Circuit Lab B/C

[ElPotato]

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.

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.

Ok, thanks. With the multimeter quality, my thought was that you never will know with some tournaments.

[Crimesolver]

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

[Schrodingerscat]

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

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.

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).

[Crimesolver]

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

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.

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).

Thank you sooooo much!!! I was trying to figure it out for 5 hours :,)

[satvik03]

Does anyone have any tips for the lab portion? Thanks.

[mdv2o5]

It looks like this question has been asked several times by several different people. Most of our answers have been fairly vague and general (e.g. "look at released exams" and "do more practice"), so I thought I would write up a more thorough answer based on what I've seen so far during the season. If you are comfortable with circuit theory, preparing for the lab portion should take no more than a couple hours of practice spread across a couple of days.

First and foremost, the biggest weakness I've seen at invitationals is a general unfamiliarity with how breadboards and multimeters work. You are almost guaranteed to see these two items appear at almost any competition, so the first step is to acquire these items to begin practicing with them. I personally dislike electronics kits since they provide very low-quality components, but in the interest of keeping costs low, I have linked to a kit that I think would work well for practice. The items you should focus on in the kit are the breadboard, jumper wires, resistors, potentiometer, and LEDs. The rest of the stuff is superfluous. I should also point out that most schools have many of the aforementioned components in the physics section of their storage closet, so it doesn't hurt to ask your science/physics teachers first.
Beginner Electronics Kit
For the multimeter, I have linked in the one used at the MIT Invite. It's very inexpensive and is certainly not a top quality meter, but I think for the purposes of Science Olympiad, it's probably better to be more familiar with a low-quality meter since that is what you are likely to find at competitions. That being said, it definitely has more than enough functionality for what you will need at competition. You can also ask your physics department if they could lend you one, but I think it would be good for SciOly to buy one yourself and bring it to competition (see one of my earlier posts regarding bringing your own meter).
Multimeter

Now that you've got your materials, probably the single most important thing you can do is get comfortable with how to use a breadboard. This doesn't mean look it up the night before and hope to figure it out on competition day. Using a breadboard takes a bit of practice, but once you've got the hang of it, it's really easy to use and should feel like second nature. Begin by making simple series and parallel circuits with your resistors and jumper wires. I would recommend the small pre-bent wires in the kit over the super long flexible wires since it makes for neater circuits and helps you develop the ability to read-off the circuits directly off the breadboard. Good breadboarding is a skill that comes with practice. After this, build up to making larger circuits. Part 1, Problem 3 of the lab section of the MIT Invite is an example of a more complex circuit on a breadboard that does not use series and parallel connections. There is a breadboard diagram included in the released exams as well as a schematic, so that can serve as practice. Once you figure out how to make series and parallel connections, you are well on your way to building most circuits you are likely to encounter.

After you can build and interpret breadboard circuits, you should get comfortable with making measurements. The first two measurements you should go for are voltage and resistance since there is a minimal chance of damaging your meter. Current measurements are more risky. Read up on how to do those properly (i.e. you almost always will have to open up the circuit and put your meter in series -- never put your meter in parallel by just poking the live circuit directly with the probes ). Ask someone if you are unsure. Resistance measurements can be made passively with the meter so try making some resistance measurements of your series and parallel circuits to see if the numbers match your theoretical values (remember most resistors you find have a 5% tolerance). Voltage measurements require you to apply some voltage across your circuit (i.e. power it on) before you measure anything. This was a common mistake I saw at tournaments which was that competitors were trying to use the voltmeter on a disconnected circuit. The kit I linked in above comes with a nice breadboard power supply that you can plug in, and it will power on the rails with 5 volts. Alternatively, you can use normal AA or AAA batteries to supply 1.5V or 3 V to your circuit. You can also try measuring the voltages of the batteries directly with your meter too. What happens when you flip the measurement you make by flipping the red and black leads? I saw this confuse several teams.

If you've gotten this far in your practice and you are truly comfortable with the two skills above, then you should be able to figure out where to go from here and what other skills you personally need to develop. Some other things to think about practicing include using LEDs in your circuits, possibly including an op-amp in your circuits (Div C), doing power calculations using voltage measurements, reading resistor values, and using a potentiometer. But even just completing the two paragraphs above will go a long way towards increasing your competitiveness in the lab section.

Some other random thoughts pertaining to the lab from observations of common problems during competition:

• LEDs are polarized. The positive side has the longer leg, but the more reliable way of determining this is to look at the cap of the LED. There should be a flat edge on the side of the negative lead. This is useful if the legs have been cut short. LEDs (and diodes in general) also do not have a single resistance value. They have a forward voltage drop. Understand what this means.
• A resistor in series with a component will allow you to make a current measurement using only a voltage measurement if you know the value of the resistor.
• Understand how to set up a multimeter to the right setting. You may be asked to pick the right setting and plug in the probes to the right sockets.
• Know how to read resistors! Spend a couple minutes to just practice this. I can't emphasize enough how many teams wasted time figuring out how color codes work during the competition rather than taking 10 minutes before the tournament to look at this. Memorizing is ideal but totally not necessary given that you have a binder. If you are trying to memorize, note that most of the colors follow the order of the rainbow, so the order goes Black, Brown, ROY G BV, Gray, White (missing Indigo). Also, the chart might be worthwhile to print out in color.

The usual advice of practicing with your partner and looking at old exams is of course very important, but hopefully this helps people get started in general.

[satvik03]

Thanks! That was very helpful.

[satvik03]

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.

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.

[ignorantcircuitboi]

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!

[Schrodingerscat]

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!

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.