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Power In A Circuit Problem

  1. Sep 13, 2007 #1
    1. Three resistors and two 10.0 volt batteries are arranged as shown in the circuit diagram. Which of the following entries in the table is correct?



    2. P= I^2 X R, P= V^2/R, P= I X V



    3. I've been working on this problem for 45 minutes and can't find a way forward. My instructor has not lectured on this material yet and my text book hasn't been much help. I think this problem involves Kirchhoff's junction rule or his loop rule. I've attached a pic of the circuit and the related table. The correct answer is highlighted but I haven't been able to solve the problem. Any assistance will be greatly appreciated. Thanks in advance!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Sep 13, 2007 #2

    learningphysics

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    The power delivered by a voltage source... is the voltage * current coming out of the positive terminal of the voltage source...

    Hint: find the voltage at the node joining the 3 resistors... then you can get all the currents..
     
  4. Sep 13, 2007 #3
    I'm not quite sure how to go about finding the voltage at the node joining the 3 resistors.

    If I use ohms law: I= V/R

    The current to the node supplied by battery 1 is equal to 10V/40 ohms which gives ".25 amps". The current to the node supplied by battery 2 is equal to 10V/10 ohms which gives "1 amp". Adding the two results together would be 1.25 amps at the node. Am I on the right track? If so, how does the 12 ohm resistor in the middle come into play?
     
  5. Sep 13, 2007 #4

    learningphysics

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    No, the current supplied by battery 1 isn't 10/40 because the 40 ohms isn't connected to ground... have you studied the node voltage method... let the voltage at the node by x... Now using KCL (sum of currents coming out = 0) at that node.... can you get an expression for x?
     
  6. Sep 13, 2007 #5
    I have not studied the node voltage method yet. However, I am looking at my text book and I see that there is a rule that states that the voltage into the node is equal to the voltage coming out of the node. Is this what you are talking about?
     
  7. Sep 13, 2007 #6

    learningphysics

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    must be current in and current out... yeah... kirchoff's current law...

    There's another way you can do this... use the principle of superposition... short out one of the voltage sources... then find the currents through the 40 ohm and 10 ohm resistors.... then turn that one back on, and short out the other voltage source... again find the currents through the 40 ohm and 10 ohm...

    The sum of the two values for current through the 40ohm resistor (careful of direction) is the actually current through the 40 ohm resistor... likewise with the 10 ohm resistor...
     
  8. Sep 13, 2007 #7
    How and when will the 12 ohm resistor come into play?
     
  9. Sep 13, 2007 #8
    Also, why is direction important?...I thought that current is constant everywhere in a closed system?
     
  10. Sep 13, 2007 #9

    learningphysics

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    No, that's not true... when there is a single loop current is the same everywhere within that one loop...
     
  11. Sep 13, 2007 #10
    Thank-you for helping me out...it's past midnight my time and my brain isn't working very well anymore. You have given me a direction to pursue in the morning so hopefully this will make more sense then. If I sleep with my book under my pillow, do you think I will wake up with the solution?!!
     
  12. Sep 13, 2007 #11

    learningphysics

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    lol! maybe... I recommend checking out any examples on "node voltage" method in your text... it gives a real quick method to solve this problem. good luck!
     
  13. Sep 14, 2007 #12
    Big thanks goes to "learningphysics"! I was able to solve this problem after another two hours of work. Way to long for me to list the steps but basically I had to use Kirchhoff's junction and loop rules to develope equations that allowed me to solve for current from batteries 1 & 2. Once I had the currents for both batteries, I plugged their values into the Power = V X I formula to solve the problem.
     
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