ENGG1203: Introduction to Electrical and Electronic Engineering
Second Semester, 2014–15
Homework 2
Due:
Mar 16, 2015, 11:55pm
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Short Questions
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Question 1
Part(a)
For the following circuit, compute (i) the voltage across each resistor, and (ii) the current flowing
through each voltage source.
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−
+
i1
100 Ω
−
+
100 mA
5V
3V
i2
1100 Ω
Using KCL,
0.1 = i1 + i2 .
Also, the voltage of each branch must be equal as they are parallel. Therefore,
5 + 100i1 = 3 + 1100i2 .
Solving these simultaneous equations, we have i1 = 90 mA and i2 = 10 mA.
The voltage across the 100 Ω resistor is (0.09)(100) = 9 V, and the voltage across the
1100 Ω resistor is (0.01)(1100) = 11 V.
ENGG1203: Introduction to Electrical and Electronic Engineering
Homework 2
Part(b)
Consider the circuit below, where we are interested to find v.
i1
+
−
+
v
4 kΩ
i3
2 kΩ
3V
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5V
i2
−
+
2 kΩ
−
• 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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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 ,
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v
5−v 3−v
+
=
2000
4000
2000
(10 − 2v) + (3 − v) = 2v
v = 2.6 V.
Part(c)
Find the voltage v across the voltage source in the following circuit.
1.8 kΩ
−
+
v
3.7 kΩ
0.25 mA
2 kΩ
2.3 kΩ
The 0.25 mA current flows through 3.7 + 2.3 = 6 kΩ, giving a voltage of 1.5 V. The same
voltage drops across the 2 kΩ resistor, such that the current flowing through that resistor
is 1.5/2 = 0.75 mA. Thus, 0.75 + 0.25 = 1 mA flows through the 1.8 kΩ resistor, giving a
voltage drop of 1.8 V. Therefore, the voltage source v = 1.8 + 1.5 = 3.3 V.
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ENGG1203: Introduction to Electrical and Electronic Engineering
Homework 2
Part(d)
Find the current i in the following circuit. Also find the power supplied or absorbed at each
component.
i
−
+
3 mA
1 kΩ
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2V
1 kΩ
The voltage drop across the “vertical” resistor is (0.003 + i)(1000) and the voltage drop
across the “horizontal” resistor is (i)(1000). Therefore,
=⇒
i = −0.5 mA.
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(3 + 1000i) + (1000i) = 2
• Power dissipated at the vertical resistor is (0.003 + i)2 (1000) = 6.25 mW.
• Power dissipated at the horizontal resistor is (i)2 (1000) = 0.25 mW.
• Voltage drop across the vertical resistor is (0.003 + i)(1000) = 2.5 V. Hence, power
generated by the current source is (0.003)(2.5) = 7.5 mW.
• Note that negative current flows out of the voltage, so power actually is absorbed
by the voltage source. Power absorbed is (i)(2) = 1 mW.
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You should always check that the total power dissipated equals to the total power generated.
Part(e)
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
−
vo
R1
v1
+
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
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and
vo − v −
v− − v2
=
R3
R4
R2
2
v1 − R1R+R
v1 − 5v2
R1 +R2 v1 − v2
2
=
R3
R4
R2
1
R1
5
v1 −
v2 =
v1 −
v2
R3 (R1 + R2 )
R3
R4 (R1 + R2 )
R4
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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,
R2
R1
1
5
=
and
=
.
R3
R4
R3
R4
Therefore, R3 = 51 R4 and R1 = 5R2 .
Part(f )
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In the following circuit, find Rf in terms of R such that vo = −15vi .
R
+
vi
Rf
R
−
vo
−
+
Rf
R
Let us label the output to the first op-amp as v1 . This op-amp acts as a non-inverting
amplifier; since v1− = v1+ = vi and v1− /R = v1 /(R + Rf ),
Rf
v1 = 1 +
vi .
R
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ENGG1203: Introduction to Electrical and Electronic Engineering
Homework 2
The second op-amp acts as an inverting amplifier; since v2− = v2+ = 0, we can invoke
KCL in the intersection next to v2− such that vi /R + v1 /R + vo /Rf = 0, and therefore
Rf
Rf
vi −
v1
R
R Rf
Rf
Rf
=−
vi −
1+
vi
R
R
R
Rf
Rf
=−
vi 2 +
vi .
R
R
Hence, we can pick Rf = 3R.
Question 2
Circuit Theorems
Superposition
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Part(a)
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vo = −
When analyzing complex circuits, it is sometimes useful to make use of the superposition theorem
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, one would turn off all the other
sources (voltage sources or current sources) in the circuit except this one. Turning off a source
can be achieved by the following rules:
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• 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 (i.e., adding) each
independent response on top of each other.
As an example, considering the following circuit, where we are interested to find v:
2 kΩ
4 kΩ
+
−
+
5V
v
2 kΩ
0.1 mA
−
To apply superposition, find v1 and v2 in the two cases below, and show that v1 + v2 gives the
correct value of v in the original circuit.
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ENGG1203: Introduction to Electrical and Electronic Engineering
Homework 2
2 kΩ
4 kΩ
+
v1
voltage source → short circuit:
0.1 mA
2 kΩ
−
4 kΩ
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2 kΩ
+
−
+
current source → open circuit:
5V
v2
2 kΩ
−
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In the first circuit, 0.1 mA is equally split between the two 2 kΩ resistors, so 0.05 mA
goes through one of them, such that v1 = (0.05)(2) = 0.1 V.
In the second circuit, 5 V is equally divided between the two 2 kΩ resistors, so
v2 = 5/2 = 2.5 V.
Therefore,
v = 0.1 + 2.5 = 2.6 V.
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To show that this value solves the original circuit, let us solve the original circuit by
KCL and KVL. Let the current flowing through the upper 2 kΩ resistor be i; by KCL,
the current through the middle 2 kΩ resistor is i + 0.1. Applying KVL on the voltage
source and these two resistors, we have (2000)(i) + (2000)(i + 0.1) = 5, giving i = 1.2 mA.
Therefore,
v = (2000)(1.2 + 0.1) = 2.6 V.
Part(b)
Thévenin’s Theorem
Another useful tool to simplify the analysis of eletrical circuits is the Thévenin’s theorem. It
converts any complex circuits with voltage sources, current sources, and resistors into one —
from the point of view of a load — with only a single voltage source in series with a resistor.
This is illustrated below:
A
A
network of sources
and resistors
vL
iL
+
Rth
−
+
+
RL
vth
vL
−
iL
RL
−
B
B
RL is the load resistor connected to the terminals A and B. We can maintain the same voltage
vL and current iL in the transformation by the following procedure:
• The Thévenin equivalent voltage, vth , is equal to the voltage across AB if there is no load,
i.e., an open circuit;
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ENGG1203: Introduction to Electrical and Electronic Engineering
Homework 2
• Calculate the current I when AB is short-ciruited. Then, the Thévenin equivalent resistance, Rth , is equal to vth /I.
Now consider the circuit below. Find the Thévenin equivalent voltage and resistance. What is
the resultant voltage across AB if they are connected by a 2 kΩ resistor?
A
4 kΩ
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2 kΩ
5V
0.1 mA
−
+
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B
When AB is an open-circuited, the 0.1 mA from the current source can only go through
the two resistors. Across the 2 kΩ resistor, it creates a voltage difference of 0.2 V; from
the direction of the current, we know this should add to the 5 V source, hence
vth = 5 + 0.2 = 5.2 V.
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On the other hand, when AB is short-circuited, 5 V must drop across the 2 kΩ resistor,
so there is a current of 2.5 mA. This is added to the 0.1 mA supplied by the current
source, giving I = 2.6 mA. Therefore,
Rth = 5.2/2.6 = 2 kΩ.
Thus, we have the following Thévenin equivalent circuit:
A
+
2 kΩ
vL
5.2 V
iL
RL
−
+
−
B
Now if the load is a 2 kΩ resistor, then half of the voltage drops across it, or
vL = 5.2/2 = 2.6 V.
Compare your answer with Part (a)!
Part(c)
Norton’s Theorem
There is a “dual” to the Thévenin’s theorem above, called the Norton’s theorem. Instead of
having a voltage source and a resistor in series, we can convert any complex circuits with voltage
sources, current sources, and resistors into one — from the point of view of a load — with only
a single current source in parallel with a resistor. This is illustrated below:
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ENGG1203: Introduction to Electrical and Electronic Engineering
Homework 2
A
A
+
network of sources
and resistors
vL
iL
+
in
RL
iL
vL
Rn
−
RL
−
B
B
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RL is the load resistor connected to the terminals A and B. We can maintain the same voltage
vL and current iL in the transformation by the following procedure:
• The Norton equivalent current, iN , is equal to the current across AB if they are shortcircuited. In other words, that’s I from the Thévenin equivalent above!
• The Norton equivalent resistance is the same as the Thévenin equivalent resistance, i.e.,
RN = Rth .
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For the same circuit as in Part(b), find the Norton equivalent current and resistance. What is
the resultant voltage across AB if they are connected by a 2 kΩ resistor?
We have computed in Part (b) that iN = 2.6 mA and RN = 2 kΩ. Thus, we have the
following Norton equivalent circuit:
+
vL
2 kΩ
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2.6 mA
A
iL
RL
−
B
Now if the load is a 2 kΩ resistor, then half of the current flows through it, causing a
voltage drop of
vL = (1.3)(2) = 2.6 V.
Compare your answer with Part (a) and (b)!
Part(d)
Power Dissipation
Find the power dissipation in a resistor RL connected across AB in Part(b) if:
1. RL = 1 kΩ;
2. RL = 2 kΩ;
3. RL = 3 kΩ.
(Bonus) If RL can be any resistance, what should be the value that would cause maximum power
dissipation in that resistor?
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ENGG1203: Introduction to Electrical and Electronic Engineering
Homework 2
The voltage across RL is given by
v=
RL
RL + 2000
(5.2),
and therefore the power dissipated in RL is
P =
RL
(RL + 2000)2
.
≈ 3 mW.
= 3.38 mW.
≈ 3.24 mW.
dissipation, we consider a general case with the Thévenin
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1. RL = 1 kΩ: We have P
2. RL = 2 kΩ: We have P
3. RL = 3 kΩ: We have P
To find the maximum power
equivalent circuit:
v2
= (5.2)2
RL
A
+
Rth
vL
−
+
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vth
iL
RL
−
B
The power dissipated in RL is given by
P (RL ) = (vth )2
RL
.
(RL + Rth )2
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To find the maximum power, we take the derivative and set to zero, i.e.,
1
−2RL
d
2
P (RL ) = (vth )
+
= 0.
dRL
(RL + Rth )2
(RL + Rth )3
This happens when RL + Rth − 2RL = 0, or,
RL = Rth .
In other words, we get maximum power dissipation when RL (often called the load resistance, or more generally, load impedence) matches the Thévenin equivalent resistance.
In this particular question, RL = 2 kΩ.
Question 3
Rotating Motor
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.
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ENGG1203: Introduction to Electrical and Electronic Engineering
Homework 2
0◦
light source
θ
RL
RR
tracker head
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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.
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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=
(1)
0
otherwise,
where km is a motor dependent constant.
When Vm is positive, the motor turns clockwise. When it is negative, the motor turns counterclockwise.
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:
RR
Vmp
motor
+−
Vmn
−
+
SO
Vdd
RL
Vdd /2
Figure 1: 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.
While the design of the circuit in Figure 1 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.
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ENGG1203: Introduction to Electrical and Electronic Engineering
Homework 2
Part(b)
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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.
Second Attempt
You search the laboratory again find some op-amps. You then modify the circuit as shown below:
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Vdd
Vdd
Vp
+
Vmp
−
RL
+ −
Vmn
−
+
RR
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Does this circuit function correctly? That is, does it correctly track the direction of the light
source? Explain your answer.
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.
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:
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ENGG1203: Introduction to Electrical and Electronic Engineering
Homework 2
Vdd
Vdd
Vdd
RR
R
+
Vp
Vmp
−
+ −
Vmn
RL
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R
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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?
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.
SO
Part(d)
Show how you may further revise the design above so it functions as desired.
Vdd
Vdd
Vdd
RR
Vdd
R
−
Vmp
+ −
Vmn
−
+
+
Vp
Vn
RL
Your Circuit
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