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What is total equivalent resistance and power dissipated?

  1. Dec 17, 2012 #1
    Another one.. I'm sorry but I will likely spend tomorrow and today here with you guys asking for help and mooching solutions. Same deal, badly explained unit. We spent 15 minutes on this. Attached questions as jpg...

    1. The problem statement, all variables and given/known data

    How do I find total equivalent resistance? What does that even mean? And power dissipated? What? :grumpy:

    2. Relevant equations

    No idea...

    3. The attempt at a solution

    I understand sometimes you add resistances and sometimes you add inverse resistances?

    Can somebody please dumb this down into simple terms and try to explain step by step? Obviously I'm willing to work this through on paper with some help, guidance and feedback. I really want to understand the concept behind this.
     

    Attached Files:

  2. jcsd
  3. Dec 17, 2012 #2

    gneill

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    Staff: Mentor

    The first problem ask to find the total equivalent resistance. That would be the net resistance "seen" by the battery. If you "cut" the circuit at the points a and b as shown in the diagram below, what is the net resistance between a and b? You'll want to simplify by combining parallel and serial resistor pairs until you're left with a single value for the total resistance.

    attachment.php?attachmentid=54027&stc=1&d=1355790348.gif

    Let's see how you do on that before moving on to the next part.
     

    Attached Files:

  4. Dec 17, 2012 #3
    Hi Gneill,

    I'm not sure if I did this correctly but here....
     

    Attached Files:

  5. Dec 17, 2012 #4
    Nevermind, it wasn't.. Tried something else like this:

    1/ (143/4300) + 10 + 10 and got 50.06. The correct answer.
    Can you explain why there's a 1 over the parallel?

    So the formula I have is

    1/parallel = 1/R1 + 1/R2 .....

    Using that, I get the whole value for the "parallel" portion of the formula. But to get the correct number I need to divide 1 by the number I got for the "parallel" part.

    :uhh:
     
  6. Dec 17, 2012 #5

    gneill

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    You didn't finish the calculation for the parallel resistance: you left off when you'd added 1/R1 + 1/R2, and failed to take the reciprocal of the sum.

    A handy way to calculate parallel resistances when there are only two involved is ##R_p = \frac{R1\times R2}{R1 + R2}##. The general formula of the reciprocal of the sum of reciprocals boils down to this when there are only two resistors involved.
     
  7. Dec 17, 2012 #6
    I'm just trying to understand why I'm supposed to take the reciprocal of the sum... Why can't it be ready as in: Parallel = 1/R1 +1/R2... :rofl: ?

    And that is an awesome formula. Now I just need to remember it :P
     
  8. Dec 17, 2012 #7

    gneill

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    It comes from the addition of conductances. Conductance is the reciprocal of resistance, so letting G stand for conductance, ##G_T = G_1 + G_2 + G_3 ...## when conductors are in parallel. Since resistance is the reciprocal of conductance, it stands to reason that ##1/R_T = 1/R1 + 1/R2 + 1/R3 ...##
    Yes, it comes in very handy.
     
  9. Dec 17, 2012 #8

    Thank you so much for explaining this. It's easier for me to remember something once I understand why it is that way.
     
  10. Dec 17, 2012 #9
    Oh my god, I completely forgot, power dissipated? What does this mean? How do I find it?
    in the 33 ohm resistor...
     
  11. Dec 17, 2012 #10

    gneill

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    Look in your notes or text for power dissipated by a current flowing through a resistor. You should find several expressions which relate the current through, or the voltage drop across the resistor to the power dissipated.
     
  12. Dec 17, 2012 #11
    I have.....

    V=IR
    P=VI

    12V/33= 0.3636A (I, current).
    12V x 0.3636 = 4.36. Wrong.
     
  13. Dec 17, 2012 #12

    gneill

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    You'll have to isolate the current flowing through the particular resistor, or the voltage across it. It's certainly not the total battery voltage across it nor is it the total current through it, since there are other current paths and potential drops in the circuit.

    You may also find the formulas P = I2R and P = V2/R for the power dissipated in a resistor to be of use as you go along. Here I is the current flowing through the resistor, and V the potential drop across the resistor.
     
  14. Dec 17, 2012 #13
    Thats what I thought but how can I isolate it? Like I said, this unit wasn't clearly explained and we certainly never talked about isolating anything.
     
  15. Dec 17, 2012 #14

    gneill

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    You'll have to apply KVL and/or KCL (Kirchhoff's laws) to analyze the circuit.

    If I may give a hint, you might find it helpful to first find the total current through the equivalent resistance, then treat the circuit like three resistances in series: the two 10 Ω resistors and the parallel resistance you found earlier (the 100 Ω in parallel with the 10 Ω + 33 Ω). The potential drop across the parallel resistance "block" can then be found, and thus the currents through the parallel branches...
     
  16. Dec 17, 2012 #15
    Just a hint?! LOL. I really cannot do this. Didn't learn/see KVL and KCL. How the heck do I find a potential drop? Are you tired yet? :/

    R total = 20.03
    V= 12V

    12/20.03 = 0.599 A total.

    Can you show me a solution?
     
  17. Dec 17, 2012 #16

    gneill

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    That's not the total resistance you found before :confused:
    Better check!

    The potential drop across a resistance is given by Ohm's law: V = I*R.
    We can only provide hints and helpful (?) suggestions, not complete answers. Sorry, them's the rules :smile:
     
  18. Dec 17, 2012 #17
    Mostly helpful. I feel horrible having to interrogate you like this though. Failure at it's most pathetic. You're right. It's 50.06. I was looking at the wrong, partial solution I scribbled on the page.

    V/R=I
    12V/50.06ohms= 0.239 A

    How do I get the 33 isolated?! :grumpy: I suspect having to substract stuff. From other stuff :rofl:
     
  19. Dec 17, 2012 #18

    gneill

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    Ha! Start by finding the potential drop across the "parallel block" resistance, Rp :

    attachment.php?attachmentid=54043&stc=1&d=1355803625.gif

    Once you have that you can narrow your focus to that portion of the circuit; two parallel branches with a known potential difference across them both.
     

    Attached Files:

  20. Dec 17, 2012 #19
    If V=IR and V is given as 12V,

    then we already have the potential drop?

    V/R = I, current
    12/30.06 (for that block) = .399 A

    Or,

    12/43 = 0.279 A
    12/100 = 0.12 A

    for each branch of that block.:uhh:
     
  21. Dec 17, 2012 #20

    gneill

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    Staff: Mentor

    That's the potential drop across the total resistance; it's the battery voltage across the net resistance of 50.06 Ω, yielding the total current ---> I = (12 V)/(50.06 Ω). The "parallel block" Rp will drop only a portion of it.

    This total current I flows through a 10 Ω resistor, then through the Rp, then through another 10 Ω resistor. Since they're in series, they all have the same current passing through them. Each resistance will produce a voltage drop due to the current flowing through it (their sum should be equal to the 12V total).

    You can work out the individual voltage drops across the resistances in this path using Ohm's law: V = I*R. What's the voltage drop across Rp?

    Can you then find the current through the branch of Rp containing the 33 Ω resistor?
     
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