16 is c, an ideal motor
Section I - Multiple Choice (3 Viewers)
- Thread starter Rafy
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someth1ng
Retired Nov '14
A is still a valid answer for Q20.
someth1ng
Retired Nov '14
What it meant was that there is no load on the motor, work is done on the motor but not BY the motor.
someth1ng
Retired Nov '14
Newton's first law of motion states that it will continue in its state of motion. There is no load and hence, it could just keep going.
in a motor back emf increases as the motor spins faster. when back emf = supply emf, the motor spins as constant speed... and since no work is being done. it is therefore C.
Yeah that makes sense, but there must have been a torque to get it moving in the first place ...
brianphamm
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I don't know if there is debate about the induction cook top MC question, but here's an article saying high resistance is better:
http://www.ledomaine.com.au/blog/2011/08/how-induction-cooktops-work/
http://www.ledomaine.com.au/blog/2011/08/how-induction-cooktops-work/
Re Q16. That's true, but for the operating conditions given (constant rotational speed) then back EMF equals applied voltage for the IDEAL dc motor. For any real motor the answer *would* have been "A", but for this question the answer is "C".
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Q20. Sorry I haven't read all the replies, but just in case there's still any uncertainty over this one the correct answer is "B" (InSb).
- Si and GaN (C and D) have too higher bandgap, so few if any carriers will be optically excited across the bandgap by the given infrared signature.
- HgCdTe (A) has such a low bandgap that many carriers would be thermally excited into the conduction band, effectively "shorting out" the device if it were operated anywhere even close to room temperature. It would also be too sensitive to background thermal radiation.
So the only correct answer is "B".
- Si and GaN (C and D) have too higher bandgap, so few if any carriers will be optically excited across the bandgap by the given infrared signature.
- HgCdTe (A) has such a low bandgap that many carriers would be thermally excited into the conduction band, effectively "shorting out" the device if it were operated anywhere even close to room temperature. It would also be too sensitive to background thermal radiation.
So the only correct answer is "B".
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Solutions. I've only just got handed a copy of the exam tonight, but these are what I believe to be the correct answers. (* = up for debate)
1. d
2. a
3. c
4. b
5. c
6. c
7. b
8. a
9. b
10. d
11. c
12. a
13. b
14. d
15. d
16. c
17. b
18. c
19. c*
20. b
* Question 19 could be debated. Too higher resistance and it certainly wont heat effectively. The induction coil will induce a voltage in the pan base but the induced currents will be insufficient to provide enough heat. This is basically a voltage driven situation, P = V^2/R, the higher the resistance the lower the power.
... up to a point. The problem with this question is that if the resistance becomes too low (lower even that that of the copper induction coil in the stove) then the cookware base gets a large induced current, so large in fact that it produces it's own flux which cancels out most of the flux produced by the induction coil. In this situation most of the power actually gets dissipated in the internal resistance of induction coil instead of the cookware, which is obviously undesirable.
So yes, Q19 is debatable. However the reasons for too lower resistance being a "fail" are quite complicated (not accessible for the average student), while the reasons for too higher resistance being a "fail" are quite straight forward. In cases such as this one, the pedagogically correct answer is always the most straight forward one. So almost certainly answer "C" will be the officially correct answer that the markers use.
1. d
2. a
3. c
4. b
5. c
6. c
7. b
8. a
9. b
10. d
11. c
12. a
13. b
14. d
15. d
16. c
17. b
18. c
19. c*
20. b
* Question 19 could be debated. Too higher resistance and it certainly wont heat effectively. The induction coil will induce a voltage in the pan base but the induced currents will be insufficient to provide enough heat. This is basically a voltage driven situation, P = V^2/R, the higher the resistance the lower the power.
... up to a point. The problem with this question is that if the resistance becomes too low (lower even that that of the copper induction coil in the stove) then the cookware base gets a large induced current, so large in fact that it produces it's own flux which cancels out most of the flux produced by the induction coil. In this situation most of the power actually gets dissipated in the internal resistance of induction coil instead of the cookware, which is obviously undesirable.
So yes, Q19 is debatable. However the reasons for too lower resistance being a "fail" are quite complicated (not accessible for the average student), while the reasons for too higher resistance being a "fail" are quite straight forward. In cases such as this one, the pedagogically correct answer is always the most straight forward one. So almost certainly answer "C" will be the officially correct answer that the markers use.
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You bloody legend!