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Math Help - find the voltage across

  1. #1
    Super Member bigwave's Avatar
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    Cool find the voltage across

    If a source of 200V is connected across the circut of (see attached image)
    (a) find the voltage across the 8\Omega resister
    (b) find the power dissipated in the  1\Omega resistence.
    Ans. (a) 40V (b) 25W



    find the voltage across-4.1.jpg

    I know that I=\frac{V}{R} but i don't see how they got the answers from this also there is parallel and series resistence in the diagram.... just want to see how they got these answers. thanks ahead.
    Last edited by bigwave; October 12th 2010 at 11:26 AM. Reason: better wording
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  2. #2
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    (a) What is the current through the 1\Omega resistor? I'm not sure I agree with the answer of 40V through the 1\Omega resistor, incidentally.

    (b) For a resistor, P=VI=(IR)I=I^{2}R. I'm not sure I agree with the book's answer here, either.

    Just to be clear, the voltage is applied across ae, correct?
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  3. #3
    Super Member bigwave's Avatar
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    what happens to the voltage when it goes thru this circut

    I see what you mean, however I am new to this so easy to get stuck

    actually I don't know exactly what happens to the voltage when it goes thru this circut, but I am assuming the the current thru

    8\Omega = \frac{200v}{8\Omega} = 25A

    but then we don't get the 40V from this.

    so something has to be happening as the circut goes into the series and parallel configuration. yes it appears that we concerned with the R_{ae}
    Last edited by bigwave; October 12th 2010 at 11:45 AM. Reason: latex
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  4. #4
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    Whoa! I think you changed the OP. That's generally discouraged in this forum, because it makes subsequent posts seem irrelevant. If you need to change the problem, do so in a later post. Generally, don't edit posts that aren't the last post in the thread. So, think of posts not in pencil, but in pen. The last post is drying ink, and all posts before the last post are dry, and can't be changed. (Ok, pretend there's no such thing as a pen erasor!)

    If you want the voltage across the 8\Omega resistor, that'll be the same as across the network of resistors that has the equivalent 8\Omega resistance. Why? Because the termination points are the same.

    If the negative side of the 200V is at e, and the positive side is at a, then the voltage across the 8\Omega resistor is just 200V, the same as the network. Resistors in parallel have the same voltage across them, and resistors in series have the same current through them.
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  5. #5
    Super Member bigwave's Avatar
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    Cool so thats what I thot

    ok, I repent no more OP sins.

    so thats what I thot 200v across the 8\Omega resistor not 40v.

    I'll be posting more ??? as they come.... hope you reply

    I attached another image as an example trying to see how they got 40v

    find the voltage across-1.27.png
    Last edited by bigwave; October 12th 2010 at 12:42 PM.
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    A Plied Mathematician
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    Ok. I'll look for them. Have a good one!
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  7. #7
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    Quote Originally Posted by bigwave View Post
    If a source of 200V is connected across the circut of (see attached image)
    (a) find the voltage across the 8\Omega resister
    (b) find the power dissipated in the  1\Omega resistence.
    Ans. (a) 40V (b) 25W



    Click image for larger version. 

Name:	4.1.jpg 
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ID:	19284

    I know that I=\frac{V}{R} but i don't see how they got the answers from this also there is parallel and series resistence in the diagram.... just want to see how they got these answers. thanks ahead.
    Hi bigwave,

    your point "b" is clearly to the right of the 16 Ohm resistor.
    If the 200V source is placed across ab rather than ae, then you will get the answers quoted,
    if the question is asking for the voltage across the 8 Ohm resistor
    and the power dissipated in the 1 Ohm resistor.

    The effective series resistance seen by the source is then 20 Ohms,
    giving a circuit current of 10 Amps.

    5 Amps will pass through each of the 8 Ohm parallel arms.
    That's 40V across the 8 Ohm resistor and 5V across the 1 Ohm resistor.

    The power dissipated in the 1 Ohm resistor is VI=25 Watts.
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  8. #8
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    Ah, very nice. I was always wondering in the back of my mind what that 16 Ohm resistor was doing, but I successfully repressed it.
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