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Subject
Electrical Engineering
Date
Dec 6, 2023
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11
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EECS 16A
Designing Information Devices and Systems I
Fall 2023
Homework 11
This homework is due November 17, 2023, at 23:00.
Self-grades are due November 20, 2023, at 23:00.
Submission Format
Your homework submission should consist of
one
file.
•
hw11.pdf
: A single PDF file that contains all of your answers (any handwritten answers should be
scanned).
Submit the file to the appropriate assignment on Gradescope.
1. Reading Assignment
For this homework, please review Notes 18.5-18.6, Notes 19.5-19.6, and Note 20 (Op-Amp Current Source
and Circuit Design).
(a) Summarize the 4-step design process outlined in the notes, and describe why each step is useful in
practice.
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2. Op-Amp in Negative Feedback
(Contributors: Adhyyan Narang, Ava Tan, Aviral Pandey, Deepshika Dhanasekar, Lam Nguyen, Panos
Zarkos, Titan Yuan, Vijay Govindarajan, Urmita Sikder, Sunash Sharma)
In this question, we analyze op-amp circuits in which the op-amp has a finite gain. Consider the circuit
shown in Figure 1. Note that this circuit is in negative feeback. This circuit is known as a non-inverting
amplifier circuit.
In parts (a) - (e), we will assume the op-amp in this circuit is ideal (i.e., gain is infinite), and then we will
consider the case of finite gain in parts (f) - (h). For all parts, assume
V
SS
=
−
V
DD
.
−
+
V
DD
V
SS
−
+
v
s
R
2
I
R
2
R
1
i
L
R
+
−
v
out
+
−
v
x
u
+
u
−
Figure 1: Non-inverting amplifier circuit
(a) Consider the circuit shown in Figure 1. What is
u
+
−
u
−
?
(b) Find
v
x
as a function of
v
out
.
(c) What is
I
R
2
, i.e. the current flowing through
R
2
as a function of
v
s
?
Hint: Find the current through R
1
first.
(d) Find
v
out
as a function of
v
s
.
(e) What is the current
i
L
through the load resistor
R
? Give your answer in terms of
v
out
.
(f) We will now examine what happens when the op-amp gain is not
∞
. To understand what happens in
this case, we will consider the equivalent circuit for an op-amp with gain
A
, shown in Figure 2. We
will use this equivalent circuit to analyze the non-inverting amplifier circuit in Figure 1.
First draw an equivalent circuit for Figure 1,
by replacing the ideal op-amp in the non-inverting
amplifier in Figure 1 with the op-amp model shown in Figure 2.
You do not need to include
V
DD
and
V
SS
in your equivalent circuit.
Then, using this setup, calculate
v
out
and
v
x
in terms of
A
,
v
s
,
R
1
,
R
2
and
R
. Assuming
A
is finite, is the
magnitude of
v
x
larger or smaller than the magnitude of
v
s
? Do these values depend on
R
?
Hint: Note
that the first golden rule still applies, i.e. the currents through the input terminals are zero.
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u
+
+
−
v
in
u
−
u
out
−
+
Av
in
+
−
v
out
V
SS
V
DD
Figure 2: Internal op-amp model
(g) Using your solution to the previous part, calculate the limits of
v
out
and
v
x
as
A
→
∞
. You should get
the same answer as in part (d) for
v
out
.
(h)
[OPTIONAL, CHALLENGE]
Now you want to make a non-inverting amplifier circuit whose gain
is nominally
G
nom
=
v
out
v
s
=
1
+
R
2
R
1
=
4. However,
G
nom
can only be achieved only if the op-amp is
ideal, i.e, if its internal gain
A
→
∞
. But, as with most considerations in the physical world, we must
account for nonidealities! In reality, because you will be working with an op-amp with finite gain
A
,
your designed circuit gain may come close to but will never quite reach
G
nom
as a result of the real
op-amp’s finite internal gain
A
.
Suppose you would like your real op-amp circuit to have a minimum error of 1% (i.e, a minimum
circuit gain of 3
.
96, i.e.
v
out
v
s
≥
3
.
96). Remember that only if your op-amp were ideal, you would have
a nominal circuit gain of
G
nom
=
v
out
v
s
=
1
+
R
2
R
1
=
4.
What is the minimum required value of
A
, called
A
min
, to achieve that specification?
Hint: Use your
expression of v
out
in part (f) to find an expression for G
min
=
v
out
v
s
when A
̸→
∞
.
3. Basic Amplifier Building Blocks
(Contributors: Ava Tan, Aviral Pandey, Lam Nguyen, Michael Kellman, Titan Yuan, Vijay Govindarajan,
Urmita Sikder)
The following amplifier stages are used often in many circuits and are well known as (a) the non-inverting
amplifier and (b) the inverting amplifier.
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+
−
v
s
+
−
v
o
R
2
R
1
R
1
+
−
v
s
+
−
v
o
R
2
(a)
(b)
(a) Label the input terminals of the op-amp with
(+)
and
(
−
)
signs in Figure (a), so that it is in negative
feedback. Then derive the voltage gain (
G
=
v
o
v
s
) of the non-inverting amplifier in Figure (a) using the
Golden Rules. Why do you think this circuit is called a non-inverting amplifier?
(b) Label the input terminals of the op-amp with
(+)
and
(
−
)
signs in Figure (b), so that it is in negative
feedback. Then derive the voltage gain (
G
=
v
o
v
s
) of the inverting amplifier using the Golden Rules.
Can you explain why this circuit is called an inverting amplifier?
4. Putting on the Pressure: Build Your Own InstantPot
(Contributors: Craig Schindler, Ava Tan, Wahid Rahman, Urmita Sikder)
Prof. Ranade had a great experience with her automatic pressure cooker, so she was inspired to try and
build her own. She’s enlisting your help! The design of the pressure cooker uses a pressure sensor and a
heating element. Whenever the pressure is below a set target value, an electronic circuit turns on the heating
element.
Pressure Sensor Resistance
The first step is designing a pressure sensor. The figure below shows your design. As pressure
p
c
is applied,
the flexible membrane stretches.
(a) You attach a resistor layer
R
p
with resistivity
ρ
= 0.1
Ω
m, width
W
, length
L
, and thickness
t
to the
pressure sensor membrane, as illustrated in the figure below. When the pressure
p
c
=
0Pa (i.e. there is
no applied pressure),
W
=
1mm,
L
=
L
0
=
1cm,
t
=
100µm
=
100
×
10
−
6
m.
R
p
0
is the value of
R
p
when there is no applied pressure. Calculate
R
p
0
.
Note that direction of
current flow in the resistor is from A to B as marked in the diagram.
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(b) When pressure is applied, the length of the resistor
L
changes from
L
0
and is a function of applied
pressure
p
c
, and is given by
L
=
L
0
+
β
p
c
,
where
L
0
is the nominal length of the resistor with no pressure applied, and
β
is a constant with units
m
/
Pa. As a result of the length change, the value of resistance
R
p
also changes from its nominal value
R
p
0
(the value of
R
p
with no pressure applied).
Derive an expression for
R
p
as a function of resistivity
ρ
, width
W
, thickness
t
, nominal length
L
0
, constant
β
, and applied pressure
p
c
, when pressure is applied.
Note:
The width and thickness of the resistor will also change with applied pressure. However, we
ignore this to keep the math simple.
(c)
Pressure Sensor Circuit Design
For this sub-part and the following sub-parts, we will use a new model for pressure-sensitive resistance
R
p
. Assume that the resistance
R
p
is a function of applied pressure
p
c
according to the relationship
R
p
=
R
o
×
p
c
p
ref
, where
R
o
=
1k
Ω
, and
p
ref
=
100kPa.
To complete our sensor circuit, we would like to generate a voltage
V
p
that is a function of the pressure
p
c
.
Complete the circuit below so that the output voltage
V
p
depends on the pressure
p
c
as:
V
p
=
−
V
o
×
p
c
p
ref
,
where
V
o
=
1V
.
Restrictions on your pressure sensor circuit design are as follows:
•
You may add at most one ideal voltage source and one additional resistor to the circuit, but
you must calculate their values and mark them in the diagram.
•
Mark the positive and negative inputs of the operational amplifier with “+” and “-” symbols,
respectively, in the boxes provided.
•
Assume op-amp supply voltages
V
DD
and
V
SS
=
−
V
DD
are already provided.
You may assume that the operational amplifier is ideal.
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−
+
R
p
+
−
V
p
(d)
Resistive Heating Element
To heat the pressure cooker, you use a heating element with resistance
R
heat
. Calculate the value of
R
heat
such that the power dissipated is
P
heat
=
1000W with
V
heat
=
100V applied across the heating
element.
(e)
Pressure Regulation
You are finally ready to complete the design of your pressure cooker.
Using all of the circuit elements below, make a circuit that will turn the heater on (i.e. will cause
a current to flow through
R
heat
) when the pressure is less than 500 kPa, and off (i.e. will cause no
current to flow through
R
heat
) when the pressure is greater than 500 kPa.
The elements are:
• A voltage source
V
s
=
10V in series with a resistance of 500
Ω
.
• A voltage source
V
p
=
V
o
×
p
c
p
ref
, with
V
o
= 1V and
p
ref
=
100kPa. (This is a voltage source whose
voltage is a function of pressure
p
c
, unrelated to any previous parts of the question.)
• A comparator that controls switch
S
0
. The switch is normally opened (i.e. an open circuit between
nodes
V
a
and
V
b
), and is closed only when
V
1
>
V
2
(i.e. a short circuit between nodes
V
a
and
V
b
).
• The heater supply (
V
heat
=
100V).
• The heater resistor
R
heat
.
• One additional resistor
R
extra
that can have
any value
.
• You may assume you have access to a ground node.
• Assume comparator supply voltages
V
DD
and
V
SS
=
−
V
DD
are already provided.
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−
+
V
s
=
10V
500
Ω
−
+
V
p
−
+
V
2
V
1
S
0
V
a
V
b
−
+
V
heat
R
heat
R
extra
(i) Since you are looking to
compare
the change in voltage associated with a change in pressure, you
decide to assign the
variable voltage
source
V
p
as one of the inputs to your comparator.
What is the value of the
variable voltage source
V
p
for
p
c
=
500kPa?
(ii) For your comparator inputs, you also need to generate a
reference voltage
which can be used to
compare against the value of the
variable voltage
source which you calculated above.
Combine the voltage source
V
s
=
10V with an associated resistance of 500
Ω
with an an additional
resistor
R
extra
to
generate a reference voltage equal to the voltage
V
p
you calculated above.
What would you choose the value of
R
extra
to be?
(iii) Label the circuit elements in the schematic below with the
circuit elements presented above and
your calculated
R
extra
value
to turn the heater on when the pressure is less than 500 kPa.
5. Island Karaoke Machine
(Contributors: Aviral Pandey, Lam Nguyen, Pangiotis Zarkos, Sashank Krishnamurthy, Titan Yuan, Urmita
Sikder, Vijay Govindarajan)
Learning Goal:
The objective of this problem is design a circuit that calculates the difference between two
signals and amplifies the result.
You’re stuck on a desert island and everyone is bored out of their minds.
Fortunately, you have your
EECS16A lab kit with op-amps, wires, resistors, and your handy breadboard. You decide to build a karaoke
machine. You recover one speaker from the crash remains and use your iPhone as your source. You know
that many songs put instruments on either the “left" or the “right" channel, but the vocals are usually present
on both channels with equal strength.
In our case, the vocals are present on both left and right channels, but the instruments are only present on
the right channel, i.e.
v
left
=
v
vocals
v
right
=
v
vocals
+
v
instrument
,
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where the voltage source
v
vocals
can have values anywhere in the range of
±
120mV and
v
instrument
can have
values anywhere in the range of
±
50mV.
What is the goal of a karaoke machine?
The ultimate goal is to
remove
the vocals from the audio output.
We’re going to do this by first building a circuit that takes the left and right audio outputs of the smartphone
and then calculates its
difference
. Let’s see what happens.
The equivalent circuit model of the iPhone audio jack and speaker is shown in Figure 3. We model the
audio signals and jack
as
v
left
and
v
right
with
equivalent source resistance
of the left/right audio channels
of
R
left
=
R
right
=
3
Ω
. The
speaker
has an equivalent resistance of
R
speaker
=
4
Ω
.
R
left
+
−
v
left
R
right
+
−
v
right
iPhone
R
speaker
Figure 3: Audio jack and speaker of an iPhone.
(a) One of your island survivors suggests the circuit in Figure 4 to do this.
Find the expression for the
voltage across the speaker
R
speaker
as a function of
v
vocals
and
v
instruments
.
Does the voltage across the speaker depend on
v
vocals
? In other words, what do you think the islanders
will hear – vocals, instruments, or both?
+
−
v
left
R
left
R
speaker
R
right
+
−
v
right
Figure 4: Circuit for part (a).
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9
(b) We need to boost the sound level to get the party going. To this end, we want a range of
±
2V across the
speaker. Design a circuit by completing the Figure 5 below that takes in
{
v
left
,
R
left
}
and
{
v
right
,
R
right
}
combos as inputs and outputs an
amplified version of
v
instrument
across
R
speaker
. Consider all op-amps
to be
ideal
for this problem.
+
−
v
left
+
−
v
right
R
left
R
right
R
speaker
Figure 5: Circuit for part (b).
Hint 1:
We need to get an output voltage with the range of
±
2V
. The input voltage v
instrument
can
have values anywhere in the range of
±
50mV
. What gain is needed from the op-amp based amplifier
circuits?
Hint 2:
Use two op-amps in the non-inverting configuration. The non-ideal voltage source
{
v
left
,
R
le ft
}
must be the input to one non-inverting amplifier and the non-ideal voltage source
{
v
right
,
R
right
}
must
be the input to the other non-inverting amplifier.
Hint 3:
Connect the speaker R
speaker
across the outputs of those two op-amps.
(c) The trouble with the approach in part (b) is that two op-amps are required. Let’s say you only have
one op-amp
with you. What would you do? One night in your dreams, you have an inspiration. Why
not combine an inverting and non-inverting amplifier into one, as shown in Figure 6!
−
+
R
1
+
−
v
1
R
3
+
−
v
2
R
2
R
4
+
−
v
o
Figure 6: The new amplifier for part (c).
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If we set
v
2
=
0V, what is the output
v
o
in terms of
v
1
? (This is the inverting path.)
(d) Consider the circuit in Figure 6 again. If we set
v
1
=
0V, what is the output
v
o
in terms of
v
2
? (This is
the non-inverting path.)
(e) Now, determine
v
o
in terms of
v
1
and
v
2
by superposing your results from the last two parts. Choose
values for
R
1
,
R
2
,
R
3
and
R
4
, such that the speaker has
v
o
=
±
2V across it for
v
2
−
v
1
=
±
50mV.
6. Wireless Communication With An LED
(Contributors: Aviral Pandey, Lam Nguyen, Pangiotis Zarkos, Sashank Krishnamurthy, Titan Yuan, Urmita
Sikder)
Learning Goal:
This problem is designed to find the response of a complex circuit by breaking it into smaller
circuit blocks and analyzing each block individually.
In this question, we are going to analyze the system shown in the figure below. It shows a circuit that can be
used as a wireless communication system using visible light (or infrared, very similar to remote controls).
+
−
V
in
R
D
1
Transmitter
Receiver
−
+
−
+
−
+
O
1
O
2
O
3
D
2
I
s
V
DD
R
1
i
1
V
0
V
1
i
2
R
2
V
2
R
3
i
3
V
3
v
out
−
+
V
ref
The element
D
1
in the transmitter is a
light-emitting diode (LED).
An LED is an element that emits light
and whose brightness is controlled by the current flowing through it. You can recall controlling the light
emitted by an LED in various labs during the course. In our circuit, the current through the LED, hence its
brightness, can be controlled by choosing the applied voltage
V
in
and the value of the resistor
R
.
The light from the element
D
1
in the transmitter is received by the element
D
2
in the receiver. In this circuit,
the LED
D
1
is used as a means for
transmitting information with light
, and the element
D
2
is used as
a
light receiver to see if anything was transmitted
.
In the receiver, the element labeled as
D
2
behaves like a
"reverse biased" solar cell
, which means that when
it receives light, it generates a current. We will denote the current generated by the solar cell by
I
S
.
Remark:
Non-idealities, such as background light, affect the performance of a system that does light mea-
surements. In this question, we will assume ideal conditions, that is, there is no source of light around except
for the LED.
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11
In our system, we define two states for the transmitter: (i) the transmitter is sending something when
they
turn on the LED
and (ii) the transmitter is not sending anything when
they turn off the LED
. On the
receiver side, the goal is to
convert the current
I
S
generated by the solar cell into a voltage and amplify it
,
so that we can read the output voltage
V
out
to see if the transmitter was sending something or not. The circuit
implements this operation through a series of op-amps and a comparator.
It might look look complicated at
first glance, but we can analyze it one section at a time.
(a) Currents
i
1
,
i
2
and
i
3
are labeled on the diagram. Assuming
op-amps are ideal
and the Golden Rules
hold, is
I
S
=
i
1
?
i
1
=
i
2
?
i
2
=
i
3
? Treat the solar cell as an ideal current source supplying
I
s
.
(b) Use the Golden Rules to find
V
0
,
V
1
,
V
2
and
V
3
in terms of
I
S
,
R
1
,
R
2
and
R
3
.
Hint: Solve for V
0
first, then use V
0
to find V
1
. Afterwards, find V
2
and V
3
in a similar manner.
(c) Now, assume that the transmitter has chosen the values of
V
in
and
R
to control the intensity of light
emitted by LED, such that when the
transmitter is sending something
,
I
S
is equal to 0
.
1A and when
the
transmitter is not sending anything
,
I
S
is equal to 0A.
The following figure shows a visual
example of how this current
I
S
might look like as time changes.
0
.
1
Time
I
S
[
A
]
For the receiver, suppose
V
ref
=
2V ,
R
1
=
10
Ω
,
R
2
=
1000
Ω
, and the supply voltages of the com-
parator are
V
DD
=
5V and
V
SS
=
−
5V. Pick a value of
R
3
such that
V
out
is
V
DD
when the transmitter
is sending something (i.e.
I
S
=
0
.
1A
) and
V
SS
when the transmitter is not sending anything (i.e.
I
S
=
0
)
.
7. Homework Process and Study Group
Who did you work with on this homework? List names and student ID’s. (In case you met people at home-
work party or in office hours, you can also just describe the group.) How did you work on this homework?
If you worked in your study group, explain what role each student played for the meetings this week.
UCB EECS 16A, Fall 2023, Homework 11, All Rights Reserved. This may not be publicly shared without explicit permission.
11
Related Questions
DEGREE: Electrical Engineering SUBJECT/COURSE: AC CircuitsTOPIC: Converter and Rectifier NOTE: Please solve in this way. 1. Please have a good handwriting, some of the answers are not readable. Thank you!2. GIVEN. (Include symbols and units)3. REQUIRED/FIND/MISSING (with symbol/s)4. ILLUSTRATION (Required).5. Step-by-step SOLUTION with Formulas and Symbols. No Shortcut, No skipping, and detailed as possible6. FINAL ANSWERS must be rounded up to three decimal places
PROBLEM:A three-anode mercury-arc rectifier supplies a 250-kw 230-volt load and is energized from a 4,600-volt three-phase source through a three-phase transformer whose primaries are in delta. Assuming an arc drop of 20-volts, calculate:(a) the d-c load current;(b) the voltage and kilovolt-ampere rating of each transformer secondary;(c) the current and kilovolt-ampere rating of the three-phase transformer.
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Don't use ai to answer I will report your answer Solve it Asap with explanation...
What is a wind rose and how is it used in wind energy analysis?
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question 1
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6) Build an Arduino code that has keypad, LCD display. In this code you are required to: (Void
Setup and Loop only)
1- Show "Hello Ardunio" blinking on LCD display with delay 1sec when pushing key I from the keypad
and the LED is OFF.
2- Make a LED blinking with delay 1sec when pushing key 2 from the keypad and the LCD display is
empty
7)An image of 250x250 elements (pixels) what is the memory size of the image if:
a)Unit 8 b) Unit 16 c) Black & Whi
8) The game port register is used for two different purposes that are:
1-
2-
9) What is the FPGA device?
10) chose the correct answer for the following:
1- What is a result of connecting two or more switches together?
a) The size of the broadcast domain is increased. b) The number of collision domains is reduced.
b) The size of the collision domain is increased. c)The number of broadcast domains is increased.
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08:32 ☑
< Block Test Assign...
96%
Q :
4. Solutions to this assignment should be submitted via Google Classroom by the due date.
5. No late submissions will be accepted.
6. A cover page should accompany your submission.
7. You will be penalized if you fail to adhere to instructions.
MARK ALLOCATION
Question
Question 1
Question 2
Marks
15
10
Question 1
a) Determine Thévenin's equivalent circuit for the circuit shown in Figure 1. given that, is the
load of interest.
1000
480 F
[15]
Question 2
10/45° V
1 kHz
33002
ли
22002
Figure 1
ZL
75 mH
Given the circuit of an ideal OPAMP, as shown in Figure 2, determine the output voltage phasor
V
[10]
V 2.50/30.0° V
f=1.00 kHz
1.00 ΚΩ
Figure 2
470 nF
OV
2/3
3/3
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Do you know what to do if the electricity in your house suddenly surges and you don't want your computer to die? Please elaborate in the box below if this is the case.
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USING: Mobile Phone (android) / Laptop (bluestacks)*Note: MacroPLC can also be accessed by laptop, mobile phone (IOS/Android) thru thiswebsite link. (https://www.macroplc.com/simulador/) No installation is required.
Create a PLC program given the following conditions:
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The project conditions are:The project functionality will be provided as per the project description suppliedbelow.You must present/demonstrate the project as a group, at the given time,andwithin a given time limit ( 20 mins ).A soft copy of all project documents is required before you
The program functionality should be as follows:• Upon entering the “run” mode all counters/timers must be reset (use the first scan bit).• When students scan RFID cards, the opening motor automatically opens the door and closing motor closes it after a 5 second wait.• The current number of students should be counted by incrementing/decrementing as students enter or leave the classroom. • The number of students entering the classroom is counted using a reflective sensor, while the number leaving is counted using a fiber optic sensor. The count should be displayed in the decimal tag "Total no of students”.• The maximum number of students in the classroom is 20, and when the classroom is occupied by the…
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Related Questions
- DEGREE: Electrical Engineering SUBJECT/COURSE: AC CircuitsTOPIC: Converter and Rectifier NOTE: Please solve in this way. 1. Please have a good handwriting, some of the answers are not readable. Thank you!2. GIVEN. (Include symbols and units)3. REQUIRED/FIND/MISSING (with symbol/s)4. ILLUSTRATION (Required).5. Step-by-step SOLUTION with Formulas and Symbols. No Shortcut, No skipping, and detailed as possible6. FINAL ANSWERS must be rounded up to three decimal places PROBLEM:A three-anode mercury-arc rectifier supplies a 250-kw 230-volt load and is energized from a 4,600-volt three-phase source through a three-phase transformer whose primaries are in delta. Assuming an arc drop of 20-volts, calculate:(a) the d-c load current;(b) the voltage and kilovolt-ampere rating of each transformer secondary;(c) the current and kilovolt-ampere rating of the three-phase transformer.arrow_forwardDon't use ai to answer I will report your answer Solve it Asap with explanation... What is a wind rose and how is it used in wind energy analysis?arrow_forwardquestion 1arrow_forward
- 6) Build an Arduino code that has keypad, LCD display. In this code you are required to: (Void Setup and Loop only) 1- Show "Hello Ardunio" blinking on LCD display with delay 1sec when pushing key I from the keypad and the LED is OFF. 2- Make a LED blinking with delay 1sec when pushing key 2 from the keypad and the LCD display is empty 7)An image of 250x250 elements (pixels) what is the memory size of the image if: a)Unit 8 b) Unit 16 c) Black & Whi 8) The game port register is used for two different purposes that are: 1- 2- 9) What is the FPGA device? 10) chose the correct answer for the following: 1- What is a result of connecting two or more switches together? a) The size of the broadcast domain is increased. b) The number of collision domains is reduced. b) The size of the collision domain is increased. c)The number of broadcast domains is increased.arrow_forward08:32 ☑ < Block Test Assign... 96% Q : 4. Solutions to this assignment should be submitted via Google Classroom by the due date. 5. No late submissions will be accepted. 6. A cover page should accompany your submission. 7. You will be penalized if you fail to adhere to instructions. MARK ALLOCATION Question Question 1 Question 2 Marks 15 10 Question 1 a) Determine Thévenin's equivalent circuit for the circuit shown in Figure 1. given that, is the load of interest. 1000 480 F [15] Question 2 10/45° V 1 kHz 33002 ли 22002 Figure 1 ZL 75 mH Given the circuit of an ideal OPAMP, as shown in Figure 2, determine the output voltage phasor V [10] V 2.50/30.0° V f=1.00 kHz 1.00 ΚΩ Figure 2 470 nF OV 2/3 3/3arrow_forwardDo you know what to do if the electricity in your house suddenly surges and you don't want your computer to die? Please elaborate in the box below if this is the case.arrow_forward
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Recommended textbooks for you
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