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Electricity Help Needed.

  1. Feb 24, 2009 #1
    Im stuck on this question.

    There is a image of a network. P-Q-R

    P goes to Q, and then splits into 2 routes into R.
    On each route there is a resistor.
    The resistor P-Q is 5 ohms, and the resistors on the 2 different routes Q-R are 3 ohms, and the other is 6 ohms.
    The p.d. across PR is 14 volts.
    Calculate

    a) the equivalent resistance of network QR.
    Wouldnt this just be 9ohms. 6+3?

    b) the current flowing through PQ.
    i did 14/5 = 2.8A

    c) the p.d across PQ.
    really confused, why is it not just 14V

    d) the current through the 3 ohm resistor.
    i did 7/3 = 2.3A

    I used the equation R = V/I
     
  2. jcsd
  3. Feb 24, 2009 #2

    Tom Mattson

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    No. You have to consider how the resistors between Q and R are connected.

    No. 14V is the potential difference across PR, not PQ.

    Because that would mean that the potential difference across QR is zero, and hence no current flows through the entire network.

    Where does 7/3 come from?
     
  4. Feb 24, 2009 #3
    No idea. Im really confused.
     
  5. Feb 24, 2009 #4

    Tom Mattson

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    OK, start with a. How are those two resistors connected?
     
  6. Feb 24, 2009 #5
    Its hard to explain. P is connected to Q with a resistor in the middle. then Q goes in two different routes to R. One resistor on each route. One resistor is 3ohms, then other is 6ohms. and the resistor between P and Q is 5 ohms.
     

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  7. Feb 24, 2009 #6

    Tom Mattson

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    No it's not, you've got 2 choices:

    1.) They are connected in series.

    2.) They are connected in parallel.

    Can you tell me which?
     
  8. Feb 24, 2009 #7
    the resistor connecting P and Q is in series.
    The two resistors connecting Q and R are in parallel.
     
  9. Feb 24, 2009 #8

    Tom Mattson

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    Right. And how do you compute the equivalent resistance of two resistors in parallel?
     
  10. Feb 24, 2009 #9
    Is the answer 2?
     
  11. Feb 24, 2009 #10

    Tom Mattson

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    Yes. Do you understand why?
     
  12. Feb 24, 2009 #11
    yeah i do. but im still confused with the other questions.
    why is 14V not across PQ as well as the whole thing, it hasnt split into parallel yet?
     
  13. Feb 24, 2009 #12

    Tom Mattson

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    Because that would mean that the potential difference across QR is zero, which means that no current flows through them. But that is impossible because of conservation of charge at junction Q. Part of the 14V is across PQ and part of it is across QR.

    So for part b tell me this: What's the equivalent resistance of the entire network?
     
  14. Feb 24, 2009 #13
    Umm, well i came out with two answers. :S
    I got 1.43. and then i tried another way and got 6.43.
     
  15. Feb 24, 2009 #14

    Tom Mattson

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    I have no idea of what you're thinking unless you show how you got those numbers (both are wrong).
     
  16. Feb 24, 2009 #15
    well on part a. i did 3 x 6 / 3 + 6. so i tried that for part b. by doing 3 x 6 x 5 / 3 + 6 + 5 = 6.43.

    and then i did 1 / (1/6 + 1/5 + 1/3) = 1.43.
     
  17. Feb 24, 2009 #16
    Im not sure how to do it because one is in series and one is in parallel.
     
  18. Feb 24, 2009 #17
    So to work out the current i need the p.d and the resistance.
    I know the resistance of PQ is 5? I just need to work out the p.d?
     
  19. Feb 24, 2009 #18
    7? wild guess.
     
  20. Feb 25, 2009 #19

    Tom Mattson

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    Yes, it's 2 ohms. Why are you guessing? Weren't you taught how to find the equivalent resistance of two resistors in series?

    Now that you have the total resistance of and the potential difference across the network, use Ohm's law to determine the total current passing through the network.
     
  21. Feb 25, 2009 #20
    No i havent been taught it. I found something on a website about it.
    So i just use the p.d the questions gave me. 14 volts?

    So 14/7 = 2A?
     
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