ENGG1203: Introduction to Electrical and Electronic Engineering
Second Semester, 2015–16
Homework 2
Due:
Mar 17, 2016, 11:55pm
Instruction: Submit your answers electronically through Moodle. In Moodle, you must submit under
the PDF Submission link with ONE PDF file containing answers to all remaining questions.
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Question 1
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*** Special Note: No late homework will be allowed. We will post the solution on March 18, so you
can use it to study for the midterm on March 22.
Short Questions
Part(a)
Find the currents i1 and i2 for the following circuit:
3Ω
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10 Ω
5Ω
−
+
12 V
10 Ω
5Ω
3Ω
20 Ω
i1
20 Ω
i2
The two 10 Ω resistors in parallel become 5 Ω; the two 20 Ω resistors in parallel become 10 Ω.
The “middle” section now has 10 Ω (with current i1 ) and the “right” section now has 15 Ω (with
current 2i2 ). These two sections are in parallel, giving an equivalent resistance of
1
1
1
=
+
R
10 15
=⇒
R = 6 Ω.
Together with the two 3 Ω resistors, the circuit has 12 Ω resistance, giving a current of 1 A going
out of the voltage source. Thus, 6 V drops across the 10 Ω resistance to give i1 , or i1 = 3/5 A;
similarly, 6 V drops across the 15 Ω resistance to give 2i2 , or i2 = 1/5 A.
ENGG1203: Introduction to Electrical and Electronic Engineering
Homework 2
Part(b)
Use KVL to compute the (i) voltage across each resistor and (ii) current flowing through each voltage
source.
+
I1
−
V R2
+
+
10 Ω
−
+
10 V
V R1
10 Ω
−
10 Ω
− 10 V
+
V R3
I2
−
10 − VR1 − VR3 = 0
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VR3 − VR2 + 10 = 0
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Using KVL we have:
Also, using KCL, we have:
VR1
VR2
VR3
=
+
10
10
10
Now,
VR2 + VR3 = 10 − VR3
(4) & (1)
(VR3 + 10) + VR3 = 10 − VR3
from (2)
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Therefore, VR3 = 0, VR2 = 10 − 2VR3 = 10 and VR1 = 10.
VR1
VR2
Also, I1 = 10
= 1, I2 = 10
= 1.
Part(c)
• Current to Voltage Converter: In the following circuit, determine Vo in terms of Iin .
R
Iin
R
−
R/2
Vo
+
r1.0
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(1)
(2)
(3)
(4)
ENGG1203: Introduction to Electrical and Electronic Engineering
Homework 2
• Voltage to Current Converter: In the following circuit, determine Io in terms of Vin .
R
RL
−
R
Vin
Io
+
N
R1
• Current to Voltage Converter: According to the ideal op-amp model, there is no current
flowing into the op-amp and v+ = v− . Therefore,
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Vo = −Iin R
• Voltage to Current Converter: According to the ideal op-amp model, there is no current
flowing into the op-amp and v+ = v. Therefore,
Io =
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Part(d)
Vin
R1
Consider the circuit below, where we are interested to find v.
i1
+
−
+
5V
v
i2
4 kΩ
i3
2 kΩ
−
+
2 kΩ
3V
−
• Find v using KVL around the two loops shown on the circuit, using the fact that i3 = i1 + i2 .
• Find v using KCL with i1 , i2 , and i3 represented in terms of v.
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ENGG1203: Introduction to Electrical and Electronic Engineering
Homework 2
Using KVL, since we have:
5 = 2000i1 + 2000(i1 + i2 ) = 4000i1 + 2000i2
3 = 4000i2 + 2000(i1 + i2 ) = 2000i1 + 6000i2
Solving these simultaneous equations, we have i1 = 1.2 mA and i2 = 0.1 mA. Thus, v = (1.2 +
0.1)(2) = 2.6 V.
We can also solve this using KCL. Since i1 + i2 = i3 ,
v
5−v 3−v
+
=
2000
4000
2000
(10 − 2v) + (3 − v) = 2v
Part(e)
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v = 2.6 V.
In the following circuit, compute the values of R1 and R3 in terms of R2 and R4 , such that vo is always
equal to v1 − 5v2 .
R4
R3
v2
−
R1
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v1
vo
+
R2
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ENGG1203: Introduction to Electrical and Electronic Engineering
Homework 2
Let v+ and v− be the voltage level at the positive and negative inputs of the op-amp, respectively.
In an ideal op-amp, v+ = v− . We can also express each of them using v1 and v2 ; thus,
v+
v1
=
R2
R1 + R2
R2
v+ =
v1 ,
R1 + R2
and
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v− − v 2
v o − v−
=
R3
R4
R2
R2
v
−
v
v
2
1 − R1 +R2 v1 − 5v2
R1 +R2 1
=
R3
R4
R2
1
R1
5
v1 −
v1 −
v2 =
v2
R3 (R1 + R2 )
R3
R4 (R1 + R2 )
R4
Therefore, R3 = 51 R4 and R1 = 5R2 .
Question 2
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The equality must hold for any v1 and v2 , so the coefficients on either side of the equation must
equal. Hence,
R1
1
5
R2
=
and
=
.
R3
R4
R3
R4
Delta Star Transformation
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When simplifying complex resistance network, it is sometimes useful to recognize and transform between
a delta topology by a star topology, which are shown below.
A
A
R
3
R1
C
Ra
R
Rc
b
B
R2
(a) Delta
B
C
(b) Star
Figure 1: Delta and star resistor networks are equivalent if the resistor values are chosen correctly.
While the 2 topologies may look quite different, by choosing the correct resistor values, it can be shown
that the two networks can be equivalent. This exercise helps you to deduce the correct resistor values.
Part(a)
Delta
Consider the delta topology in Figure 1(a). Define Rab , Rbc , and Rca as the equivalent resistances
between terminal A and B, B and C, and C and A respectively. Express Rab , Rbc , and Rca in terms of
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ENGG1203: Introduction to Electrical and Electronic Engineering
Homework 2
the 3 resistors R1 , R2 and R3 .
Using the notation Rx k Ry to means a parallel combination of Rx and Ry , then
Rab = (R3 + R2 ) k R1
=
(R3 + R2 )R1
R1 + R2 + R3
Rbc = (R1 + R3 ) k R2
(R1 + R3 )R2
R1 + R2 + R3
Rca = (R1 + R2 ) k R3
Part(b)
Star
(R1 + R2 )R3
R1 + R2 + R3
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=
N
=
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Now, consider the star topology in Figure 1(b). Similarly define Rab , Rbc , and Rca as the equivalent
resistances between terminal A and B, B and C, and C and A respectively. Express Rab , Rbc , and Rca
in terms of the 3 resistors Ra , Rb and Rc .
The resistances can be simply be obtained by the series combination of the resistors, i.e.:
Rab = Ra + Rb
Rbc = Rb + Rc
Rca = Ra + Rc
Part(c)
Transformation
For the delta and star topology to be equivalent, the values for Rab , Rbc , and Rca must be the same in
both cases. Using your results from above, express Ra , Rb , Rc in terms of R1 , R2 , R3 .
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ENGG1203: Introduction to Electrical and Electronic Engineering
Homework 2
Combining the results from Part (a) and (b), equating on the values of Rab , Rbc and Rca gives:
(R3 + R2 )R1
= Ra + Rb
R1 + R2 + R3
(R1 + R3 )R2
=
= Rb + Rc
R1 + R2 + R3
(R1 + R2 )R3
=
= Ra + Rc
R1 + R2 + R3
Rab =
(5)
Rbc
(6)
Rca
(7)
Subtracting (6) from (7) and adding the result to (5):
(R1 + R2 )R3 − (R1 + R3 )R2 + (R3 + R2 )R1
R1 + R2 + R3
R1 R3
Ra =
R1 + R2 + R3
Similarly, we can obtain:
and
Rc =
Question 3
R1 R2
R1 + R2 + R3
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Rb =
N
2Ra =
R2 R3
R1 + R2 + R3
Superposition Theorem
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When analyzing complex circuits, it is sometimes useful to make use of the superposition theorem. Using
this theorem, it is possible to decompose a circuit with multiple input sources into multiple circuits with
only 1 source. By doing so, it simplifies the analysis and may help to understand the behavior of a
circuit.
In particular, the superposition theorem states that the response of a circuit with multiple input sources
is the sum of the responses from each independent source when acting alone. In practice, to obtain the
response due to one particular source S, one would turn off all the sources (voltage sources or current
sources) in the circuit except S. Turning off a source can be achieved by:
• For voltage sources, replace the source with a short circuit.
• For current sources, replace the source with an open circuit.
The overall response of the circuit can then be obtained by superpositioning each independent response
on top of each other. This question illustrates how superposition theorem works and you will use it to
analyze some complex op-amp circuits.
Part(a)
Circuit Analysis without using Superposition
Compute the current I1 , I2 and I3 in the following circuit.
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ENGG1203: Introduction to Electrical and Electronic Engineering
I1
Homework 2
1Ω
I2
I3
−
+
1Ω
−
+
10 V
5V
Figure 2: Circuit with 2 voltage sources


I1 + I2 + I3 = 0
10 + I3 = 0


I2 = 5 + I3
Solving the equations yield
Part(b)
I2 = −5 A,
I3 = −10 A
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I1 = 15 A,
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Based on KVL and KCL, we have:
Circuit Analysis with Superposition
It is possible to analyze the same circuit in Figure 2 by decomposing it into the following 2 circuits:
I2a
I1b
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I3a
1Ω
−
+
10 V
1Ω
(a) Decomposition A – Turn off right voltage source
I2b
1Ω
I3b
1Ω
5V
−
+
I1a
(b) Decomposition B – Turn off left voltage source
Figure 3: Two decompositions of circuit in Figure 2 with each of the voltage source replaced by a short
circuit.
In Figure 3(a), the 5 V voltage source is replaced by a short circuit while keeping the 10 V source intact.
On the other hand in Figure 3(b), the 10 V is replaced by a short circuit, leaving the original 5 V source
intact. Calculate the values I1a , I2a and I3a from Figure 3(a), and I1b , I2b and I3b from Figure 3(b).
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ENGG1203: Introduction to Electrical and Electronic Engineering
Homework 2
I1a = 20 A
I2a = −10 A
I3a = −10 A
I1b = −5 A
I2b = 5 A
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I3b = 0 A
Part(c)
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According to the superposition theorem, the overall response of the circuit can be obtained by summing
the contribution from each independent sources. Compute the 3 values I1 = I1a + I1b , I2 = I2a + I2b ,
I3 = I3a + I3b . Are they the same as the results you have obtained from Part (a)?
I1 = I1a + I1b = 20 + (−5) = 15
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I2 = I2a + I2b = −10 + 5 = −5
I3 = I3a + I3b = −10 + 0 = −10 The resulst are the same as Part (a).
Part(d)
In the following circuit, calculate the values of R1 and R4 in terms of R2 and R3 such that Vo = V1 −10V2 .
To take advantage of the superposition theorem, set V1 = 0 and V2 = 0 in turn.
R1
R2
V2
−
R3
Vo
V1
+
R4
Vo = V1 − 10V2
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ENGG1203: Introduction to Electrical and Electronic Engineering
Homework 2
Replace the voltage source V1 with a short circuit, i.e., set V1 = 0, then:
V2 −0
0−Vo
R2 = R1
Vo = −10V2
Solving the above gives:
R1 = 10R2
(8)
Now, replace voltage source V2 with a short circuit, i.e., set V2 = 0, then:
(
R4
R4
V1
V −Vo
0− R +R
3
4
= R3 +RR4 1
R2
Vo = V1
Solving the above yields:
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R1 R4 = R2 R3
Now, combining (8) and (9) results in
Question 4
R1 = 10R2
R3 = 10R4
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(
(9)
Rotating Motor
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One key component of your project is the light tracker. In this question, you will explore some of its
circuit function.
Your light tracker is mount on top of a motor and is able to rotate left or right such that it is always
facing directly toward the light source. To detect the direction of light, it uses 2 light-sensitive resistors
that are placed at ±45◦ with respect to centerline of the device as shown below.
0◦
light source
θ
RL
RR
tracker head
The resistance of the light-sensitive resistor decreases when light is shined on it and increases when there
is no light. Therefore, when the light is on the left of the centerline, RL decreases and RR increases.
Similarly, when the light is on the right of the centerline, RR decreases and RL increases.
In addition, the motion of the tracker head is controlled by the motor it is attached to. The rotational
speed of the motor, ω, depends linearly on the voltage across its two input terminals Vm = Vmp − Vmn ,
provided that it exceeds a certain threshold. That is,
(
km Vm if Vm > 2
ω=
(10)
0
otherwise,
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ENGG1203: Introduction to Electrical and Electronic Engineering
Homework 2
where km is a motor dependent constant.
When Vm is positive, the motor turns clockwise. When it is negative, the motor turns counter-clockwise.
Part(a)
First Attempt
Armed with the above information about the light tracker head, your project partner has designed the
following circuit to control the light tracker:
Vdd
RR
Vmp
motor
+−
Vmn
−
+
Vdd /2
N
RL
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Figure 4: First Attempt Motor Control Circuit
Your partner’s idea is that when the light is on the right of the centerline, RR decreases, making Vm
increases, turning the motor clockwise and toward the light. Similarly, when the light is on the left of
the centerline, RL decreases, making Vm decreases, turning the motor counter-clockwise so it will point
back to the light source.
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While the design of the circuit in Figure 4 may seem fine, it does not work as expected when you test
it in the lab. Explain why it does not work according to the design by considering the voltage Vm when
light is on left and on right of the centerline. Assume the resistance of the light-sensitive resistor is 100 Ω
when there is light and 10 kΩ when there is no light. Also assume the voltage sources has 0 resistance,
and the internal resistance of the motor is 4 Ω. Vdd is 5 V.
When the light is on the right, RR ≈ 100, RL ≈ 10000. Therefore, Vmp ≈ 2.6 V and Vm ≈ 0.1 V.
When the light is on the left, RR ≈ 10000, RL ≈ 100. Therefore, Vmp ≈ 2.4 V and Vm ≈ −0.1 V.
As a result, the motor may work according to what is expected but its speed is slow. In fact, the
smaller the resistance of motor is, the smaller Vm is going to be.
Part(b)
Second Attempt
You search the laboratory again find some op-amps. You then modify the circuit as shown below:
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ENGG1203: Introduction to Electrical and Electronic Engineering
Homework 2
Vdd
Vdd
RR
+
Vp
Vmp
+ −
Vmn
−
RL
−
+
N
Does this circuit function correctly? That is, does it correctly track the direction of the light source?
Explain your answer.
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This circuit functions correctly. The introduction of a buffer using the op-amp allows Vp to
change as expected – When light is on right, RR RL , therefore, Vp ≈ Vdd . When light is on
left, RL RR , therefore Vp ≈ 0. Using the op-amp as a voltage follower, Vmp = Vp . Therefore,
light is on right, Vm > 0 and the motor turns clockwise toward the light. When light is on the
left, Vm < 0 and the motor turns counter-clockwise toward the light.
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Part(c)
Armed with your experience in constructing potential divider using resistors, you try to produce the
voltage Vdd /2 using two identical resistors R as shown below:
Vdd
Vdd
Vdd
RR
Vp
+
Vmp
R
+ −
Vmn
−
RL
R
Unfortunately, the circuit is no longer behaving the same as before. Explain why the motor does not
rotate the same way as before. Hint: What is the volgate Vmn in the circuit?
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ENGG1203: Introduction to Electrical and Electronic Engineering
Homework 2
Because of the lack of buffer on the negative end of the motor, the voltage at Vmn tends to follow
that of Vmp . As a result, there’s still not enough voltage (power) delivered into the motor.
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Part(d)
RR
Vp
+
Vmp
+ −
RL
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R
Vn
Vmn
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−
Vdd
Vdd
−
Vdd
+
Vdd
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Show how you may further revise the design above so it functions as desired.
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Your Circuit
R