This problem is a bit tricky. Start by drawing a diagram, and note that the question asks for the power delivered to the cleaner, which would exclude power that heats the extension cord.
Also, see my latest post about the need to calculate intermediate results to an accuracy much better than 1%. You should be safe with six significant digits as you are working.
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13 comments:
Here is a student's comment and my response that were posted before I broke this chapter up into sections
Anonymous said...
I'm having trouble with problem 12 part (a). It says to assume that the cleaner has constant resistance and that the two conductors of the extension cord have a given resistance each.
To start, I found the constant resistance without the cord in place by R = V^2/P, using the rated power and the 120V.
Then, I modified my drawing to include the two resistances of the extension cord in series with the constant resistance, and calculated the observed power with P=V^2/Rtotal, summing the resistors for Rtotal. However, this yields the wrong answer, and I'm quite confused at where I'm going wrong.
Professor Stephens said...
Looks like you have the right start. But in the second step, the power you calculate as V^2/Rtotal will be the total power dissipated in the vacuum cleaner and the cord. You want to find the power dissapated in the cleaner alone, which will be lower because less than 120 Volts appear across it.
For this problem I found the current using voltage = 120V and Power = 520W. Then I used that current value and .950 Ohms which was given for the first resistor to get how much power was lost after passing through the first resistor. I used P=RxI^2. Next I found the amount of power left by taking 520W - power lost in the first resistor. I used that new value to find the new amount of current with the original voltage of 120V in the equation P[after first resistor] = 120V x I. With the new current value, I found the power lost after passing through the second resistor the same way as with the first resistor. Finally I subtracted the power lost in the second resistor from the power after the first resistor.
Am I doing this correctly? Because I keep getting an answer that's close but not exact and it doesn't look like what the person before me posted.
Was 520 Watts the number you were given in the problem? With the extra resistance in the extension cord, that will be neither the power drawn from the plug nor the power delivered to the cleaner. I suggest you find the resistance of the cleaner from the fact that it draws a load of 520 Watts when connected to 120 Volts (or whatever your numbers are). P=V^2/R will give you that resistance, call it Rvac. Now you have a circuit with total resistance Rvac+2*Rwire. If you connect that to the 120V outlet, what current will flow? Proceed from there.
So I understand now the current flow is constant and use that to find part a. How do you get part b and c if you do not have the resistance values for Rwire to find the current with?
For parts b and c, you are given the desired power to the vacuum cleaner. Since you know its resistance, you can find the required current. That lets you find the resistance in the wires, and consequently their thickness (given resistivity and length).
Dr. Stephens,
Can you please explain why in an earlier post you said the total resistance would be the Rvac+ 2*Rwire? I don't understand why it is 2*Rwire. I was previously trying with just the Rwire=2(resistance of each conductor), which was obviously wrong. I then followed your advice about multiplying the 2*Rwire to get the answer for part (a), but I must admit that I do not understand why that is.
Thank you in advance.
When I wrote Rvac + 2*Rwire, I meant Rwire to be the resistance of each conductor. As printed in the book, each conductor has a resistance of 0.900 Ohm, so the additional resistance in the circuit would be Rvac + 1.800 Ohm. Does that clarify?
Thank you Dr. Stephens,
Funny thing is, I got the question correct by using 2*1.8 ohms, which is what I thought you meant by 2*Rwire. Previously, I was using 2*0.9 ohms and was getting the answer wrong.
Thank you for your time.
The total added resistance really is 2*0.9, so if that gave you the wrong answer and 2*1.8 gave you the correct numerical value, then I think there is another step that you are getting wrong. Are you calculating the energy dissipated by the vacuum (correct) or the energy delivered from the power source (incorrect)?
I possibly did do another step wrong. I will try to make it to your help room hours today, if you do not mind, to go over the question (I am still having trouble with parts b and c).
Thanks again, Dr. Stephens.
I'm having some trouble getting the part b and c. I know that I'm supposed to get the current by using the given value of Power and using the resistance of the vac. I divided the difference between the two values of power by I^2. I used that R value and divided it with the resistivity value then I divided (2*Length) with that to find the area. I divided that with pi and took the square root and multiplied it by 2.
I don't know what I'm doing wrong!
The 11:30 post seems too indirect. I would start from scratch: What current must flow through the vacuum cleaner in order to have the desired power? (Remember that the resistance of the vacuum cleaner is constant throughout.) What must be the total resistance in order that that current flow? (Ohm's law) Subtracting the resistance of the vacuum cleaner, what is the resistance of each wire? Now use resistivity etc. to find the area, therefore the diameter, of each wire.
Another tip for this problem. In parts b and c, you wind up calculating the lead resistance as the relatively small (order 0.1 Ohm) difference between two much larger numbers (order 20 or 30 Ohms). To get the difference accurate to 1% (which WebAssign requires), you have to know each of the two larger resistances to an accuracy of something better than 0.001 ohm. So you should carry your calculation from the start to, play it safe, 6 significant digits.
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