Test Type Name: Company: Lab name: Test date: Geometry: Specimen Specimen name: Width: Thickness: Length:
Compression Test with Secondary Parallel Sample Fall 2003-Group 2 AUS Material Sc. Lab 10/19/2003 Rectangular area=.000625 1 Wood 25 mm 25 mm 1
CHAPTER 11 KEY EQUATIONS
Equation Number Key Equation v o (t ) = Avv i (t ) 11.1 i 11.2 Ai = o 11.3 11.4 11.5 11.6 11.7 11.8 11.9
ii v R i R Ai = o = o L = Av i ii v i Ri RL P G = o Pi R P VI G = o = o o = Av Ai = (Av )2 i Pi Vi I i RL Av = Av 1Av 2
CHAPTER 13 KEY EQUATIONS
Equation Number Key Equation 13.1 v iE = I ES exp BE V T iE = iC + iB 13.2
- 1
13.3 13.6 13.9
=
iC iE
13.10 13.20 13.21 13.23 13.24 13.35 13.38 13.39 13.40 13.43 13.44 13.45 13.46
v iC I s exp BE V T i
CHAPTER 17
Solutions for Exercises
E17.1 From Equation 17.5, we have
Bgap = Kia (t ) cos( ) + Kib (t ) cos( - 120 ) + Kic (t ) cos( - 240 )
Using the expressions given in the Exercise statement for the currents, we have
Bgap = KI m cos(t ) cos( )
CHAPTER 16
Solutions for Exercises
E16.1 The input power to the dc motor is Pin = Vsource I source = Pout + Ploss Substituting values and solving for the source current we have 220I source = 50 746 + 3350 I source = 184.8 A Also we have
50 746 Pou
CHAPTER 15
Solutions for Exercises
E15.1 If one grasps the wire with the right hand and with the thumb pointing north, the fingers point west under the wire and curl around to point east above the wire. If one places the fingers of the right hand on
CHAPTER 14
Solutions for Exercises
E14.1
(a) iA =
vA RA
iB =
vB RB
vA RA
iF = iA + iB =
+
v A vB + RA RB
v o = -RF iF = -RF
(b) For the vA source, RinA (c) Similarly RinB = RB .
vB RB v = A = RA . iA
(d) In part (a) we found that th
CHAPTER 13
Solutions for Exercises
E13.1 given by the Shockley equation: - 1 v For operation with iE > I ES , we have exp BE > 1 , and we can write V T v iE I ES exp BE V T Solving for v BE , we have The emitter current is v iE = I
CHAPTER 12
Solutions for Exercises
E12.1 (a) vGS = 1 V and vDS = 5 V: Because we have vGS < Vto, the FET is in cutoff. (b) vGS = 3 V and vDS = 0.5 V: Because vGS > Vto and vGD = vGS - vDS = 2.5 > Vto, the FET is in the triode region. (c) vGS = 3 V an
CHAPTER 11
Solutions for Exercises
E11.1 (a) A noninverting amplifier has positive gain. Thus v o (t ) = Avv i (t ) = 50v i (t ) = 5.0 sin(2000t ) (b) An inverting amplifier has negative gain. Thus v o (t ) = Avv i (t ) = -50v i (t ) = -5.0 sin(2000t
CHAPTER 10
Solutions for Exercises
E10.1 Solving Equation 10.1 for the saturation current and substituting values, we have
Is =
iD exp(vD / nVT ) - 1
10 -4 = exp(0.600 / 0.026) - 1 = 9.502 10 -15 A Then for vD = 0.650 V, we have
= 0.6841 mA
iD =
CHAPTER 9
Solutions for Exercises
E9.1 The equivalent circuit for the sensor and the input resistance of the amplifier is shown in Figure 9.2 in the book. Thus the input voltage is
v in = v sensor
Rsensor + Rin
Rin
We want the input voltage with
MCE 230 L
Materials Science
Corrosion Test
Gagan Gururaj ID: 8526 Group 2 Date of Experiment: 9/11/03
2
Table of Contents
1) Abstract.3 2) Part A: Introduction..3 3) Theory.3 4) Experimental procedure.4 5) Data..5 6) Calculations.6 7) Part B: Intr
CHAPTER 10 KEY EQUATIONS
Equation Number Key Equation 10.1 v iD = I s exp D nV T
10.2 10.5 10.10 10.12 10.15
kT q VSS = RiD + v D IT C = L Vr IT C = L 2r V VT =
di rd D dv D
- 1
10.22
rd =
nVT I DQ
Q
-1
Specimen Aluminum Aluminum Brass Brass Aluminum Aluminum Aluminum
Notch type V U V U V V V
Temperature (degrees celsius) 25 25 25 25 100 200 300
Width (cm) 0.5 0.49 0.5 0.49 0.5 0.5 0.5
Depth (cm) 0.49 0.41 0.49 0.41 0.49 0.49 0.49
Net Impact Wo
SUMMARY:
Fatigue is a form of failure that occurs in structures subjected to dynamic and fluctuating stresses (e.g., bridges, aircraft, and machine component); in this experiment we see that four different metals aluminum, copper, brass and steel, wh
Test Type Name: Company: Lab name: Test date: Geometry: Specimen Specimen name: Width: Thickness: Support span: Span ratio: Fixture type: Comment:
Static Bending Test Fall 2003 Gourp 2 AUS Materials Science 11/16/2003 Rectangular 1 Wood 25 mm 25 mm
Test Type Name: Company: Lab name: Test date: Geometry: Specimen Specimen name: Length: Diameter:
Tensile Test Without Extensometer Group2 AUS Material Sc. Lab 9/28/2003 Circular 1 aluminum 95 mm 6.1 mm Extension mm 0.000478316 0.01973053 0.03683033
MCE 230 L
Materials Science
Charpy Impact Test
Gagan Gururaj ID: 8526 Group 2 Date of Experiment: 5/10/03
Abstract:
Aluminum and brass samples were tested using the Charpy pendulum impact test at room temperature and aluminum under different temper
CHAPTER 17 KEY EQUATIONS
Equation Number Key Equation 17.13 s = P 2 120f 17.14 ns =
17.16 17.17 17.23 17.24 17.25
P s - m ns - nm s = = s ns slip = s
Pdev = 3
1-s
Pr = 3Rr (I r )2
s
Rr (I r )2
17.26 17.27
Ps = 3Rs I s2 Pin = 3I sVs cos(
CHAPTER 15 KEY EQUATIONS
Equation Number Key Equation f = qu B 15.1 f = quB sin( ) 15.2 15.3 15.4 15.5 15.6 15.7 15.8 15.9 15.10 15.11
df = idl B f = ilB sin( )
=
15.12 15.15
= BA = N d e= dt e = Blu B = H 0 = 4 10 -7 Wb Am r = 0
Hl =
A
Measurement results for Fatigue Experiment Type of material: Aluminum Number of load cycles (Sample without polishing) (Sample without polishing) (load cycles)
Applied Load
CHAPTER 12 KEY EQUATIONS
Equation Number Key Equation iD = 0 for v GS Vto 12.1 12.2
2 iD = K [2( GS - vto )v DS - v DS ] v W KP K = L 2
12.3
12.4 12.6 12.9 12.24
iD = K ( GS - vto ) v
2
2 iD = Kv DS v DD = RD iD (t ) + v DS (t )
gm = 2 KI D
CHAPTER 8
Solutions for Exercises
E8.1 The number of bits in the memory addresses is the same as the address bus width, which is 20. Thus the number of unique addresses is 220 = 1,048,576 = 1024 1024 = 1024K. (8 bits/byte) (64 Kbytes) = 8 64 1024
CHAPTER 7
Solutions for Exercises
E7.1 (a) For the whole part we have: Quotient Remainders 23/2 11 1 11/2 5 1 5/2 2 1 2/2 1 0 1/2 0 1 Reading the remainders in reverse order we obtain: 2310 = 101112 For the fractional part we have 2 0.75 = 1 + 0.5 2