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Finding EMF and resistance

  1. Apr 19, 2013 #1
    1. The problem statement, all variables and given/known data

    In the circuit shown in the figure http://s3.amazonaws.com/answer-board-image/ffa3225e-0730-480c-8ed0-f417490d6b96.jpeg[/COLOR] , the current in the 20.0-V battery is 5.00A in the direction shown and the voltage across the 8.00-Ω resistor is 16.0V , with the lower end of the resistor at higher potential. Find the emf of the battery .
    Part B
    Find the polarity of the battery .

    the upper terminal -
    the upper terminal +

    Part C
    Find the current through the 200.0- battery.
    Part D
    Find the direction of the current through the 200.0- battery.

    The current through the 200.0 battery is in the direction from the - to the + terminal.
    The current through the 200.0 battery is in the direction from the + to the - terminal.

    Part E
    Find the resistance .


    2. Relevant equations

    Kirrchoff's Laws

    3. The attempt at a solution

    I know how to use the laws well. I thought that since there is a pd of 16 across the 8 ohm resistor so there is a 2A moving up this line. When it came to the junction I'm lost. In the SM, they say that we can combine the unknown R as parallel ones. We can also combine the 30 and 20 ohm resistors as parallel. HOW IS THIS POSSIBLE? ... We can never assume they are parallel. I need clarification!
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Apr 19, 2013 #2

    SammyS

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    attachment.php?attachmentid=58033&stc=1&d=1366406536.jpg

    The SM (whatever that is) is correct.

    The conductor common to an end of all of the resistors can be thought of as one big junction -- which you redraw any way you want, as long as you continue to make contact with the same end of each of the conductors.

    There is zero voltage drop anywhere in that piece of conductor.
     

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    Last edited by a moderator: May 6, 2017
  4. Apr 19, 2013 #3

    gneill

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    Are you given the polarity of the voltage across the 8.00 Ω resistor? Otherwise, how are you to know the direction of the current flowing through it?

    Does "X" represent a third battery for which you're to find the EMF?
     
    Last edited by a moderator: May 6, 2017
  5. Apr 19, 2013 #4
    Can't really get ur point!?? Please clarify
     
  6. Apr 19, 2013 #5

    gneill

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    What's an "ur point"? Please clarify. :devil:
     
  7. Apr 20, 2013 #6

    SammyS

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    I think you find it at the very tip of an "ur" ... don't 'cha think ?
     
  8. Apr 20, 2013 #7

    CWatters

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    These two circuits are the same..
     

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  9. Apr 20, 2013 #8
    I am sorry about my previous reply. I was a bit asleep and tired. I hope you aren't sad about it. Anyway, you said here we can think the ends to be one big junction. I don't really get this part. How is it possible. There are more than one emf source. That's why we use Kirrchoff's Rules!
     
  10. Apr 20, 2013 #9
    That was a good attachment. But HOW can we say they are the same. What are the roles of the junctions then?
     
  11. Apr 20, 2013 #10

    gneill

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    Junctions attach components to nodes. Nodes are contiguous zero-resistance conducting paths, all at the same potential. It makes no difference where on a given node a component is connected.

    Connections to a node can be moved anywhere, so long as they remain connected to that node. It is the mathematical topology that determines the circuit, not artistic license :smile:
     
  12. Apr 20, 2013 #11

    CWatters

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    Compare a circuit diagram with an actual piece of equipment. The lines on a circuit represent ideal conductors. Their shape need not be the same as the wires or tracks on the actual printed circuit board.

    Good example is a car. The whole chassis is used as a 0V node. Anywhere you need 0V you can drill a hole in the chassis and fix a wire. The wiring diagram doesn't need to inclued a 3D drawing of the car chassis. Electrically all points on the chassis behave similarly. On the circuit you can just show a sybmol representing a connection to the chassis.
     
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