Having trouble

Discussion in 'Homework Help' started by arthur112, Dec 15, 2014.

  1. arthur112

    Thread Starter New Member

    Dec 15, 2014
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    There's two questions I need help on, the first one is simple but the diagonal is throwing me off, and the second I just have no idea.

    First one:

    Is the top problem, just need Rth and Vth

    Second one:

    1) voltage drop across the 30 ohm resistor

    2) power dissipated across the 15 ohm resistor

    3) current through the 5 ohm resistor

    4) current through the 15 ohm resistor


    I really appreciate the help, and if you wouldn't mind just including the formula for the second part(s) would be great

    Thanks for taking the time time
     
  2. WBahn

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    Mar 31, 2012
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    So... do you plan to put in ANY effort on YOUR homework problems?
     
  3. arthur112

    Thread Starter New Member

    Dec 15, 2014
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    It's only two of them, out of 25. So yes I do plan to put effort into it. Just got stumped on those two, the first one is easy but the diagonal confused me - and the second one I tired but I'm pretty sure my answers are wrong.

    1) 60v
    2) 327w
    3) 8A
    4) 4.67A

    And the first one

    Rth= 8.63 ohm
    Vth= ?

    Got that much
     
  4. WBahn

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    You need to SHOW your work. That way we can see where you went right and where you went wrong and then try to steer you back onto a valid path.
     
  5. WBahn

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    Also, your picture for the second one doesn't say what the voltage is for the left-most voltage source.
     
  6. arthur112

    Thread Starter New Member

    Dec 15, 2014
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    It's 50v,

    I don't have the work sheet on me now, but the work for the second one is:

    1) 50+20-10=60
    2) 4.67^2 * 15= 327
    3) Vr1/R1 = 40/5= 8
    4) Vr3/R3 = 70/15 = 4.67
     
  7. arthur112

    Thread Starter New Member

    Dec 15, 2014
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    The Rth= ((R1 parallel R2) in parallel with (R4 + R5) ) + R3

    Just don't remember how to find Vth
     
  8. WBahn

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    Always, always, ALWAYS track your units throughout your work.

    Which side of the 30Ω resistor is more positive? If, as you claim, the voltage across the resistor is 60V, then which side of the 30Ω resistor is more positive?

    What are the voltages (relative to the indicated ground reference) of the four nodes, Vo, V1, V2, and V3?

    Observing the indicated polarities, what are the voltages across the three resistors, Vr1, Vr2, and Vr3?
     
  9. arthur112

    Thread Starter New Member

    Dec 15, 2014
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    I'm pretty sure I'm wrong but +50v, +20v, -10v
     
  10. WBahn

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    What's R1, R2, R3, R4, and R5? I don't see any such labels on that circuit.

    Always, always, ALWAYS ask if the answer makes sense. Between points 'a' and 'b' there is a 5Ω resistor and then another path with a bunch more. Isn't the parallel combination of two resistances always less than the resistance of the lowest resistance path individually? If so, then what is the largest value that the resistance between points 'a' and 'b' can be?

    If asked to find the voltage Vab (the voltage at 'a' relative to the voltage at 'b') could you do it?
     
  11. WBahn

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    I asked for four voltages and you gave me three? Which three? In what order? Stop being sloppy and make yourself clear and unambiguous. Say, V0=1.287V, V1=3.98V, V2=.... using whatever values you determined. Also, there's no need to keep saying that you are pretty sure that you are wrong, at least not unless you indicate WHY you think you are wrong.
     
  12. arthur112

    Thread Starter New Member

    Dec 15, 2014
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    R1= 8 R2= 8 (diagonal) R3= 8 (cb) R4= 6 R5= 5
     
  13. arthur112

    Thread Starter New Member

    Dec 15, 2014
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    Vr1= 50-10 = 40
    Vr2= 40+20 =60
    Vr2 = 50+20 = 70
     
  14. WBahn

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    '8' is not a resistance, 8Ω is a resistance. Always track units!

    So let's go back to your description:

    The Rth= ((R1 parallel R2) in parallel with (R4 + R5) ) + R3

    Thank you for using parens to make it clear what your groupings are.

    Explain your reasoning for your expression. For instance, you indicate that R4 is in series with R5. Does this make sense? Remember, we are looking for the effective resistance between 'a' and 'b'. For R4 and R5 to be in series, that means that any path that goes from 'a' through R4 must also go through R5 in order to get to 'b'. Is that true?
     
  15. WBahn

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    Mar 31, 2012
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    UNITS!

    I've said it enough times. I am not going to respond to any further posts in which you don't use units properly.
     
  16. WBahn

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    FYI: I'm going to bed, so don't be alarmed if I don't respond for some time. I have two dives that I'm doing in the morning and a night dive tomorrow night, so I might or might not get back online before tomorrow evening.
     
  17. arthur112

    Thread Starter New Member

    Dec 15, 2014
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    Sorry on my phone, and writing out ohms every time is not productive, plus they are not in kilo form I thought it would be save to assume ohms since it's a R statement.

    As for the first one, since it is thevenins the 5 ohn gets dropped since it's Rl, as well as the 6 ohm, since it doesn't have a connection?
     
  18. WBahn

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    I understand the hassle factor, but tracking units is important (though I admit that I am one of the few people that emphasize it and that I am a self-confessed Units Nazi). You can access the symbol list by clicking on the S at the end of the tool bar above the Reply dialog box, which might make things easier (and then copying and pasting might speed things along, too).

    Most mistakes you make will mess up the units and if you track your units -- religiously -- then you will catch those mistakes almost immediately. Often those mistakes happen during the initial set up of a problem and I can't even begin to count the time I have seen students (and practicing engineers) make such a mistake and, because the units weren't there to be messed up, proceed to spend pages (and hours) of effort chasing an answer that was guaranteed to be wrong from the very beginning and then still not realizing they had wasted all that time and effort once they got their wrong answer.

    I don't see any indication that any of the resistors is supposed to be treated as a load, let alone that a specific resistor is. But I'll certainly allow that it is ambiguous and that I wondered if the 5Ω resistor was supposed to be treated as a load or if you were supposed to find the equivalent for the entire circuit given. I'm guessing that you are supposed to find it for the entire circuit because notice how in the next problem below it the terminals are open circuits which are more suggestive of being the terminals at which a load is connected. As a grader, unless there were other instructions given to clarify things, I would accept the solution either way PROVIDED it was clear what the assumption was, which is best either stated explicitly or via a circuit diagram of the subcircuit that is being used for the equivalent computation.

    In your case, you provided an equation -- namely Rth= ((R1 parallel R2) in parallel with (R4 + R5) ) + R3 -- that used all five resistors so I could only conclude that you found the Thevenin equivalent circuit for the entire circuit given. Now you are saying that you omitted both the 5Ω (R5) and the 6Ω (R4) resistors, so how you only used three resistors; so why are they in the expression you gave for Rth? Also, the expression you gave results in a different answer:

    Rth= ((R1 parallel R2) in parallel with (R4 + R5) ) + R3
    Rth = ((8Ω || 8Ω) || (6Ω+5Ω)) + 8Ω
    Rth = (4Ω || 11Ω) + 8Ω
    Rth = 2.93Ω + 8Ω
    Rth = 10.93Ω

    You need to show your work. Otherwise about all I can do is say that your answer is wrong and not given much help as to why your answer is wrong.

    The first thing you need to do is make it clear whether you are partitioning the circuit into a 'load' and 'everything else'. Then draw the circuit that you are finding Rth for that reflects both removing the load (if any) and zeroing all of the independent sources. And keep in mind that Rth is the resistance seen at the terminals 'a' and 'b'. Even if you remove R5, if you connect an ohmmeter between 'a' and 'b', isn't R4 part of the circuit? In fact, isn't it in series with everything else?
     
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