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Determining the power delivered by a battery -image included

  1. Jul 3, 2016 #1
    1. The problem statement, all variables and given/known data
    See attachment for the question and the picture

    2. Relevant equations
    Pe= E x Is
    Is = V/R

    3. The attempt at a solution
    I figured that in order to find the power delivered by the source, I would have to use the equation: Pe= E x Is.
    E is already given to me.. 60.
    That means I just have to find Is. I used the formula Is= V/R and applied it to each resistor to find the current. Here is what I got:
    I1= 12A
    I2= 30 A
    I3= 7.5 A
    I4= 3 A
    I5= 5 A
    I added these values up to find Is and got 57.5 A. Now I apply the formula Pe= E x Is (Pe= 60 x 57.5) and got 3450 Watts for the power delivered by the battery.

    However.. The correct answer is 1.56 kW

    Am I using the wrong formula completely? Where did I go wrong?
    Thanks!
     

    Attached Files:

  2. jcsd
  3. Jul 3, 2016 #2

    nsaspook

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    Use ohms law to check your resistance to current for voltage calculations.
     
  4. Jul 3, 2016 #3

    The Electrician

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    You are calculating the current through the 2 and 8 ohm resistors as though they were each separately across the 60V battery. Actually the 2 and 8 ohm resistors are in series and the two of them could be replaced by a single 10 ohm resistor across the battery.
     
  5. Jul 4, 2016 #4
    I checked the math, and everything checked out. Am I going about the problem in the right way?
     
  6. Jul 4, 2016 #5
    I tried replacing the 8 and 2 ohms with a 10 ohm resistor. After I added them up, I got Is= 23. I threw that into the formula.. 60 (23) and got 1380 watts.
    Still didn't work out.

    Is that what you meant to do?
     
  7. Jul 4, 2016 #6

    The Electrician

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    Current in:

    5 ohm = 12A
    10 ohm = 6A
    20 ohm = 3A
    12 ohm = 5A

    12+6+3+5 = 26A

    60V*26A = 1560 watts
     
  8. Jul 4, 2016 #7
    I see now.. But how do you know to combine the 2 and 8 ohm resistors? Arent the 5 ohm and 2 ohm resistors in series?
     
    Last edited: Jul 4, 2016
  9. Jul 4, 2016 #8

    The Electrician

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    Perhaps you are thinking that the wire btween the 2 and 8 ohm resistors is connected to the wire between the + terminal of the battery and the top of the 20 ohm resistor. The 2 and 8 ohm resistors are in series only and the wire between them doesn't connect to anything else.

    What 1 ohm resistor are you talking about? I don't see any such resistor.
     
  10. Jul 4, 2016 #9
    I apologize, I meant the 5 ohm resistor.
     
  11. Jul 4, 2016 #10

    The Electrician

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    The left end of the 5 ohm resistor is connected to ground; the negative terminal of the battery is also connected to ground. Therefore the 5 ohm resistor is connected across the battery. The 2 and 8 ohm resistor are connected in series across the battery, but with respect to the battery, the 5 and 2 ohm resistors are not connected in series.
     
  12. Jul 4, 2016 #11
    Aaaahhhh.. Makes sense now. Thanks alot!
     
  13. Jul 4, 2016 #12

    The Electrician

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    This circuit is shown in such a way as to trick you. I usually helps to redraw the circuit. Then you can more clearly see how things are hooked up:

    Trick.jpg
     
  14. Jul 4, 2016 #13
    Alrighty! Ill keep that in mind. It's much easier to see that way.
     
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