determining unknown resistance by ohms law

Discussion in 'Homework Help' started by electronicsnewb, Nov 8, 2013.

  1. electronicsnewb

    Thread Starter New Member

    Dec 8, 2012
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    Here's another question I'm not quite sure about. Its asking to determine it unmarked resistor's resistance value using Ohm's Law. Show calculations that result in the VI ratio of 600 ohms. However, do not exceed the power rating of 10 W.

    With the given info of 600 ohms and 10 W, I can calculate for V using V=√PR=√(10)(600)=77.4V

    I=V/R=77.4/600=129 mA

    I guess with the given info on ohms and power that my answer is correct, but practically, wouldn't all I would need is an ohmmeter to measure the resistance or connect a voltage source across the resistor, with an ammeter in series to calculate the VI ratio for resistance. Any input is greatly appreciated. Thanks!
     
  2. WBahn

    Moderator

    Mar 31, 2012
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    This question makes no sense.

    You are asked to determine an unmarked resistor's resistance but then told to show calculations that result in a value of 600 ohms?

    What does the power rating have to do with it?

    You are basically correct. In practice you would use either an ohmeter or, for better accuracy, you would apply but it in series with a known resistor and then put a known voltage across the pair, monitoring the current by measuring the voltage across the known resistance. In doing so, you would also make sure that you weren't exceeding the power rating of the unknown resistor.
     
  3. electronicsnewb

    Thread Starter New Member

    Dec 8, 2012
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    0
    The question didn't make sense to me either. I just solved it by what I was given in the problem. The method using the 2 resistors in series that you mentioned is another way. So I place a known voltage across the 2 resistors and measure the voltage across the known resistor and divide this voltage by the resistance to get i. Then I just subtract the voltage across the known resistor from the voltage source to get the voltage across the unmarked resistor and divide it by the current to find the unmarked resistor's resistance value.

    My question is that since it is an unmarked resistor, how can I know if I'm exceeding it's power rating? Do I just guess by looking at the physical size of the resistor and go by what material it's made of? Thanks!

    anks for your response!
     
  4. WBahn

    Moderator

    Mar 31, 2012
    17,769
    4,804
    Looking at the physical size of the resistor and the material that it's made from may give you a hint as to the power rating, but you already know the power rating. It won't give you the faintest clue regarding the current and voltage levels that correspond to that power level.

    But can you size the current sensing resistor and the output of the voltage source such that you are guaranteed not to be able to exceed the power rating of the unknown resistor no matter what its value is?

    Given a voltage source Vo, a sensing resistor Ro, and an unknown resistor Rx all placed in series, what is the power dissipated in the unknown resistor?

    What value of Rx would result in the highest possible power dissipated in it?

    What does the relationship between Pmax, Vo, and Ro have to be to ensure that the highest possible power in Rx does not exceed the Pmax rating?
     
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