# CS计算机代考程序代写 Family Name ……………………………….. Given Name ………………………………… Student No. …………………………………. Signature ……………………………………..

Family Name ……………………………….. Given Name ………………………………… Student No. …………………………………. Signature ……………………………………..
THE UNIVERSITY OF NEW SOUTH WALES
School of Electrical Engineering & Telecommunications
MID-SEMESTER EXAMINATION
Summer Semester 2017-2018
ELEC1111
Electrical and Telecommunications Engineering
TIME ALLOWED: 1.5 hours (90 minutes) TOTAL MARKS: 100
TOTAL NUMBER OF QUESTIONS: 4
THIS EXAM CONTRIBUTES 25% TO THE TOTAL COURSE ASSESSMENT
This paper contains 6 pages.
Candidates must ATTEMPT ALL questions.
Marks for each question are indicated beside the question.
This paper MAY NOT be retained by the candidate.
Print your name, student ID and question number on the front page of each answer book. Authorised examination materials:
Candidates should use their own UNSW-approved electronic calculators.
This is a closed book examination.
All answers must be written in ink. Except where they are expressly required, pencils may only be used for drawing, sketching or graphical work.
For the numerical solutions, you can use either fraction form or floating-point form (maximum 2 digits after decimal point is enough)
Page 1 of 6

QUESTION 1 [25 marks]
(i) [12 marks] For the circuit shown in Figure 1,
a. (10 marks) Calculate the equivalent resistance 𝑅eq as seen from terminals a-b. b. (2 marks) Find the current 𝑖 through the network using the result of part (a).
i
a 2.5Ω
30 Ω
20Ω
25 Ω
10 Ω
60 V
Req
b
15 Ω
Figure 1
(ii) [13marks]ForthecircuitshowninFigure2,
a. (10 marks) Calculate all the powers absorbed and/or supplied by the elements. b. (3 marks), Verify the law of conservation of energy.
5 mA 10 kΩ vo
0.01vo 5 kΩ 20 kΩ
Figure 2
12 Ω
60 Ω
Page 2 of 6

QUESTION 2 [30 marks]
(i) [14 marks] For the circuit shown in Figure 3,
a. (10 marks) Apply nodal analysis and show that the nodal equations are given as below,
{ 9𝑣1 − 5𝑣2 = 120 𝑣1 − 5𝑣2 = −40
b. (4 marks) Given the values of node voltages as 𝑣1 = 20 V and 𝑣2 = 12 V, calculate the total power supplied by the sources.
v1 8Ω v2 3Ω
3 A 10 Ω
6 Ω
15 V
Figure 3
(ii) [16 marks] For the circuit shown in Figure 4,
a. (12 marks) Apply mesh analysis and show that the mesh equation are given as
below,
𝑖1 + 3𝑖2 − 2𝑖3 = 8 { 𝑖1 + 𝑖2 − 4𝑖3 = −1
𝑖1 − 𝑖2 = −4
b. (4 marks) Given the values of mesh currents as 𝑖1 = −0.5 A, 𝑖2 = 3.5 A, and
𝑖3 = 1 A, find the voltage 𝑣 across 4-A current source.
v4A 400 V i1
i2 100 Ω 50 Ω
50 Ω i3
100 Ω
50 V
Figure 4
Page 3 of 6

QUESTION 3 [25 marks]
(i) [15 marks] In the circuit of Figure 5,
a. (12 marks) Use only source transformation to reduce the circuit into a single resistor in series with a single voltage source as seen from terminals a-b, and then determine Thevenin voltage 𝑉Th and Thevenin resistance 𝑅Th at the terminals a-b.
b. (3 marks) Find the value of load resistance 𝑅𝐿 for maximum power transfer, and then calculate the maximum power that can be delivered to 𝑅𝐿.
20 Ω
20 V
5Ω 10 V
5 Ω
3 A
20 Ω
10 Ω
a
b
RL
Figure 5
(ii) [10marks]ForthecircuitshowninFigure6,obtaintheNortonequivalentcircuitas seen from terminal a-b and draw the Norton equivalent circuit.
4Ω 1Ω
a ix
10ix 2Ω
Figure 6
b
Page 4 of 6

QUESTION 4 [20 marks]
(i) [8 marks] In circuit shown in Figure 7, the switch has been in position A for a long
time. At 𝑡 = 0, the switch moves to position B.
a. (4 marks) Find the voltage 𝑣𝐶(𝑡) across the capacitor immediately after the switch changes to position B, 𝑣𝐶(0+), and its final voltage when 𝑡 → ∞, 𝑣𝐶(∞).
b. (3 marks) Derive an expression for the capacitor voltage 𝑣𝐶(𝑡) for 𝑡 > 0.
c. (1 marks) Find the current 𝑖(𝑡) through 3-kΩ resistor for 𝑡 > 0.
A t=0 6kΩ B
3 kΩ
Figure 7
i(t)
10 mA 60 V
2 mF vC(t)
(ii) [12 marks] In the circuit of Figure 8, the switch has been closed for a long time before being opened at 𝑡 = 0.
a. (10 marks) Derive an expression for the inductor current 𝑖𝐿(𝑡) for all time (i.e., for both 𝑡 < 0 and 𝑡 > 0) and sketch 𝑖𝐿(𝑡) as a function of time showing all critical points in the sketch.
b. (2 marks) Find the total energy stored or released by the inductor for 𝑡 ≥ 0, i.e., from 𝑡 = 0 to 𝑡 → ∞.
t=0
20 Ω

v(t) 6Ω
iL(t) 0.5 H
20 V
2 A
12 Ω
Figure 8
END OF PAPER
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