Kirchoffs Law

Discussion in 'Homework Help' started by sunny1982, Mar 24, 2013.

  1. sunny1982

    Thread Starter Member

    Dec 27, 2012
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    Hey Folks
    I was wondering if someone can check to see if I have done this task correctly I am posting it in pdf format for file size issues. Basically I have to calculate the value of load (RL) using Kirchoffs.
     
  2. WBahn

    Moderator

    Mar 31, 2012
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    First, you have given no indication what the goal of the problem is. RL is just a resistor and without any constraints you can make it whatever you want.

    Second, because you are so sloppy with your units you set a resistance equal to a current and don't even spot that your final answer has units of amps instead of ohms.

    This is after adding 15V to 0.293A/15V and, not knowing how to add V to A/V, deciding just to drop units completely and got 15.02 instead of recognizing that you CAN'T add V to A/V and stopping right there.

    The point of carrying the units throughout your work (and I commend you on the fact that you at least started out doing that) is because most mistakes you make will screw up the units. But you are just going to ignore screwed up units, then what's the point?
     
  3. sunny1982

    Thread Starter Member

    Dec 27, 2012
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    I have to calculate what the value for RL is and then calculate the maximum power transfer for RL, You say I cant calculate V+A/V so where have i gone wrong? I was under the impression that what ever was flowing through the other resistors will be the same as RL.
     
  4. WBahn

    Moderator

    Mar 31, 2012
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    Okay, now you have provided a piece of critical information. You are trying to calculate the value of RL given the CONSTRAINT of achieving maximum power transfer to it. It would have been useful to have included that in your original post. We are NOT mind readers!

    If x=2kg amd y=5m, what is x+y?

    It's a meaningless question because you can't add kg to m. Well, you can't add volts to 1/ohms and come up with a meaningful result.

    So you need to work backward and ask why you came up with the sum of two dimensionally inconsistent terms in the first place.

    The claim that "what ever was flowing through the other resistors will be the same as RL" makes no sense. The current in RL is a single current. There are many other resistors and they don't all have the same current flowing in them, so what does it mean that the current in RL will be the same as what is flowing in the others?
     
  5. sunny1982

    Thread Starter Member

    Dec 27, 2012
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    Another peice of information I forgot to give you was I have to calculate the value of load, so im guessing i have to find what the resistance of RL is and then use that figure to go on to find out what the maximum power transfer is.
     
  6. WBahn

    Moderator

    Mar 31, 2012
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    Umm..... in your first post you said, "Basically I have to calculate the value of load (RL)," so you didn't leave out that piece of information unless you were talking about something other than the load resistance. But if that's the case, you should have worded it differently to indicate just what it is you are trying to find the value of.

    If you are trying to find the resistance of RL that results in max power transfer, what do you know that is special about that particular value?
     
  7. sunny1982

    Thread Starter Member

    Dec 27, 2012
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    Basically this was the question I was trying to answer, so i'm guessing I need to find the resistance in RL, so would that answer 15.02ohms be right instead of amps?
     
  8. WBahn

    Moderator

    Mar 31, 2012
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    How can it be right? It came from adding 15V to something that has units of 1/Ω. You know it is not correct (except possibly by pure coincidence).

    Look carefully at all of your work and make sure that the units work properly at each step. Also, check your math. Does it make sense that 25*0.293 comes out to something less than 1? What would 25*0.2 be?

    Always, always, always track your units and always, always, always as if the answer (including intermediate answers) make sense.
     
  9. sunny1982

    Thread Starter Member

    Dec 27, 2012
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    How about if I said that I need to find the resistance of the circuit and what ever the resistance of the circuit is the load (RL) is the same does that make sense?

    Then I use that rsitance to calculate maximum power transfer.

    Is that an better explanation?
     
  10. WBahn

    Moderator

    Mar 31, 2012
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    Yes. Assuming you have been through the proof that setting the load resistance equal to the effective source resistance results in maximum power transfer to the load (for that source resistance). If that's the case, then a slightly revised wording might be: "I need to set the load resistance equal to the effective resistance looking back into the driving circuit as seen by the load." But I'm pretty sure that anyone reading your wording will understand what you are saying. The only real nitpick is that it's a bit ambiguous whether "the circuit" includes the load or not, but most people will realize pretty quickly that you almost have to be exclusing the load.
     
  11. sunny1982

    Thread Starter Member

    Dec 27, 2012
    41
    0
    So from what you are saying my friend is that this task is completely wrong lol and if i need to find the resistance of the circuit which RL is equivalent to then this is more complexed, but can you answer me this please

    Do I need the resistance or current in the load to calculate maximum power transfer?
    Can I use kirchoffs to calculate the circuit resistance which will effectively give the value of RL?
     
  12. WBahn

    Moderator

    Mar 31, 2012
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    There are several ways at getting at the source impedance. The "normal" way is to turn off all the sources and just calculate the effective impedance seen between the load terminals (with the load removed). This doesn't work if there are dependent sources involved. Another way is to calculate the open circuit voltage where the load would go and then calculate the short circuit current if the load were to be replaced with a short circuit. The ratio of the two is the equivalent impedance.
     
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