initial current) (current value from 0 to t)

21

21

:parallelin inductors 2For

LLLLLeq


54

DC Circuits

Example

55

DC Circuits

Transient Response

RC Transients

Constant TimeCircuit RC

(t = t = t2 t1)

Capacitor Voltage (V is battery voltage if present)

seconds) in 36.8V 100V 100V, V (If

VR


56

DC Circuits

Example

57

DC Circuits

Example/Contd.

i (t) actual

58

DC Circuits

Example/Contd.

stssts

sst

e

V

VeVetvc

sst

sst

sst

C

73.430.2657.3

10ln1

658.35ln

10lnln58.35ln

10ln34.58Veln

equation theof sidesboth of logarithm natural theTake

101058.34)( )(

16

1s6s-t

16

16

59

DC Circuits

RL Transients

Constant TimeCircuit RL

VR(t)

VL(t)


60

DC Circuits

Example Find the voltage at point A at the instant the switch is closed. The switch has been open for a long time, and there is no initial current in the inductor.

(Inductor behaves like a short at t >> 0)

61

DC Circuits

Example Consider the following circuit with R1= 3 , R2= 4 , R3= 2 , C = 1 F, and E = 12 volts. Assume that the switch has been in the close position for a long time and at t = 0 the switch is open. Write the equation for vc(t) and ic(t) for t > 0. Calculate vc(t) and ic(t) at t = 3 sec and at t = 6 sec.

ampseRRtv

ti

voltsevtv

VVERRR

RRv

FCRRRC

RCt

CC

RCt

cC

C

32

321

32

32

)()(

)(

81296)0(

sec616)(

Ic(t)

amptivoltstvamptivoltstv

ampsetivoltsetv

C

C

C

C

stC

stC

49.0sec)6(94.2sec)6(81.0sec)3(

85.4sec)3(33.1)(

8)(sec6

6/

6/

62

DC Circuits

Example Consider the following circuit with R1= 6 k, R2= 12 k, L = 2 H, and E = 12 volts. Assume that the switch has been in the close position for a long time and at t = 0 the switch is opened. Write the equation for iL(t) and vL(t) for t > 0. Calculate iL(t) and vL(t) at t = 100 sec and at t = 200 sec.

voltseRRitv

mAeiti

mAmAkki

RRRi

mAkV

RREi

mkH

RRL

RL

t

LL

t

LL

EL

E

))(0()(

)0(

13186)0(

3412

||

sec11.0182

21

21

1

21

21

voltstvmAtivoltstv

mAtivoltsetv

mAeti

L

L

L

L

stL

stL

97.2sec)200(16.0sec)200(

73.0sec)100(41.0sec)100(

18)(

1)(sec110msec 11.0

110/

110/

63

DC Circuits (Sampling Theorem)

Sampling is the process of converting a signal (for example, a function of continuous time or space) into a numeric sequence (a function of discrete time or space). A sample refers to a value or set of values at a point in time and/or space. If a function x(t) contains no frequencies higher than B Hertz, it is completely determined by giving its ordinates at a series of points spaced 1/(2B) seconds apart. The theorem shows that a band-limited analog signal that has been sampled can be perfectly reconstructed from an infinite sequence of samples if the sampling rate exceeds 2B samples per second, where B is the highest frequency in the original signal. If a signal contains a component at exactly B hertz, then samples spaced at exactly 1/(2B) seconds do not completely determine the signal. A signal or function is band-limited if it contains no energy at frequencies higher than some band-limit or bandwidth B. A signal that is band-limited is constrained in how rapidly it changes in time, and therefore how much detail it can convey in an interval of time. The sampling theorem asserts that the uniformly spaced discrete samples are a complete representation of the signal if this bandwidth is less than half the sampling rate.

64

DC Circuits

Sampling Frequency Determines the sampling rate to reproduce accurately in the discrete time system. Sampling Rate or Frequency: {where fN is the maximum frequency component of g(t) and fN is the Nyquist frequency} Sampling Frequency: (Sample rate to recover signal g(t) exactly from samples)

NS ff 2

TS = 1/2fN

Spectrum of band-limited signal g(t)

NS ff 2

SS Tf 1

g(t)


65

DC Circuits

Sampled Messages A low-pass message m(t) can be exactly reconstructed from uniformly spaced samples taken at a sampling frequency of fs = 1/Ts fs 2fN where M(f ) = 0 for f > fN The frequency fN is called the Nyquist frequency. Sampled messages are typically transmitted by some form of pulse modulation. The minimum bandwidth B required for transmission of the modulated message is inversely proportional to the pulse length . B 1 / Frequently, for approximate analysis B 1 / is used as the minimum bandwidth of a pulse of length .

66

DC Circuits

Therefore, a band-limited signal can be reconstructed exactly if it is sampled at a rate at least twice the maximum frequency component in it. T(t) is the sampling signal with fs = 1/Ts > 2fN

Original Signal Spectrum G()

Spectrum T()

67

DC Circuits

Sampled Signal s = 2N Ts = 1/2fN

Spectrum Gs() Sampled Signal gs(t)

68

DC Circuits

Aliasing What happens if we sample the signal at a frequency that is lower that the Nyquist rate? When the signal is converted back into a continuous time signal, it will exhibit a phenomenon called aliasing. Aliasing is the presence of unwanted components in the reconstructed signal. These components were not present when the original signal was sampled. In addition, some of the frequencies in the original signal may be lost in the reconstructed signal. Aliasing occurs because signal frequencies can overlap if the sampling frequency is too low. Frequencies "fold" around half the sampling frequency - which is why this frequency is often referred to as the folding frequency.

69

DC Circuits

Aliasing Aliasing is a phenomenon where the high frequency components of the sampled signal interfere with each other because of inadequate sampling s < 2m. Aliasing leads to distortion in recovered signal. This is the reason why sampling frequency should be at least twice the bandwidth of the signal.

70

DC Circuits

Oversampling In practice, signals are oversampled, where fs is significantly higher than Nyquist rate to avoid aliasing.

Oversampled signal avoids aliasing

fs >> 2fN

71

DC Circuits

Example

or fs = 1/Ts

72

DC Circuits

Example/Contd. The sampling frequency must be greater than the Nyquist rate for accurate reproduction. fs 2fN The greatest frequency that can be reproduced at this sampling rate is Therefore, the correct is (C).

HzHzHzf

f SN66

6

10 x 16 10 x 7.16210 x 33

2


73

DC Circuits

Example What is the minimum sampling frequency that can be used to avoid aliasing with the following analog signal? x(t) = Cos 100t + Sin 200t + Sin 60t (A) 30Hz (B) 60Hz (C) 200Hz (D) 400Hz Solution: To avoid aliasing, the analog signal must be sampled at least at the Nyquist rate, which must be at least twice the highest frequency contained in the signal. The highest frequency contained in the given signal is 200 rad/s or 100 Hz. Therefore, the minimum frequency that can be used to avoid aliasing is 200 Hz.

= 2f

f = /2

f = 200/2 Hz

= 100 Hz

74

DC Circuits

Instrumentation Transducers Transducer sensitivity the ratio of change in electrical signal magnitude to the change in magnitude of the physical parameter being measured.


75

DC Circuits

Instrumentation Resistance Temperature Detector (RTD) a device used to relate change in resistance to change in temperature. Typically made from platinum, the controlling equation for an RTD is given by: where: RT is the resistance of the RTD at temperature T (measured in C) R0 is the resistance of the RTD at the reference temperature T0 (usually 0 C) is the temperature coefficient of the RTD

)(1 00 TTRRT


76

DC Circuits

Example

77

DC Circuits

Example/Contd. The error in the measurement is: Error = 383.1 C - 400 C = -16.9 C

78

DC Circuits

Resistance Example A cube with an edge length of 0.01 m has resistivity of 0.01 m. What is the resistance from one side to the opposite side? (A) 0.0001 (B) 0.001 (C) 0.1 (D) 1 Therefore, the answer is (D).

)( ALR

1)01.0(

)01.0)(01.0(2m

mmALR

0.01m

0.01m

0.01m

A = (0.01) m

= 0.01 m


79

DC Circuits

Strain Gauges Metal or semiconductor foils that change resistance linearly with the strain. Strain Gauge a device whose electrical resistance varies in proportion to the amount of strain in the device. Gauge Factor (GF) the ratio of fractional change in electrical resistance to the fractional change in length (strain): Where: R is the nominal resistance of the strain gauge at nominal length L. R is the change in resistance due to the change in length L. is the normal strain sensed by the gauge. The gauge factor for metallic strain gauges is typically around 2.

RRLLRRGF /

//


80

DC Circuits

Example

81

DC Circuits

Wheatstone Bridges/Instrumentation


balanced is bridge theand 0V If 03241 VRRRR

82

DC Circuits

Example

R1 = 10.00k + 0.24k = 10.24k

83

DC Circuits Measurement Uncertainty/Instrumentation

Kline-McClintock Equation: A method for estimating the uncertainty in a function that depends on more than one measurement. Suppose that a calculated result R depends on measurements whose value are x1 w1, x2 w2, x3 w3, etc, where R = f(x1, x2, x3,xn), xi is the measured value, and wi is the uncertainty in that value. The uncertainty in R, wR, can be estimated using the Kline- McClintock equation:


22

22

2

11 ...

!

!

!

!

!

!

nnR x

fwxfw

xfww

84

DC Circuits

Example

85

III. AC Circuits

Circuit Analysis

86

AC Circuits

Frequency and Period of a Sinusoidal Waveform Frequency (Linear and Angular) Period = T

Tf

Tf "

" 22 2

1


87

AC Circuits

Average Value Average and Effective Values

T

t XMAX

2

31

:Sawtoothor TriangularFor

MAXAVEMAXRMS

XXXX


88

AC Circuits

Average Value Example

AmpstqI

dtdqti

sC

)(

FsA 96,485 electrons

of mole one of charge of Magnitude Faraday 1Constant sFaraday' x eCapacitancq

Handbook)in 19 (Pg. 485,96

:Constant sFaraday'

FsA

VMAX = 170V

(q)

89

AC Circuits

Effective or Root-Mean-Square (RMS) Value

VRMS %)70( 707.0

707.02

1

MAXRMS VV

*

*

* Not in Handbook

T=1/f

T

t


90

AC Circuits

Effective or Root-Mean-Square (RMS) Value Example

120.2V = 170V

= 3610 W 3.61kW

91

AC Circuits


92

AC Circuits


93

AC Circuits


VRMS VRMS = VDC = steady-state voltage

94

AC Circuits

Phasor Transforms Phase Angle = as the angle between the reference and voltage as the angle between the reference and current If the phase angle is positive, the signal is called lagging or inductive. If the phase angle is negative, the signal is called leading of capacitive. If the phase angle is zero, the signal is called in phase.

t

Ref

i

v

(+) (-)

ELI the ICE man

If = + Lagging (Inductive)

If = - Leading (Capacitive)

95

AC Circuits

Impedance Definitions

XL

XC

R 90

-90

0

0

901j:Inductor

9011:Capacitor

#

# jj


96

AC Circuits

Example

VMAXSin (wt + )

160V0 w = 400 rads/sec

ZT

R jwL 1/jwC

ZT = ZR + ZL + ZC + ZE

An Element

ZE = RCos + jRSin

Z = R jX ZT = |Z|

VL

i(t) = V/Z = |i| = Tan (x/y)

VL

97

AC Circuits

Example/Contd.

j

Z = R + jwL j/wC + 940

R = RCos + jRSin

)F 10 6x

98

AC Circuits

Example/Contd.

(Voltage Across Inductor)

99

AC Circuits

Example

i1 i2

1800 1800

- j4

i1 =

i2 =

- j0.25 = 0.25-90 - j = 1- 90

100

AC Circuits

Complex Power Lagging (Inductive) Complex Power Triangle

Cos = P/S

I = a + jb

I* = a - jb

KW = p.f. x KVA

S = P + jQ

= kW/kVA


101

AC Circuits Complex Power Example

RIRVVIP 2

2

3714.02.68..2.68387736001440

0

0

$

#

CosfpVarjWS

102

AC Circuits

Complex Power

S = P + jQ = I* V Inductive

ELI Q = - 3600 VAR

If S were Capacitive

ICE

P = 1440 W

S = 3877 VA

103

AC Circuits

Resonance

CL

LCf

00

00

1

)(Resonance 12

"

"

"

R L C

R C L

Z

RLRC

= w0 (RC/2)

Z

= w0 (R/2L) Resistance = Reactance

PARALLELSERIES Q

Q 1

X = XL + XC


104

AC Circuits

Example

Z

At Resonance:

XL = - XC

Therefore, Z = R

105

AC Circuits

Example/Contd.

= W0 (R/L)

At Resonance VC = - VL

Peak Voltage across each component.

VL

VC

- 90

+90

106

AC Circuits

Example

52

10NEWR

2

2

121

1212 12

OLDNEW

NEWOLD

OLDNEW

RR

RRRR

RCRC

BWBW

(ROLD)

(RNEW)

10kHz 20 kHz BW1 BW2

107

AC Circuits

Analog-to-Digital Conversion Voltage Resolution The range from a high voltage, VH, and a low voltage, VL, is divided up into the 2 ranges For example, if all the bits are 1 then the discrete value is somewhere between VH and VH v. To calculate the discrete value from the digital value use


number of bits Voltage Resolution =

Discrete Value =

Typical values for n:

4, 8, 10, 12, 16 bits

Key Design Parameter

N: { 0, 2 - 1 }

digital value VH

VL

Highest measurable voltage = VH - VL

108

AC Circuits

Analog-to-Digital Conversion Example

LHVn

LH

HLnLH

V

VVVV

VVVV

2 221

2

109

AC Circuits

Transformer Equivalent Circuit with Secondary Impedance

IS

VS

Zeq

P

S

S

P

SSPP

SECPRI

II

VV

IVIVPP

PPRI

PSEC


110

AC Circuits

Transformer Example

V2

111

AC Circuits

Transformer

Example

3 + j1 (50/100)(12 + j4)

10050

2

1

NNa

Ideal Transformer has no losses

= 6 + j2

112

AC Circuits

Rotating Machines

Angular Velocity = = 2w/p = 4f/p w = (p)/2

= 120w/2p = 9.5 x

Synchronous Speed (RPM):

p = # of poles n = rotor operating speed 120

ABC = BCA = CAB


113

AC Circuits

Example Armature: The structure (rotor or stator) where a voltage is induced by a magnetic field. (Motor armature: rotor; Generator armature: stator.)

= 125.67 rads/s

pnS " 2

602

f = p/4 p = 4f

1 pair = 2 poles

114

AC Circuits

Rotating Machines

snn

snn

nns

speedssynchronounspeedoperatingn

S

S

S

S

1

1

1


115

AC Circuits

Rotating Machines Example

p = 4 poles

60 Hz = 1800 RPM

50 Hz = 1500 RPM 5%

116

AC Circuits

Rotating Machines

(Kf is a design constant)

Lf

If

Rf

Vf

(Webers)

Stator

Rotor

Motor


117

AC Circuits

Rotating Machines DC Generators

A device that produces DC potential

B

VOUT

VRMS Vpeak

or VMAX

%)64(6366.02

%)70(707.02

MAXMAX

AVG

MAXMAX

RMS

VVV

VVV

118

AC Circuits

Rotating Machines DC Generators

Va = Ka n

Power MechanicalxTPTorque T

Speedngular ConstantDesign

peedOperatingFlux

m

AK

Sn

a

%

(Heat Loss)

Pa = IaRa + IaEg = PHEAT + Pm


119

AC Circuits

Rotating Machines Example

RPMRPMAA

RPMxn

n

nn

nKnK

nn

II

a

a

150018001012

15006

5180056

1012 1800

1012

VV

2

2

2

1

2

1

2

1

2

1

2

1

%

%

120

AC Circuits

Rotating Machines

Example/Contd.

RPMxn

n

nn

II

15006

5180056

1012 1800

1012

2

2

2

1

2

1

121

AC Circuits

Rotating Machines

Va = Kan Pm = Ta x = VaIa

Handbook)in 198 (Pg.2

1

2

1

%

%

TT

= 2ns/60

Ph = Heat Loss

(Nm)

122

AC Circuits

Rotating Machines

Example

123

IV. Computers

Terminology, Spreadsheets, and

Structured Programming

124

Computers

Terminology CPU stands for Central Processing Unit. This is where all the computer's data processing is handled - all the

data manipulation, calculations and formatting data for output. When you buy a computer you will become more focused on the CPU and its capabilities.

The execution of the instructions within the computer system is extremely fast and is measured in cycles of time and referred to as megahertz. For this reason the MHz of a computers processor is sometimes referred to as its clock speed. This pulsing is expressed as "MHz" e.g. 2400Mhz.

The CPU pulsing is turning electrical current on and off. An electrical circuit can be either open or closed, and the power is either running through or not; that is, it is either on or off. This two-state situation is called binary, and the two states are controlled by binary digits or bits. The two bits of a computer are zero and one. For you to talk to your computer your message must be translated into binary form, a series of ones and zeros.

The CPU is located usually around the center of the motherboard, and under a giant fan employed to keep it cool. The motherboard is the main piece of circuitry of your computer. It houses the CPU, the ROM and RAM, a variety of computer cards for receiving signals from other input devices, the power supply, hard drive and so on.

The term CPU is sometimes used to refer to the case that houses the computers components, but technically the CPU is only one chip inside the computers case. The case is sometimes described as the "tower" which means the case stands on the floor on its narrow side. Whereas a "desktop" machine sits on your desk and generally has the monitor placed on it.

125

Computers

Terminology

RAM RAM means Random Access Memory" (or sometimes "ready access memory"). It is like a temporary notepad where your computer sends information it has processed before writing it to disk, or where instructions from other input devices (keyboard, mouse, floppy drive) are stored. The words "random access" indicate that memory locations in RAM are accessible in any order unlike sequential access of a data cassette tape, for example.

ROM ROM means Read Only Memory. Your computers ROM is a chip on the motherboard that stores a permanent set of start-up instructions for your computer. The familiar term for ROM is "BIOS", short for "basic input output system". ROM is sometimes referred to as "firmware" (as opposed to software) because it is permanently housed inside the hardware. Instructions in ROM remain intact when the power is off. Generally you can not alter the information in ROM; it is fixed at the time of manufacture. Some manufacturers provide updates to the ROM of their chips. The process of installing these updates is called "flashing the bios". This is only recommended for advanced users and only after all other avenues of repair are explored. DRAM Dynamic random-access memory (DRAM) is a type of random-access memory that stores each bit of data in a separate capacitor within an integrated circuit. The capacitor can be either charged or discharged; these two states are taken to represent the two values of a bit, conventionally called 0 and 1.

126

Computers

Terminology Bit, Bytes and Binary A bit stands for Binary Digit. A byte is 8 bits. A kilobyte (kB) is 1,024 bytes. A megabyte (MB) is approximately 1,024,000 bytes. A gigabyte (GB) 1,024,000,000 bytes - thats a lot of bits!!! A bit is the smallest information parcel on a computer. A computer file can be made up of many hundreds of bytes and many thousands of bits. The bits are like switches, they are represented by 0s and 1s. Characters on a keyboard are therefore represented by 1s and 0s, so a "P" may be 01010000 to a computer - it knows when it sees those on/off instructions to output a "P". Hardware Hardware is the term given to all the physical parts of a computer system. Hardware includes the monitor, the keyboard, the mouse, the main case which houses the RAM, CPU and the motherboard. Operating System Every computer needs a set of programs called the operating system to run the system and make all the other programs work. The word processor, database or spreadsheet programs can not operate unless the operating system is present. Programs written for one operating system will not work on a different operating system. Common operating systems include: Windows 7, Vista, Macintosh OSX, and Linux. There are many versions aimed at different uses.

127

Computers

Terminology Hard Drive A computers hard disk drive is like an audio CD, except the computer can read and write to it. In other words, a computer can take data off its hard drive (to process it in the CPU or place it in RAM to work with) or it can record the results of the work it does back to the disk, this is called "writing to disk". The abbreviation HDD means "hard disk drive". If the HDD is opened up (not recommended as this usually voids the manufacturer's warranty), a pancake stack would be found on double sided disks. Each side of each disk has an arm that holds a "head". The head is responsible for writing data to the disk. When one handles a disk, carry a computer, or laptop, one should be careful not to bang the case as this can force the heads to collide with the surface of the disk and potentially damage it. Hard disks are "formatted" to make them writable. Through this process "clusters", "sectors" and a "file allocation table" are created. With these mechanisms the computer writes information to the disk and can retrieve it later.

128

Computers

Terminology

access time - The performance of a hard drive or other storage device - how long it takes to locate a file. application - a program in which you do your work. ASCII (pronounced ask-key ) - American Standard Code for Information Interchange. a commonly used data

format for exchanging information between computers or programs. bit - the smallest piece of information used by the computer. Derived from "binary digit". In computer

language, either a one (1) or a zero (0). backup - a copy of a file or disk you make for archiving purposes. boot - to start up a computer. bug - a programming error that causes a program to behave in an unexpected way. bus - an electronic pathway through which data is transmitted between components in a computer. byte - a piece of computer information made up of eight bits. cartridge drive - a storage device, like a hard drive, in which the medium is a cartridge that can be removed. Chooser - A desk accessory used to select a printer, or other external device, or to log onto a network. Clock Rate (MHz) - The instruction processing speed of a computer measured in millions of cycles per second

(i.e., 200 MHz). compiler - a program the converts programming code into a form that can be used by a computer. compression - a technique that reduces the size of a saved file by elimination or encoding redundancies (i.e.,

JPEG, MPEG, LZW, etc.) CPU - the Central Processing Unit. The processing chip that is the "brains" of a computer. crash - a system malfunction in which the computer stops working and has to be restarted. daisy chaining - the act of stringing devices together in series (such as Small Computer System Interface

(SCSI). database - an electronic list of information that can be sorted and/or searched.

129

Computers

Terminology driver - a file on a computer which tells it how to communicate with an add-on piece of equipment (like a

printer). Ethernet - a protocol for fast communication and file transfer across a network. expansion slot - a connector inside the computer which allows one to plug in a printed circuit board that

provides new or enhanced features. file - the generic word for an application, document, control panel or other computer data. folder - an electronic subdirectory which contains files. gig - a gigabyte = 1024 megabytes. hard drive - a large capacity storage device made of multiple disks housed in a rigid case. head crash - a hard disk crash caused by the heads coming in contact with the spinning disk(s). high density disk - a 1.4 MB floppy disk. installer - software used to install a program on a hard drive. kilobyte - 1024 bytes. Measurements

*a bit = one binary digit (1 or 0) * 1 byte (or Octet) = 8 bits

*1024 bytes = one kilobyte *K = kilobyte *Kb = kilobit *MB = megabyte *Mb = megabit *MB/s = megabytes per second *Mb/s = megabits per second *bps = bits per second i.e., 155 Mb/s = 19.38 MB/s

megabyte - 1024 kilobytes. memory - the temporary holding area where data is stored while it is being used or changed; the amount of

RAM a computer has installed. nibble - a piece of computer information made up of four bits. It is equal to half a byte.

130

Computers

Terminology operating system - the system software that controls the computer. RAM - acronym for Random-Access Memory. RISC - acronym for Reduced Instruction Set Computing; the smaller set of commands used by the PowerPC

and Power Mac. ROM - acronym for Read Only Memory; memory that can only be read from and not written to. server - a central computer dedicated to sending and receiving data from other computers (on a network). software - files on disk that contain instructions for a computer. 32 bit addressing - a feature that allows the Mac to recognize and use more than 8MB of memory. vaporware - "software" advertised, and sometimes sold, that does not yet exist in a releasable for. virtual memory - using part of your hard drive as though it were "RAM". WORM - acronym for Write Once-Read Many; an optical disk that can only be written to once (like a CD-

ROM).

131

Computers

Spreadsheets A B C D

1

2

3

4

Functions

=SUM(A4:D4), INT(n) = m where m n, SIN(), PMT(rate, nper, pv, fv, n) where rate = interest, nper = total # of payments for loan, n = 0 (end of period) or 1

(beginning of period).


132

Computers

Spreadsheets

Example

C3 = $A$4 + B$2 + B2

C4 = $A$4 + B$2 + B3

C5 = $A$4 + B$2 + B4

D5 = $A$4 + C$2 + C4

A B C D E F

1

2

3

4

5

$A$4


133

Computers

Spreadsheets Example

A

4

5

6

7

6

A4 + $A$4

A5 + $A$4

A6 + $A$4

12

18

24

Relative Addressing

Absolute Addressing

134

Computers

Algorithm Flowcharts and Symbols

(To a different page)

(Connects to algorithm or subroutine)

135

Computers

Algorithm Flowcharts

Example

to main program

(= is replaced by)

X = 4

T = 5

136

Computers

Structured Programming Sequence of operations from left to right in the following hierarchy: Brackets and Parentheses, then Exponentiation, then multiplication and division, then addition and subtraction.

- 6 5

/ 4* 3

**or ^ 2) ( ] [ 1

Hierarchy lOperationa

th

th

th

rd

nd

st

137

Computers

Structured Programming Example

12224

3 - 56 4

x

138

Computers


Example

7X 8X 4T 734X 4 82X 4T 824 X 4 2

4222X 4 2 2 T 3 4T X

&

'

&

139

Computers


Example (Contd.)

140

Computers


Example

6X 615X 514X61 - 7T 71 - 8 T

Pass 2 Pass 1 ndst

return to main program

141

Computers


142

Computers


8X ENDWHILE

826X 624XXT 42 - 6T 62 - 8T

No Yes Yes Pass 3 Pass 2 Pass1 rdndst

'

143

Computers


Example (Contd.)

144

Computers


145

Computers

Structured Programming Example (Contd.)

146

Computers


Example What is most nearly Z after structured programming is executed?

If not DO

147

Computers


Example (Contd.)

3.7429

IY Z

413I 312 I Endwhile 292)(2 25Y 25 5)(50Y 211I 2X 5X 3X

3I No, 3I Yes, 3I Yes, 3I Yes,Pass4 Pass3 Pass2 Pass1 thrdndst

(

(

&

148

V. Ethics and Business Practices

Professional Engineering Ethics

149

Ethics and Business Practices

Hints for Exam Questions Page 121 from Supplied Reference Handbook

150


Obligations to Society Page 121 from Supplied Reference Handbook

151


Obligation to Society

152


Obligations to Employers and Clients

153



154



Example

155



Example

156



157


Obligations to Other Registrants

158


Obligations to Other Registrants

159


Expert Witness

160


Expert Witness

Example

161


Consultants

162


Consultants

Example

Documents

Electricity CPUs Ethics Refresher