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tuftedtitmouse12 wrote:

andrewwski wrote:Not very difficult at all. Assuming it's an electrical resistance heater and supplied 120V (standard outlet voltage), just use the simple equation . P=1500 W and V=120 V. Solve that and you get R=9.6 ohms.

but it didn't tell us the voltage that was supplied....

If that's the case you have to assume that the voltage is 120V, since that's standard AC voltage, otherwise the problem is impossible to do, perhaps state your assumption. Or you have to express it in variable form.

"Electrical plugin heater" Does imply to me that it plugs into the wall though, which if you're in america (or Japan/some other places) is approximately 120V

sean9keenan wrote:

tuftedtitmouse12 wrote:

andrewwski wrote:Not very difficult at all. Assuming it's an electrical resistance heater and supplied 120V (standard outlet voltage), just use the simple equation . P=1500 W and V=120 V. Solve that and you get R=9.6 ohms.

but it didn't tell us the voltage that was supplied....

If that's the case you have to assume that the voltage is 120V, since that's standard AC voltage, otherwise the problem is impossible to do, perhaps state your assumption. Or you have to express it in variable form.

"Electrical plugin heater" Does imply to me that it plugs into the wall though, which if you're in america (or Japan/some other places) is approximately 120V

AC??? but im div b..so wouldn't it b dc??

Yeah, household outlets are AC, however we can assume they are DC and that won't change anything that we need.

Also, thanks once again for your response! When you have a Wheatstone Bridge, will the movement point to the left or to the right when there is a higher potential on the left junction than on the right one? I am thinking the current flows to the right in this case thus making the movement on the galvanometer go to the right.

As to the AC voltages you can assume that it's DC voltage because the value for AC voltage we normally chose to write down is the value that allows us to treat it just like DC voltage. It's a special value called the "RMS value" which is actually different then the peak of the AC voltage, which is around 170V if I remember correctly.

If I understand your question correctly then if there is a higher voltage on the left junction as compared to the right junction then the current through the middle resistor will be to the right, which depending on how you have your galvanometer hooked up will register as a positive current or negative curent.

Also a good problem related to wheatstone bridges: If all the resistors are 10 ohms, INCLUDING the middle resistor (that's labeled null in your diagram) what is the equivalent resistance of the circuit, without using delta-wye transformations. (I did try and explain it in my last post, dunno if it got through though)

Hmm, this is probably wrong but this is what came to mind at once when I read this problem:

We have the same potential at both the left and right junctions that connect the resistor in the middle, this means no current will go through the middle resistor. Therefore we only need to account for the other four resistors by pretending the middle one isn't there. This is just two parallel branches which makes a total resistance of .

In addition, will the Shock Value test involve transformers and/or transistors?

Yup, that's right, that's pretty much the point that I was trying to get across. Also I don't think transistors are covered in the rules. As to transformers, I would argue that the AC applications of transformers are outside the scope of this event, but the DC applications of transformers (although limited) might show up. Did you know that a lot of cars have transformers in them in order to start their cars? But the voltage of the car battery isn't AC, what does the transformer do?

sean9keenan wrote:As to the AC voltages you can assume that it's DC voltage because the value for AC voltage we normally chose to write down is the value that allows us to treat it just like DC voltage. It's a special value called the "RMS value" which is actually different then the peak of the AC voltage, which is around 170V if I remember correctly.

RMS means Root-Mean Square, which is the peak voltage divided by . So if we take 120V as the RMS value, we get 170. If AC is out of the scope of this event though, so is that calculation.

If I understand your question correctly then if there is a higher voltage on the left junction as compared to the right junction then the current through the middle resistor will be to the right, which depending on how you have your galvanometer hooked up will register as a positive current or negative curent.

Best way to approach this problem would be to draw the currents arbitrarily - whichever way you wish. Then, if you get a negative current, you know it actually goes the other way. Naturally, though, the current flows from higher to lower potential, so yeah, left-to-right.

sean9keenan wrote:Yup, that's right, that's pretty much the point that I was trying to get across. Also I don't think transistors are covered in the rules. As to transformers, I would argue that the AC applications of transformers are outside the scope of this event, but the DC applications of transformers (although limited) might show up. Did you know that a lot of cars have transformers in them in order to start their cars? But the voltage of the car battery isn't AC, what does the transformer do?

Highly, highly doubt anything like this would be within the scope of the event. The Kettering ignition system would be the only example of this that I can think of, and I doubt one would be expected to know it for the event. But then again, I'm not the one that writes the events.

andrewwski wrote: Highly, highly doubt anything like this would be within the scope of the event. The Kettering ignition system would be the only example of this that I can think of, and I doubt one would be expected to know it for the event. But then again, I'm not the one that writes the events.

Perhaps, but over preparing never hurt. Not to mention the principles behind the fast change in voltage inducing a current and creating a higher voltage across the spark plug isn't something that I think it outside of the scope of this event. Maybe not directly under the DC circuits section but a combination of the circuit theory and the magnetism that they have to know.

Could you tell me more about...

sean9keenan wrote:...the principles behind the fast change in voltage inducing a current and creating a higher voltage across the spark plug...

please? You've caught my interest.

sean9keenan wrote:I do believe that the rules say you dont need to know much about inductors as they apply in circuits, but you do need to know about electromagnets. For instance you probably won't be asked to solve for the current through an inductor as a function of time. They can definitely ask you about the strength of the magnetic field or to draw the field lines.

I would recommend reading this: http://hyperphysics.phy-astr.gsu.edu/hb ... lemag.html It covers a lot of the material you'd need! You might also want to do some reading up about how toroids are used as cores of electromagnets

Please may you explain some of the information in simpler terms than what on the website.