ELEC1111  Midterm test revision Solution
1.
What is the potential difference between the points x and y in the network?
SOLUTION
_
2.
Find current i.
SOLUTION
i = 0, as all the current reaching point a from source, will be diverted through the short circ
ELEC1111  Midterm test revision Solution
1.
What is the potential difference between the points x and y in the network?
SOLUTION
_
2.
Find current i.
SOLUTION
i = 0, as all the current reaching point a from source, will be diverted through the short circ
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THE UNIVERSITY OF NEW SOUTH WALES
FINAL EXAMINATION,
SESSION II  2008
ELEC1111
Electrical and Telecommunications Engineering
1.
Time allowed: 3 hours.
2.
PART A: 10 Multiple Choice Questions (Value = 30%).
3.
Ques
THE UNIVERSITY OF NEW SOUTH WALES
SAMPLE EXAMINATION
ELEC1111
Electrical and Telecommunications Engineering
SOLUTIONS
Question 1: (6 marks)
Consider the circuit shown in Figure 1.
(a)
Calculate the voltage V1. (V1 = 15 2x2 = 11V)
(b)
Calculate the power
CHAPTER 8
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 = 524,288 bits Startin
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THE UNIVERSITY OF NEW SOUTH WALES
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Sample Examination
Electrical and Telecommunications Engineering
TIME ALLOWED:
TOTAL MARKS:
TOTAL NUMBER OF QUESTI
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Lecture 7: Capacitors and RC
Transients
ELEC1112 Electrical Circuits
Capacitors
Capacitor: Constructed using two parallel conducting plates
separated by an insulating material (e.g. air)
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Capacitance

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THE UNIVERSITY OF NEW SOUTH WALES
SAMPLE EXAMINATION
ELEC1111
Electrical and Telecommunications Engineering
Time allowed: 3 hours.
1. CANDIDATES should attempt ALL QUESTIONS
2. Questions are of different values.
3.
THE UNIVERSITY OF NEW SOUTH WALES
School of Electrical Engineering & Telecommunications
SAMPLE FINAL EXAMINATION
ELEC1111
Electrical and Telecommunications Engineering
TIME ALLOWED:
TOTAL MARKS:
TOTAL NUMBER OF QUESTIONS:
3 hours
100
6
Reading Time: 10 mi
The University of New South Wales
School of Electrical Engineering and Telecommunications
ELEC1111
Electrical and Telecommunications Engineering
Practical Exam  Sample paper
Family name:
Student number:
Q1. Set the signal generator to give a sine wave wi
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 an intern
CHAPTER 11
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 ) E11.2
A= v
RL Vo 7
CHAPTER 10
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 = I s exp(vD / nVT ) 1 = 9.502
CHAPTER 13
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 ES exp BE V T
i v BE VT ln E I ES
v BC
CHAPTER 15
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 the periphery of th
CHAPTER 16
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 Pout 100% = = 91.76% 50 7
CHAPTER 14
Exercises
E14.1
(a) iA =
vA RA
iB =
vB RB
iF = iA + iB =
vA RA +
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 the output voltage is independent of th
CHAPTER 17
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( ) + KI m cos(t 240 ) cos( 12
CHAPTER 12
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 and vDS = 6 V: Because
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Student No. .
Student Signature .
THE UNIVERSITY OF NEW SOUTH WALES
SAMPLE EXAMINATION
ELEC1111
Electrical and Telecommunications Engineering
Time allowed: 3 hours.
1. CANDIDATES should attempt ALL QUESTIONS
2. Questions are of different values.
3.
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Student No. .
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THE UNIVERSITY OF NEW SOUTH WALES
School of Electrical Engineering & Telecommunications
MIDSESSION EXAMINATION
S1 2015
ELEC1112
Electrical Circuits
TIME ALLOWED:
TOTAL MARKS:
TOTAL NUMBER OF QUESTIONS:
Determine the current i
Find the equivalent resistance between terminals A and B, Req
Find the equivalent resistance between terminals C and D, Req
Find the power provided by or dissipated by each element in the circuit, and verify that the law of
conserv
Lecture 15: Digital Circuits
ELEC1112 Electrical Circuits
Analogue versus Digital
Twolevel logic
Almost universal voltage
encoding Level values:
technology
dependent
power supply
dependent
Typical electronic systems
Why use digital
Good noise rejection
PROGRAMMING ORGANIC MATERIALS
MIT researchers put bacteria to work producing
conducting biofilms
"What we've been able to achieve here is use living
cellular communities to organize and make nonliving
material systems. My thought is that the tools of
s
Midsession exam
Thursday 12:001:00 in Week 7
During normal lecture class ChemSc M18
Please be on time 55 minutes duration
May cover material up to the end of Week 6
Lecture 8: Inductors and RL
Transients
ELEC1112 Electrical Circuits
Inductors
Inductors
NOTE: For each of the questions, enter the answer values and click on the check icon
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Page 1 of 4
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Explanation (Solution)
Eapp
2kp
z3
Also,
Eact
Where,
2kpz
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2
p qd
is the electric dipole moment and
k
So,
Eapp
Eact
2kp ( z 2 d
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Academic English
Listening
Final Examination
Time Allowed: 20 minutes
This booklet consists of 8 pages including the covers
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(+)C Fin 08
UNSW Foundation Year
Background
Engineers are increasingly seeking to use their expertise to assist development in disadvantaged and
developing communities. By partnering with Engineers Without Borders (EWB) and the United Nations High
Commissioner for Refugees (UNHCR) Zambia