That looks very much like the nichrome igniter for model rockets. Does anyone in your family work with that type of stuff?Okay, so I'm not in circuit lab. But I found this electronic* component on the floor of my garage today. I was wondering, what is it?
https://drive.google.com/file/d/1LucJUe ... sp=sharing
*maybe, unsure
Oh yeah! That's what it was! I completely forgot!That looks very much like the nichrome igniter for model rockets. Does anyone in your family work with that type of stuff?Okay, so I'm not in circuit lab. But I found this electronic* component on the floor of my garage today. I was wondering, what is it?
https://drive.google.com/file/d/1LucJUe ... sp=sharing
*maybe, unsure
Rule 2d addresses this issue. It appears that it's up to the supervisor if you can use your own, so it might be worthwhile to bring one and ask. That being said, I highly doubt that the multimeter provided to you will be bad enough to be an issue in solving the problem. The voltage and resistance measurements are fairly standard, and I don't think any ES would be bold enough to let students use an ammeter to probe the circuit.Quick question, are event supervisors allowed to restrict the multimeter you use in labs or prevent you from using your multimeter altogether? I was thinking about practicing with worse multimeters in case I had to use one in competition.
Ok, thanks. With the multimeter quality, my thought was that you never will know with some tournaments.Rule 2d addresses this issue. It appears that it's up to the supervisor if you can use your own, so it might be worthwhile to bring one and ask. That being said, I highly doubt that the multimeter provided to you will be bad enough to be an issue in solving the problem. The voltage and resistance measurements are fairly standard, and I don't think any ES would be bold enough to let students use an ammeter to probe the circuit.Quick question, are event supervisors allowed to restrict the multimeter you use in labs or prevent you from using your multimeter altogether? I was thinking about practicing with worse multimeters in case I had to use one in competition.
The first problem works because it is an ideal current source on the left. Because it is parallel to a short, R1 has zero voltage and thus zero current (given non-zero resistance), so it can be excluded from the current analysis. After removing it, you have two current loops in opposite directions, one is limited to 8A in a downward direction by the current source, and the other from the voltage source is limited to 5A in the upwards direction by R2. The net result is a downwards (positive current) of 3A.When the question asks you to show your work, but there's no work on the answer key
Could anyone please explain these?
https://drive.google.com/file/d/1_jkQbi ... sp=sharing I'm sure this one is probably not possible since it's a short circuit, but the answer key says 3 amps soooo
https://drive.google.com/file/d/1cTjnsH ... sp=sharing The answer for this one is 0.625A, 21.875V, I just don't know how to approach it
Thank you sooooo much!!! I was trying to figure it out for 5 hours :,)The first problem works because it is an ideal current source on the left. Because it is parallel to a short, R1 has zero voltage and thus zero current (given non-zero resistance), so it can be excluded from the current analysis. After removing it, you have two current loops in opposite directions, one is limited to 8A in a downward direction by the current source, and the other from the voltage source is limited to 5A in the upwards direction by R2. The net result is a downwards (positive current) of 3A.When the question asks you to show your work, but there's no work on the answer key
Could anyone please explain these?
https://drive.google.com/file/d/1_jkQbi ... sp=sharing I'm sure this one is probably not possible since it's a short circuit, but the answer key says 3 amps soooo
https://drive.google.com/file/d/1cTjnsH ... sp=sharing The answer for this one is 0.625A, 21.875V, I just don't know how to approach it
For the second one, I'd personally start with the Thevenin equivalence theorem (for simplicity, you could do it without) to replace the right side of the circuit of the terminals with a downwards 5V and a 43 ohm resistor. Then I'd apply KCL through node A or B. If you get node B to 0V as reference ground and A to V, flowing into A you have 5A and out of V/5 A and (V--5)/43=(V+5)/43 A. Thus if current in=current out, 5=V/5+(V+5)/43. Multiplying by 43*5 gives 1050=43V+5V+25, or V=1050/48=21.875V. Finally note that the Thevenin equivalent will have the same current loop as desired in the final answer, so you can calculate (V+5)/43=(21.875+5)/43=0.625A. (I just noticed you are in B division and apparently not expected to be able to use Kirchhoff current analysis by the rules, so that may have been more of a C level problem than B).
Yeah, they can tell you to not use your multimeter. This is because there are lab stations/questions in which you have to find the resistance of a mystery resistor only using a voltmeter. With a multimeter, you could measure resistance, which would be unfair. In competitions, the multimeters, voltmeters, and ohmeters are usually good quality, but sometimes can be analog and not digital.Quick question, are event supervisors allowed to restrict the multimeter you use in labs or prevent you from using your multimeter altogether? I was thinking about practicing with worse multimeters in case I had to use one in competition.
Your method as I read it sounds like it should work, so unless I am overlooking something, you are measuring wrong, or your circuit is not connected correctly. Also, if you are not required to measure all of them in a single circuit, I would advise measuring one at a time for greater sensitivity.I have a question. How do I solve the following question (like step by step process)?
I am given the following materials:
a voltmeter
1 known resistor with resistance of 1,000 ohms
5 unknown resistors
a breadboard
alligator clips
a 3 V battery
How do I find the resistance value of each of the unknown resistors?
This was my approach, which turned out to be blatantly wrong:
First, I connected all the resistors in series. My thought process was that since I knew the resistance of one resistor, I could just use the voltmeter to find the find the voltage across the known resistor, and then I could find the current across that resistor. That current should be the same as the total current and the current across all the resistors because the circuit was in series. Using this, I could then measure the voltage across all the unknown resistors. Since I knew the current value, if I knew the voltage across the resistors, I could find all the resistance values.
Then, I put the leads of the voltmeter on the ends of the resistors (so I connected it in parallel). But the reading for the voltage was 0, so I could not find the value of the resistors. I think my thought process is wrong, does anyone know what I should do? Thanks!
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