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Question 1  (1 point) |
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Question 5-1a
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Which of the five following circuits is a non-inverting amplifier?
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Student response: |
Percent
Value |
Correct
Response |
Student
Response |
Answer Choices |
|---|
| 0.0% |
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a. |
 |
| 100.0% |
 |
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b. |
 |
| 0.0% |
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c. |
 |
| 0.0% |
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d. |
 |
| 0.0% |
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e. |
 |
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General feedback: |
See experiment 4 or class notes for these circuits. |
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Score: | 1 / 1
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| Question 2 (1 point) |
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Question 5-1b
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Which of the five following circuits is an inverting amplifier?
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Student response: |
Percent
Value |
Correct
Response |
Student
Response |
Answer Choices |
|---|
| 100.0% |
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a. |
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| 0.0% |
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b. |
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| 0.0% |
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c. |
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| 0.0% |
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d. |
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| 0.0% |
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e. |
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General feedback: |
See experiment 4 or class notes for these circuits. |
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Score: | 1 / 1
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| Question 3 (1 point) |
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Question 5-1c
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Which of the five following circuits is a differential amplifier?
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Student response: |
Percent
Value |
Correct
Response |
Student
Response |
Answer Choices |
|---|
| 0.0% |
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a. |
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| 0.0% |
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b. |
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| 100.0% |
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c. |
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| 0.0% |
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d. |
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| 0.0% |
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e. |
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General feedback: |
See experiment 4 or class notes for these circuits. |
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Score: | 1 / 1
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| Question 4 (1 point) |
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Question 5-1d
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Which of the five following circuits is a buffer?
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Student response: |
Percent
Value |
Correct
Response |
Student
Response |
Answer Choices |
|---|
| 0.0% |
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a. |
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| 0.0% |
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b. |
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| 0.0% |
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c. |
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| 0.0% |
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d. |
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| 100.0% |
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e. |
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General feedback: |
See experiment 4 or class notes for these circuits. |
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Score: | 1 / 1
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| Question 5 (1 point) |
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Question 5-1e
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Which of the five following circuits is an adder?
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Student response: |
Percent
Value |
Correct
Response |
Student
Response |
Answer Choices |
|---|
| 0.0% |
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a. |
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| 0.0% |
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b. |
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| 0.0% |
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c. |
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| 100.0% |
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d. |
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| 0.0% |
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e. |
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General feedback: |
See experiment 4 or class notes for these circuits. |
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Score: | 1 / 1
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| Question 6 (1 point) |
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Question 5-2a
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What is the transfer function (Vout/Vin) for the following circuit?
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Student response: |
Percent
Value |
Correct
Response |
Student
Response |
Answer Choices |
|---|
| 100.0% |
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a. |
Vout/Vin = 1 |
| 0.0% |
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b. |
Vout/Vin = -Rf/Ri |
| 0.0% |
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c. |
Vout = -Rf(V1/R1 + V2/R2) |
| 0.0% |
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d. |
Vout/Vin = -1/(jwRiC) |
| 0.0% |
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e. |
Vout/Vin = -jwRfC |
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General feedback: |
These five circuits are in the background for experiment 8. You should be able to recognize and derive the transfer functions for all five. |
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Score: | 1 / 1
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| Question 7 (1 point) |
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Question 5-2b
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What is the transfer function (Vout/Vin) for the following circuit?
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Student response: |
Percent
Value |
Correct
Response |
Student
Response |
Answer Choices |
|---|
| 0.0% |
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a. |
Vout/Vin = 1 |
| 100.0% |
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b. |
Vout/Vin = -Rf/Ri |
| 0.0% |
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c. |
Vout = -Rf(V1/R1 + V2/R2) |
| 0.0% |
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d. |
Vout/Vin = -1/(jwRiC) |
| 0.0% |
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e. |
Vout/Vin = -jwRfC |
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General feedback: |
These five circuits are in the background for experiment 8. You should be able to recognize and derive the transfer functions for all five. |
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Score: | 1 / 1
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| Question 8 (1 point) |
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Question 5-2c
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What is the transfer function (Vout/Vin) for the following circuit?
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Student response: |
Percent
Value |
Correct
Response |
Student
Response |
Answer Choices |
|---|
| 0.0% |
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a. |
Vout/Vin = 1 |
| 0.0% |
|
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b. |
Vout/Vin = -Rf/Ri |
| 100.0% |
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c. |
Vout = -Rf(V1/R1 + V2/R2) |
| 0.0% |
|
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d. |
Vout/Vin = -1/(jwRiC) |
| 0.0% |
|
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e. |
Vout/Vin = -jwRfC |
|
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General feedback: |
These five circuits are in the background for experiment 8. You should be able to recognize and derive the transfer functions for all five. |
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Score: | 1 / 1
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| Question 9 (1 point) |
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Question 5-2d
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What is the transfer function (Vout/Vin) for the following circuit?
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Student response: |
Percent
Value |
Correct
Response |
Student
Response |
Answer Choices |
|---|
| 0.0% |
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a. |
Vout/Vin = 1 |
| 0.0% |
|
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b. |
Vout/Vin = -Rf/Ri |
| 0.0% |
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c. |
Vout = -Rf(V1/R1 + V2/R2) |
| 100.0% |
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d. |
Vout/Vin = -1/(jwRiC) |
| 0.0% |
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e. |
Vout/Vin = -jwRfC |
|
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General feedback: |
These five circuits are in the background for experiment 8. You should be able to recognize and derive the transfer functions for all five. |
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Score: | 1 / 1
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| Question 10 (1 point) |
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Question 5-2e
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What is the transfer function (Vout/Vin) for the following circuit?
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Student response: |
Percent
Value |
Correct
Response |
Student
Response |
Answer Choices |
|---|
| 0.0% |
|
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a. |
Vout/Vin = 1 |
| 0.0% |
|
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b. |
Vout/Vin = -Rf/Ri |
| 0.0% |
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c. |
Vout = -Rf(V1/R1 + V2/R2) |
| 0.0% |
|
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d. |
Vout/Vin = -1/(jwRiC) |
| 100.0% |
|
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e. |
Vout/Vin = -jwRfC |
|
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General feedback: |
These five circuits are in the background for experiment 8. You should be able to recognize and derive the transfer functions for all five. |
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Score: | 1 / 1
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| Question 11 (1 point) |
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Question 5-3
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Which of the following is NOT true about an Op-Amp?
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Student response: |
Percent
Value |
Correct
Response |
Student
Response |
Answer Choices |
|---|
| 100.0% |
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a. |
Positive feedback stabilizes the output. |
| 0.0% |
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b. |
In a stable op-amp, the output attempts to do whatever is necessary to make the voltage difference between the two inputs zero. |
| 0.0% |
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c. |
Theoretically, the inputs draw no current. |
| 0.0% |
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d. |
The slew rate of an op-amp is related to the amount of time the output takes to adjust when there is a change in the input. |
| 0.0% |
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e. |
The output voltage can never be larger than the +Vcc or smaller than -Vcc. |
| 0.0% |
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f. |
It can be used to amplify a signal. |
| 0.0% |
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g. |
Its intrinsic gain is very, very large. |
| 0.0% |
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h. |
In all op-amps the theoretical design rules ("golden rules") will be obeyed only if the op amp is in the active region. |
| 0.0% |
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i. |
It can be used to perform mathematical operations, such as addition, subtraction, integration, and differentiation. |
| 0.0% |
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j. |
An ideal op amp looks like an ideal voltage source. A real op amp has a real voltage source as an output. |
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General feedback: |
See notes for experiment 4 and op amp description in your book (or on the internet). Know the difference between the gain of a configuration and the intrinsic op-amp gain. Know what saturation is. Know the rules for analyzing op amps. |
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Score: | 1 / 1
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| Question 12 (1 point) |
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Question 5-4a
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When an electronics device inverts an AC signal it...
(Note: http://www.twysted-pair.com/dicta.htm may be helpful)
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Student response: |
Percent
Value |
Correct
Response |
Student
Response |
Answer Choices |
|---|
| 0.0% |
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a. |
shifts the phase by +90 degrees (turns a sine into a cosine) |
| 0.0% |
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b. |
shifts the phase by -90 degrees (turns a sine into a -cosine) |
| 0.0% |
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c. |
makes the signal zero (ground) |
| 100.0% |
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d. |
makes all the positive values negative and all the negative values positive. (This is also equivalent to shifting the phase by 180 degrees). |
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General feedback: |
If you want to view a site and the question at the same time, bring up another instance of your web browser and cut and paste the site name. The index for the dictionary http://www.twysted-pair.com/dicta.htm is at the bottom of the page. |
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Score: | 1 / 1
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| Question 13 (1 point) |
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Question 5-4b
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When electrical engineers talk about "gain" they are usually referring to...
(Note: http://www.twysted-pair.com/dicta.htm may be helpful)
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Student response: |
Percent
Value |
Correct
Response |
Student
Response |
Answer Choices |
|---|
| 0.0% |
|
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a. |
making the voltage higher |
| 0.0% |
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b. |
making the power higher |
| 0.0% |
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c. |
making the frequency higher |
| 0.0% |
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d. |
making the current higher |
| 0.0% |
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e. |
a, b and c above |
| 0.0% |
|
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f. |
a, c and d above |
| 100.0% |
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g. |
a, b and d above |
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General feedback: |
If you want to view a site and the question at the same time, bring up another instance of your web browser and cut and paste the site name. The index for the dictionary http://www.twysted-pair.com/dicta.htm is at the bottom of the page. |
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Score: | 1 / 1
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| Question 14 (1 point) |
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Question 5-5a
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What is the gain for the amplifier pictured below if R1=2K ohms and R2=15K ohms?
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Student response: |
-7.5 |
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Correct answer: |
-7.50000 |
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General feedback: |
Use the gain formula for this type of amplifier in the experiment 4 handout. Remember the gain for an inverting amplifier is always negative and for a non-inverting amplifier, it is always positive. |
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Score: |
1 / 1
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| Question 15 (1 point) |
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Question 5-5b
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What is the gain for the amplifier pictured below if R1=11K ohms and R2=3K ohms?
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Student response: |
4.67 |
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Correct answer: |
4.66667 |
|
General feedback: |
Use the gain formula for this type of amplifier in the experiment 4 handout. Remember the gain for an inverting amplifier is always negative and for a non-inverting amplifier, it is always positive. |
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Score: |
1 / 1
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| Question 16 (1 point) |
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Question 5-6a
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In the circuit below, find the amplitude of the output if the amplitude of V1 is 100 mV, R1 = 5 kOhm and R2 = 3 kOhm. Specify your answer in volts.
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Student response: |
0.167 |
|
Correct answer: |
0.16667 |
|
General feedback: |
Use the gain equation for this type of amplifier. Notice that the amplitude is never negative. Your answer must be in VOLTS. |
|
Score: |
1 / 1
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| Question 17 (1 point) |
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Question 5-6b
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In the circuit below, find the amplitude of the output if the amplitude of Vin is 400 mV, R1 = 5 kOhm and R2 = 3 kOhm. Specify your answer in volts.
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Student response: |
1.067 |
|
Correct answer: |
1.06667 |
|
General feedback: |
Use the gain equation for this type of amplifier. Notice that the amplitude is never negative. Your answer must be in VOLTS. |
|
Score: |
1 / 1
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| Question 18 (1 point) |
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Question 5-7a
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In the circuit given calculate the output voltage if V1 = 3 volts, V2 = 2 volts, R1 = R4 = 4 KOhms and R2 = R3 = 1 KOhms. Express your answer in volts.
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Student response: |
-4 |
|
Correct answer: |
-4.00000 |
|
General feedback: |
This is a difference amplifier. Remember to subtract the input into the negative and add the input into the positive. The result can be either negative or positive. |
|
Score: |
1 / 1
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| Question 19 (1 point) |
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Question 5-7b
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In the following circuit, what is the value of the output voltage, Vout, if V1 = 300 mV, V2 = 200 mV, Rf = 17K ohms, R1 = 2K ohms and R2 = 3K ohms? Assume the inputs are DC and express your answer in VOLTS.
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Student response: |
-3.68 |
|
Correct answer: |
-3.68333 |
|
General feedback: |
This is a weighted adder. See the experiment. Note that the inputs are given in millivolts, but the output should be in volts. Also note that the sign of the output is important. |
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Score: |
1 / 1
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| Question 20 (1 point) |
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Question 5-8
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In the following circuit, what is the voltage over the load if the input voltage is 5 volts and the load resistor is 192 Kohms. Express your answer in volts.
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Student response: |
5 |
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Correct answer: |
5 |
|
General feedback: |
A buffer works to match a poor voltage source to a load. See your book p.78 |
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Score: |
1 / 1
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| Question 21 (1 point) |
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Question 5-9a
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Which is NOT a characteristic of an op-amp differentiator circuit?
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Student response: |
Percent
Value |
Correct
Response |
Student
Response |
Answer Choices |
|---|
| 0.0% |
|
|
a. |
Ideally, the output is out of phase with the input by -90 degrees. |
| 0.0% |
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b. |
Ideally, it has a transfer function of H(jw) = -jwRC |
| 0.0% |
|
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c. |
Ideally, the output is the negated differentiation of the input. |
| 0.0% |
|
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d. |
In reality, it works best at frequencies below the frequency, 1/RC. |
| 0.0% |
|
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e. |
It can be used in an analog computer. |
| 100.0% |
|
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f. |
Ideally, if the input is a sine wave, the output is a positive cosine wave. |
| 0.0% |
|
|
g. |
Ideally, if the input is a triangle wave, the output is a square wave. |
| 0.0% |
|
|
h. |
The non-ideal version (with some resistance in series with the input capacitor) works like a high pass filter.
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|
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General feedback: |
You should be familiar with the characteristics of differentiators and integrators. |
|
Score: | 1 / 1
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| Question 22 (1 point) |
|
Question 5-9b
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Which is NOT a characteristic of an op-amp integrator circuit?
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Student response: |
Percent
Value |
Correct
Response |
Student
Response |
Answer Choices |
|---|
| 100.0% |
|
|
a. |
Ideally, the output is out of phase with the input by -90 degrees. |
| 0.0% |
|
|
b. |
Ideally, it has a transfer function of H(jw) = -1/(jwRC) |
| 0.0% |
|
|
c. |
Ideally, the output is the negated integration of the input. |
| 0.0% |
|
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d. |
In reality, it works best at frequencies above the frequency, 1/RC. |
| 0.0% |
|
|
e. |
It can be used in an analog computer. |
| 0.0% |
|
|
f. |
Ideally, if the input is a sine wave, the output is a positive cosine wave. |
| 0.0% |
|
|
g. |
Ideally, if the input is a square wave, the output is a triangle wave. |
| 0.0% |
|
|
h. |
The non-ideal version (with some resistance in parallel with the feedback capacitor) works like a low pass filter. |
| 0.0% |
|
|
i. |
Although the ideal version is studied in theory, it does not function well on any realisitic signal because it integrates the DC offset.
|
|
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|
|
General feedback: |
You should be familiar with the characteristics of differentiators and integrators. |
|
Score: | 1 / 1
|
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| Question 23 (1 point) |
|
Question 5-9c
|
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What is NOT a characteristic of a voltage follower?
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Student response: |
Percent
Value |
Correct
Response |
Student
Response |
Answer Choices |
|---|
| 0.0% |
|
|
a. |
It is a way to configure an op-amp. |
| 0.0% |
|
|
b. |
It is also called a voltage buffer. |
| 0.0% |
|
|
c. |
It has a theoretical feedback resistance of 0. |
| 0.0% |
|
|
d. |
It has a theoretical input resistance of infinity. |
| 0.0% |
|
|
e. |
It has a theoretical gain of 1. |
| 0.0% |
|
|
f. |
The output and input voltages of the voltage follower are the same. |
| 0.0% |
|
|
g. |
The output is in phase with the input. |
| 100.0% |
|
|
h. |
It pretty much gives out the same voltage that you put in, so although it is interesting to study in theory, it is of no practical use. |
| 0.0% |
|
|
i. |
It can solve the problem of poor voltage transfer by effectively transforming a poor voltage source (with a high resistance) to an ideal voltage source. |
|
|
|
|
General feedback: |
You should look in your book for information about op-amp voltage followers, integrators and differentiators. |
|
Score: | 1 / 1
|
|
| Question 24 (1 point) |
|
Question 5-10a
|
Given the following input,
which the following signals can be an output of a buffer?
|
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Student response: |
Percent
Value |
Correct
Response |
Student
Response |
Answer Choices |
|---|
| 100.0% |
|
|
a. |
|
| 0.0% |
|
|
b. |
 |
| 0.0% |
|
|
c. |
 |
| 0.0% |
|
|
d. |
 |
|
|
|
|
General feedback: |
You should be aware of the way buffers, integrators, differentiators and inverting amplifiers change the phase of the input. Remember that differentiators and integrators also negate the signal. |
|
Score: | 1 / 1
|
|
| Question 25 (1 point) |
|
Question 5-10b
|
Given the following input,
which the following signals can be an output of an inverting amplifier?
|
|
Student response: |
Percent
Value |
Correct
Response |
Student
Response |
Answer Choices |
|---|
| 0.0% |
|
|
a. |
|
| 100.0% |
|
|
b. |
|
| 0.0% |
|
|
c. |
|
| 0.0% |
|
|
d. |
|
|
|
|
|
General feedback: |
You should be aware of the way buffers, integrators, differentiators and inverting amplifiers change the phase of the input. Remember that differentiators and integrators also negate the signal. |
|
Score: | 1 / 1
|
|
| Question 26 (1 point) |
|
Question 5-10c
|
Given the following input,
which the following signals can be an output of an Op-Amp integrator?
|
|
Student response: |
Percent
Value |
Correct
Response |
Student
Response |
Answer Choices |
|---|
| 0.0% |
|
|
a. |
|
| 0.0% |
|
|
b. |
|
| 100.0% |
|
|
c. |
|
| 0.0% |
|
|
d. |
|
|
|
|
|
General feedback: |
You should be aware of the way buffers, integrators, differentiators and inverting amplifiers change the phase of the input. Remember that differentiators and integrators also negate the signal. |
|
Score: | 1 / 1
|
|
| Question 27 (1 point) |
|
Question 5-10d
|
Given the following input,
which the following signals can be an output of an Op-Amp differentiator?
|
|
Student response: |
Percent
Value |
Correct
Response |
Student
Response |
Answer Choices |
|---|
| 0.0% |
|
|
a. |
|
| 0.0% |
|
|
b. |
|
| 0.0% |
|
|
c. |
|
| 100.0% |
|
|
d. |
|
|
|
|
|
General feedback: |
You should be aware of the way buffers, integrators, differentiators and inverting amplifiers change the phase of the input. Remember that differentiators and integrators also negate the signal. |
|
Score: | 1 / 1
|
|
| Question 28 (1 point) |
|
Question 5-11a
|
|
The following two circuits are correct for the analysis of which op-amp configuration?
|
|
Student response: |
Percent
Value |
Correct
Response |
Student
Response |
Answer Choices |
|---|
| 100.0% |
|
|
a. |
inverting amplifier |
| 0.0% |
|
|
b. |
non-inverting amplifier |
| 0.0% |
|
|
c. |
differential amplifier |
| 0.0% |
|
|
d. |
voltage follower |
| 0.0% |
|
|
e. |
integrator |
| 0.0% |
|
|
f. |
differentiator |
|
|
|
|
General feedback: |
To analyze an op-amp circuit you must take the op-amp out of the circuit. This is possible because one of the op amp rules assumes that no current is going into the op amp. You can create two circuits, one for the + input and another for the - input. Then, you can use the other rule (V+ and V- should be equal) to figure out what the op amp configuration is doing. See class notes for experiment 5. |
|
Score: | 1 / 1
|
|
| Question 29 (1 point) |
|
Question 5-11b
|
|
The following two circuits are correct for the analysis of which op-amp configuration?
|
|
Student response: |
Percent
Value |
Correct
Response |
Student
Response |
Answer Choices |
|---|
| 0.0% |
|
|
a. |
inverting amplifier |
| 100.0% |
|
|
b. |
non-inverting amplifier |
| 0.0% |
|
|
c. |
differential amplifier |
| 0.0% |
|
|
d. |
voltage follower |
| 0.0% |
|
|
e. |
integrator |
| 0.0% |
|
|
f. |
differentiator |
|
|
|
|
General feedback: |
To analyze an op-amp circuit you must take the op-amp out of the circuit. This is possible because one of the op amp rules assumes that no current is going into the op amp. You can create two circuits, one for the + input and another for the - input. Then, you can use the other rule (V+ and V- should be equal) to figure out what the op amp configuration is doing. See class notes for experiment 5. |
|
Score: | 1 / 1
|
|
| Question 30 (1 point) |
|
Question 5-11c
|
|
The following two circuits are correct for the analysis of which op-amp configuration?
|
|
Student response: |
Percent
Value |
Correct
Response |
Student
Response |
Answer Choices |
|---|
| 0.0% |
|
|
a. |
inverting amplifier |
| 0.0% |
|
|
b. |
non-inverting amplifier |
| 100.0% |
|
|
c. |
differential amplifier |
| 0.0% |
|
|
d. |
voltage follower |
| 0.0% |
|
|
e. |
integrator |
| 0.0% |
|
|
f. |
differentiator |
|
|
|
|
General feedback: |
To analyze an op-amp circuit you must take the op-amp out of the circuit. This is possible because one of the op amp rules assumes that no current is going into the op amp. You can create two circuits, one for the + input and another for the - input. Then, you can use the other rule (V+ and V- should be equal) to figure out what the op amp configuration is doing. See class notes for experiment 5. |
|
Score: | 1 / 1
|
|
| Question 31 (1 point) |
|
Question 5-11d
|
|
The following two circuits are correct for the analysis of which op-amp configuration?
|
|
Student response: |
Percent
Value |
Correct
Response |
Student
Response |
Answer Choices |
|---|
| 0.0% |
|
|
a. |
inverting amplifier |
| 0.0% |
|
|
b. |
non-inverting amplifier |
| 0.0% |
|
|
c. |
differential amplifier |
| 100.0% |
|
|
d. |
voltage follower |
| 0.0% |
|
|
e. |
integrator |
| 0.0% |
|
|
f. |
differentiator |
|
|
|
|
General feedback: |
To analyze an op-amp circuit you must take the op-amp out of the circuit. This is possible because one of the op amp rules assumes that no current is going into the op amp. You can create two circuits, one for the + input and another for the - input. Then, you can use the other rule (V+ and V- should be equal) to figure out what the op amp configuration is doing. See class notes for experiment 5. |
|
Score: | 1 / 1
|
|
| Question 32 (1 point) |
|
Question 5-11e
|
|
The following two circuits are correct for the analysis of which op-amp configuration?
|
|
Student response: |
Percent
Value |
Correct
Response |
Student
Response |
Answer Choices |
|---|
| 0.0% |
|
|
a. |
inverting amplifier |
| 0.0% |
|
|
b. |
non-inverting amplifier |
| 0.0% |
|
|
c. |
differential amplifier |
| 0.0% |
|
|
d. |
voltage follower |
| 0.0% |
|
|
e. |
integrator |
| 100.0% |
|
|
f. |
differentiator |
|
|
|
|
General feedback: |
To analyze an op-amp circuit you must take the op-amp out of the circuit. This is possible because one of the op amp rules assumes that no current is going into the op amp. You can create two circuits, one for the + input and another for the - input. Then, you can use the other rule (V+ and V- should be equal) to figure out what the op amp configuration is doing. See class notes for experiment 5. |
|
Score: | 1 / 1
|
|
| Question 33 (1 point) |
|
Question 5-11f
|
|
The following two circuits are correct for the analysis of which op-amp configuration?
|
|
Student response: |
Percent
Value |
Correct
Response |
Student
Response |
Answer Choices |
|---|
| 0.0% |
|
|
a. |
inverting amplifier |
| 0.0% |
|
|
b. |
non-inverting amplifier |
| 0.0% |
|
|
c. |
differential amplifier |
| 0.0% |
|
|
d. |
voltage follower |
| 100.0% |
|
|
e. |
integrator |
| 0.0% |
|
|
f. |
differentiator |
|
|
|
|
General feedback: |
To analyze an op-amp circuit you must take the op-amp out of the circuit. This is possible because one of the op amp rules assumes that no current is going into the op amp. You can create two circuits, one for the + input and another for the - input. Then, you can use the other rule (V+ and V- should be equal) to figure out what the op amp configuration is doing. See class notes for experiment 5. |
|
Score: | 1 / 1
|
|
| Question 34 (1 point) |
|
Question 5-12a
|
|
How do you derive the transfer function, H(jw) = Vout/Vin for an op-amp configuration like the one below?
|
|
Student response: |
Percent
Value |
Correct
Response |
Student
Response |
Answer Choices |
|---|
| 100.0% |
|
|
a. |
1] V+ = 0
2] (Vin - V- ) / Zi = (V- - Vout) / Zf
3] V+ = V- = 0
4] Substitute and solve |
| 0.0% |
|
|
b. |
1] V+ = 0
2] Vin / Zi= Vout / Zf
3] Substitute and solve |
| 0.0% |
|
|
c. |
1] V+ = 0
2] (Vin - V-) / Zf = (V- - Vout) / Zi
3] V+ = V- = 0
4] Substitute and solve |
| 0.0% |
|
|
d. |
1] V+ = 0
2] Vin / Zf = Vout / Zi
3] Substitute and solve |
|
|
|
|
General feedback: |
Keep in mind that the current going into the op amp inputs is zero and the voltage at V+
and V- are equal. See your notes for experiment 5. |
|
Score: | 1 / 1
|
|
| Question 35 (1 point) |
|
Question 5-12b
|
|
How do you derive the transfer function, H(jw) = Vout/Vin for an op-amp configuration like the one below?
|
|
Student response: |
Percent
Value |
Correct
Response |
Student
Response |
Answer Choices |
|---|
| 100.0% |
|
|
a. |
1] V+ = [Z3 / (Z2 + Z3)] * V2
2] (V1 - V-) / Z1 = (V- - Vout) / Zf
3] V+ = V-
4] Substitute and solve |
| 0.0% |
|
|
b. |
1] V+ = [Z2 / (Z2 + Z3)] * V2
2] (V1 - V-) / Z1 = (V- - Vout) / Zf
3] V+ = V-
4] Substitute and solve |
| 0.0% |
|
|
c. |
1] V+ = [Z2 / (Z2 + Z3)] * V2
2] V1/Z1 - Vout/Zf = V-
3] V+ = V-
4] Substitute and solve |
| 0.0% |
|
|
d. |
1] V+ = [Z3 / (Z2 + Z3)] * V2
2] V1/Z1 - Vout/Zf = V-
3] V+ = V-
4] Substitute and solve |
|
|
|
|
General feedback: |
Keep in mind that the current going into the op amp inputs is zero and the voltage at V+
and V- are equal. See your notes for experiment 5. |
|
Score: | 1 / 1
|
|
| Question 36 (1 point) |
|
Question 5-13a
|
Giving the following op amp configuration and the input shown, which picture below best represents the output? 
|
|
Student response: |
Percent
Value |
Correct
Response |
Student
Response |
Answer Choices |
|---|
| 0.0% |
|
|
a. |
 |
| 0.0% |
|
|
b. |
 |
| 100.0% |
|
|
c. |
 |
| 0.0% |
|
|
d. |
 |
|
|
|
|
General feedback: |
Be able to recognize a real differentiator and integrator and know what these circuits do to square and triangular waves. |
|
Score: | 1 / 1
|
|
| Question 37 (1 point) |
|
Question 5-13b
|
Giving the following op amp configuration and the input shown, which picture below best represents the output? 
|
|
Student response: |
Percent
Value |
Correct
Response |
Student
Response |
Answer Choices |
|---|
| 100.0% |
|
|
a. |
|
| 0.0% |
|
|
b. |
|
| 0.0% |
|
|
c. |
|
| 0.0% |
|
|
d. |
|
|
|
|
|
General feedback: |
Be able to recognize a real differentiator and integrator and know what these circuits do to square and triangular waves. |
|
Score: | 1 / 1
|
|
| Question 38 (1 point) |
|
Question 5-13c
|
|
Giving the following op amp configuration and the input shown, which picture below best represents the output?
|
|
Student response: |
Percent
Value |
Correct
Response |
Student
Response |
Answer Choices |
|---|
| 0.0% |
|
|
a. |
|
| 100.0% |
|
|
b. |
|
| 0.0% |
|
|
c. |
|
| 0.0% |
|
|
d. |
|
|
|
|
|
General feedback: |
Be able to recognize a real differentiator and integrator and know what these circuits do to square and triangular waves. |
|
Score: | 1 / 1
|
|
| Question 39 (1 point) |
|
Question 5-14a
|
|
For the circuit below, which of the following equations expresses the relation between the output and the inputs? [Hint: Review the analysis of the op-amp adder circuit.]
|
|
Student response: |
Percent
Value |
Correct
Response |
Student
Response |
Answer Choices |
|---|
| 0.0% |
|
|
a. |
Vout = - jwR1C1 V1 - jwR2C1 V2 |
| 100.0% |
|
|
b. |
Vout = - (1/jwR1C1) V1 - (1/jwR2C1) V2 |
| 0.0% |
|
|
c. |
Vout = - (R1/R2) V2 - jwR1C1 V1 |
| 0.0% |
|
|
d. |
Vout = - (R2/R1) V2 - (1/jwR1C1) V1 |
|
|
|
|
General feedback: |
Use the general form of the adder circuit except replace the resistances with impedances where appropriate. |
|
Score: | 1 / 1
|
|
| Question 40 (1 point) |
|
Question 5-14b
|
|
For the circuit below, which of the following equations expresses the relation between the output and the inputs? [Hint: Review the analysis of the op-amp adder circuit.]
|
|
Student response: |
Percent
Value |
Correct
Response |
Student
Response |
Answer Choices |
|---|
| 0.0% |
|
|
a. |
Vout = - jwR1C1 V1 - jwR2C1 V2 |
| 0.0% |
|
|
b. |
Vout = - (1/jwR1C1) V1 - (1/jwR2C1) V2 |
| 100.0% |
|
|
c. |
Vout = - (R1/R2) V2 - jwR1C1 V1 |
| 0.0% |
|
|
d. |
Vout = - (R2/R1) V2 - (1/jwR1C1) V1 |
|
|
|
|
General feedback: |
Use the general form of the adder circuit except replace the resistances with impedances where appropriate. |
|
Score: | 1 / 1
|
|
| Question 41 (1 point) |
|
Question 5-15
|
|
Many of you are already familiar with binary numbers. They are important because the binary number system is used by digital computers. In experiment 6, we will begin our coverage of binary logic devices. These are based on signals that have only two states: ON and OFF. In a binary number system OFF=0 and ON=1. If you are not familiar with binary numbers or if you need a reminder, the following web site will be helpful: http://electronics.howstuffworks.com/bytes2.htm]
What is the value in decimal of the following binary number: 1000?
|
|
Student response: |
8 |
|
Correct answer: |
8 |
|
General feedback: |
To create the decimal equivalent of a binary number, b3 b2 b1 b0, you need to multiply each bit by a power of two (bn is multiplied by 2n) and add them together. |
|
Score: |
1 / 1
|
|
|
|
Total score: |
41 / 41 = 100.0% |