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    pop000's Avatar
    pop000 Posts: 352, Reputation: 6
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    #1

    Nov 15, 2011, 12:35 PM
    Find the electrical potential difference
    Given electrical network (in the picture).

    I asked to find the electrical potential difference between A and B. UAB
    I tried and not find the answer.
    I really need help.

    Thanks.
    ma0641's Avatar
    ma0641 Posts: 15,681, Reputation: 1012
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    #2

    Nov 15, 2011, 12:43 PM
    Where is the picture? I don't see one.
    pop000's Avatar
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    #3

    Nov 15, 2011, 12:47 PM
    Oh sorry.

    Here is it
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    ebaines's Avatar
    ebaines Posts: 12,130, Reputation: 1307
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    #4

    Nov 15, 2011, 01:12 PM
    Start by finding the equivalent resistance that can replace the 4-ohm, 3-ohm, and two 6-ohm resistors. What do you get for that resistor? That equaivalent resistor splits the battery voltage in conjunction with the 3 ohm reistor on the left. Post back with what you get for an answer.
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    #5

    Nov 15, 2011, 01:52 PM
    well I find the equivalent resistance that can replace the 6 ohm and the 3 ohm (they are parallel) so I get:
    6*3/(6+3)=2ohm.

    and I get a new electrical network (in the picture).

    my big problem is how to keep from this point, I know there is a law that helps in this kind of problem (I not sure how is call in english), maybe is call a Converting triple star.

    thanks.
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    ebaines's Avatar
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    #6

    Nov 15, 2011, 02:02 PM
    Remember that resistors in parallel don't add - you must use the formula:



    Hence the 3-ohm and 6-ohm resistors in parallel are equivalent to:

    ohms.


    Next step is that this 2-ohm resistor plus the 4-ohm in series adds to 6 ohms, which is in parallel with the remaining 6-ohm resistor.

    Can you take it from here?
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    #7

    Nov 15, 2011, 02:17 PM
    yes I get 6 ohms which is in parallel with the remaining 6-ohm resistor so is equal to 6*6(12)=3ohms
    and this equivalent resistance is series to the last 3 ohms resistor and I get RT=6 ohms

    then I do E=6V/RT so I get 6/6=1v Uab=1v
    do I correct ?
    ebaines's Avatar
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    #8

    Nov 15, 2011, 02:28 PM
    You have RT = 6 ohms which is correct. But voltage divided by resistance is current, not voltage. From Ohm's Law:

    V=IR,
    6 volts = I * 6 ohms,
    I = 1 amp.

    Now you can use use Ohm's Law again to find the voltage drop across the 3-ohm resistor that spans points a and b:

    V = IR = 1 amp x 3 ohms = ? Volts
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    #9

    Nov 15, 2011, 02:39 PM
    Oh yes you correct.
    But I have to ask why u took the 3-ohm resistor voltage that spans points a and b:
    And not the 6 ohms resistor that also spans points a and b ?

    ebaines's Avatar
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    #10

    Nov 15, 2011, 03:01 PM
    Yes - I made an error - sorry! The progression of how we got from your original circuit diagram to two 3-ohm resistors in series is shown in the figure below. The current around the circuit is 1 amp. But only half of that goes through each of the 6 ohm resistor equivalents. So the voltage drop across ab should be V = IR = 1/2 amp x 2 ohms = 1 volt.
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    #11

    Nov 15, 2011, 03:06 PM
    wow you made is clear now :) now I understand it.
    so as u told beofoe V = IR = 1 amp x 3 ohms = 3 volts

    so this is the final answer Uab=3v ?

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    #12

    Nov 15, 2011, 03:07 PM
    *made it
    ebaines's Avatar
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    #13

    Nov 15, 2011, 03:34 PM
    Quote Originally Posted by pop000 View Post
    wow you made is clear now :) now i understand it.
    so as u told beofoe V = IR = 1 amp x 3 ohms = 3 volts

    so this is the final answer Uab=3v ?
    After I made my previous post I realized I had an error, so was in the midst of correcting it when you posted. Sorry for the confusion.
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    #14

    Nov 15, 2011, 03:41 PM
    that OK :) .so you can tell me if the final answer is Uab=3v?

    thanks so much for your help and for your time.
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    #15

    Nov 16, 2011, 09:18 AM
    Quote Originally Posted by pop000 View Post
    that ok :) .so you can tell me if the final answer is Uab=3v?
    No - 1/2 amp of current flowing through a 2 ohm resistor yields? Voltage drop across that resistor.
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    #16

    Nov 17, 2011, 01:01 AM
    Ok now I got it.

    Thank you very much.

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