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Homework Help: Coils arrangement so the hotplate operates at 3 diff. powers

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  1. Oct 8, 2016 #1
    1. The problem statement, all variables and given/known data
    A certain electric hotplate, designed to operate on a 250 V supply, has two coils of nichrome wire of resistivity 9.8 * 10-7 Ω m. Each coil consists of 16 m of wire of cross-sectional area 0.20 mm2.

    (a) For one of the coils calculate (i) its resistance, (ii) the power dissipation when 250 V supply is connected across the coil, assuming its resistance does not change with temperature.

    (b) Show, by means of diagrams, how these coils may be arranged so that the hotplate may be made to operate at three different powers. In each case, calculate the power rating.

    (c) The hotplate is connected to the 250 V supply by means of cable of total resistance 3.0 Ω. (i) Calculate the power loss in the connecting cable when the hotplate is being used on its middle power rating. (ii) Comment qualitatively on any change in power loss in the cable when the hotplate is operating at each of its other power ratings.

    (d) Different connecting cables are available for use with the hotplate. The maximum safe current which can be used in any one of the cables is 1 A or 3 A or 6 A or 12 A. State which is the most appropriate cable to use and briefly explain one possible danger of using cable with a lower maximum safe current.

    Answers: (a) (i) 78 Ω, (ii) 8 * 102 W, (c) (i) 28 W.

    2. The attempt at a solution
    (a) (i) ρ = (A R) / L → R = (ρ L) / A = (9.8 * 10 -7 * 16) / (2 * 10-7) = 78.4 Ω.

    (a) (ii) P = V2 / R = 2502 / 78.4 = 797 W.

    (b) Not even sure where to start. Am I required to make a circuit? What are coils in a circuit? Resistors? What is a hotplate in a circuit? A cell?

    (c) (i) As I understand we'll have RTotal = RCoil + RCable = 78 Ω + 3 Ω = 81 Ω. Then P = 2502 / 81 = 771.6 W. Difference between 800 W and 771.6 W is 28.4 W.

    (c) (ii) Not sure on this question either. So we need to say whether there will be any power losses in the cable when the hotplate is operating at what? What does "each of its other power ratings" actually mean?

    (d) Maybe the higher is the maximum safe current the better it is? So in case more current starts going through the circuit the cables will not overheat and start buring? Just a guess, also not sure on this part.

    Any suggestions please?
     
  2. jcsd
  3. Oct 8, 2016 #2

    mfb

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    The coils act as resistors, right. The diagrams mentioned in the problem statement are circuit diagrams. The heating pad is irrelevant - just the electrical part matters.

    You need (b) here (e.g.: what about the other coil?), but you picked the right circuit. Why did you calculate the difference to 800 W? The coil cannot have 800 W, that would be more than the power in wire plus coil, and it would give a negative power loss in the cable.
    Your answer is not far away from the right answer by accident here, but it is not right.
    That needs an answer to (b). Depending on how you connect the coils, you get different heating power values.
    Well, the 12 A cable is fine, but more expensive than others. Can you also use cables rated for lower currents?
    Overheating is indeed the issue if the current rating is too low.
     
  4. Oct 8, 2016 #3

    gneill

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    Yes, you're being asked to draw three configurations where one or more of the coils is being supplied power.

    For purposes of this exercise the coils can be represented as resistors.


    [edit: I see that @mfb got there first]
     
  5. Oct 8, 2016 #4
    Let's then finish (b) first.

    Maybe it should be something like this:
    82611fcc6c65.jpg
     
  6. Oct 8, 2016 #5

    gneill

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    There is no circuit element "hotplate". A hotplate is just a metal disk with heating coils below (or inside). The interesting part as far as this problem is concerned is the two resistive coils. So your circuit diagrams should contain only power supply and resistors.

    You need to determine how to apply voltage to the resistors to get three different power settings.
     
  7. Oct 8, 2016 #6
    I can only think of: (i) power supply in series with two resistors in series and (ii) power supply in series with two resistors that are in parallel.
     
  8. Oct 8, 2016 #7

    gneill

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    Nowhere does it say that both coils have to be in use at the same time :wink:
     
  9. Oct 8, 2016 #8
    Maybe then:
    and (iii) power supply in series with just a one resistor.
     
  10. Oct 8, 2016 #9

    gneill

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    Yes.
     
  11. Oct 8, 2016 #10
    And this means
    that we need to calculate the power (like P = V I or P = V2 / R) for each of the graphs?
     
  12. Oct 8, 2016 #11

    gneill

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    Yes.
     
  13. Oct 8, 2016 #12
    Then we would have:
    1. One resistor: P = 2502 / 78 = 801 W.
    2. Two resistors in series: P = 2502 / (78 + 78) = 401 W.
    3. Two parallel resistors: P = 2502 / 39 = 1603 W. (1 / R = 1 / 78 + 1 / 78 → R = 39 Ω.)
     
  14. Oct 8, 2016 #13

    gneill

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    Looks good.
     
  15. Oct 8, 2016 #14
    OK, so for the next part we have the middle power 801 W or (797 W if we use 78.4 Ω). So, roughly 800 W.

    Then the power loss will be:
    What's wrong with this one?
     
  16. Oct 8, 2016 #15

    gneill

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    I don't see anything wrong, per se. Does it answer the question posed?
     
  17. Oct 8, 2016 #16
    I would say it's correct, but:
    Maybe @mfb meant that I got the right answer by accident withouth considering (b) first?
     
  18. Oct 8, 2016 #17

    gneill

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    I think it's a matter of interpretation as to what "power lost" means. Lost as compared to what benchmark?

    If the cable was ideal (no resistance) then the single coil produces about 800 W of heat. What heat does that coil produce when the cable is not ideal but has a resistance of 3 Ohms? The difference between "ideal" and "not ideal" could be considered power lost.

    On the other hand, you could simply calculate the power consumed by the cable directly under operating conditions and call that the power lost (since it doesn't contribute to heating up your soup).

    I think that both approaches should lead to similar, but not identical answers. And I also suspect that the latter approach s what is intended in the question: find the power dissipated by the cable.
     
  19. Oct 8, 2016 #18
    I did that actually. But P = 2502 / 3 = 20 833 W.

    While if we do 2502 / (78 + 3) = 771.6 W. 800 - 771.6 = 28.4 W.
     
  20. Oct 8, 2016 #19

    gneill

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    But the 250 V is not directly across the cable resistance! It and the hotplate resistance are in series. Better to first find the current, then apply ##P = I^2 R## for the resistances in a series connection.
    Yes, but then you need to justify the 800 W value (as I did in post #17). It's easier to just calculate the actual power lost during operation, as above.
     
  21. Oct 8, 2016 #20

    mfb

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    I think the problem statement is unambiguous, it asks for the power dissipation in the cable. If it would ask about power differences in the hot plate, it would be "due to the connecting cable" or something like that.
     
  22. Oct 8, 2016 #21
    Hm, so: I = V / R = 250 / (78 + 3) = 3.086 A. P = I2 R = 3.0862 * 3 = 28.6 W. Should be correct now?
     
  23. Oct 8, 2016 #22

    gneill

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    Yes, that's fine.
     
  24. Oct 8, 2016 #23
    Alright, so (c) (ii) and (e) are left.

    What does this actually mean?

    If we have two resistors in series I would say that the power loss is larger (more cable is used). And when they are connected in parallel less power is lost, but still more than when there is one resistor. Like this?

    Maybe the higher is the maximum safe current the better it is? So in case more current starts going through the circuit the cables will not overheat and start buring? Just a guess, also not sure on this part.
     
  25. Oct 8, 2016 #24

    gneill

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    You should consider the length of the cable to be fixed. The interconnections and switching of the coils would be handled inside the hotplate package, while the cable would be like an extension cord that plugs into an electric outlet. The cable length doesn't change when you turn a switch on the hotplate.

    You could calculate the power loss for each case and compare the amounts as well as compare the lost amount to the "delivered" amount in each case. It's up to you to think of what points might be of interest since they invite you to "Comment qualitatively", but you might get extra credit if you can put it in terms of "energy efficiency".
    Well, it's definitely not a good idea if using an appliance creates a fire hazard :smile: Can the thinnest cable offered operate the hotplate safely on all three settings? What do think an appropriate safety margin might be for cable rating versus actual maximum current?
     
  26. Oct 9, 2016 #25
    Energy losses with one resistor and cable: 28.6 W.

    Energy losses with two resistors and cable: I = 250 / (78 + 78 + 3) = 1.57 A → P = 1.572 * 3 = 7.4 W.

    Energy losses with two parallel resistors and cable: I = 250 / (39 + 3) = 5.95 A → P = 5.952 * 3 = 106.3 W.

    So if I got the parallel one correct (calculated the resistance of the resistors first and then added the cable resistor) the most efficient way would be to put two resistors in series and the least efficient way would be putting them in parallel.

    Well, our currents are: 3.08 A, 1.57 A and 5.95 A for the three settings, if I calculated them correctly.

    The most appropriate would be the 6 A cable in order to use it for the three settings. 12 A cable would be more expensive. The danger of using a cable with a lower current is the risk of fire.

    Though I don't know how to calculate a safety margin.
     
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