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Find the current and Voltage of the following pieces

  1. Jan 23, 2017 #1
    1. The problem statement, all variables and given/known data
    upload_2017-1-23_20-34-10.png

    Above is the circuit I have to deal with. We are given the Vin = 6V, Vout = 3.367 V and all the R values. We are asked to find the remaining values (The I's and V's)

    2. Relevant equations

    Kirchhoff Voltage and current laws
    V = IR
    R's in series add
    Voltage from a source has the same value on a resistor if they are in parallel

    3. The attempt at a solution
    At the moment I did not really start plugging in numbers, but i tried thinking of how to get certain values:

    I2: looks like V on it is 6 since parallel to V in so 6/20000
    I5: same procedure as stated for ^
    V3: Use a ratio between the two resistors in that node line to see how much voltage goes to that as opposed to R4
    V1: confused because when I use the KVL i get Vin - I2R2 + V1 = 0 but the first 2 arguments in that phrase would already = 0 unless I2 is not as simple as stated above.
    V6 : completely unsure

    ** Also to add, we have not learned current or voltage division yet and I feel like that may play a role because I am not sure how to find Req of this nor understand what Vout does to help.
     

    Attached Files:

  2. jcsd
  3. Jan 23, 2017 #2
    How can that be unless there is no voltage drop across R1?

    Edit: Clarification: Even if there was no voltage drop across R1, it would still not be true that R2 is in parallel with Vin, if that is what you were saying.

    Edit 2: Is there any current flowing through R6? If so, where does it go?
     
    Last edited: Jan 23, 2017
  4. Jan 23, 2017 #3
    Just reading that alone makes complete sense. I was just reading in my notes that voltage in parallel remains the same but I'm seeing that does not apply to this case. I feel like my issue is that I can't really figure out where to properly start? I can not directly find V1 since I do not know the current coming from the 6v source and I can't figure out how to get that without finding Req
     
  5. Jan 23, 2017 #4
    Consider my "Edit 2" above where I ask about current flowing through R6.
     
  6. Jan 23, 2017 #5
    Sorry if I seem uneducated in this topic. I know it's not an excuse but we get prelab assignments due before the actual lectures and we don't have text I can really learn from. To answer your question, I believe there is a current that flows through R6 and out to the output?

    Edit: this is the first time I have ever seen a Vout
     
  7. Jan 23, 2017 #6
    Okay, that is not an unreasonable answer. But what is at the output that will allow current to flow? Is there anything?
     
  8. Jan 23, 2017 #7
    Nope, the circuit is broken at that point, so despite there being some Vout, there is no current at the output?
     
  9. Jan 23, 2017 #8
    That's how I saw it too - well, unless we consider that there was a voltmeter to measure the voltage. But that voltmeter will have a very high resistance and therefore, negligible current. So if we know (at least we think we know) that no current flows through R6, then what is the voltage at the other side of R6?
     
  10. Jan 23, 2017 #9
    Hmmm. My intuition is telling me 3.367 V since that Vout value must come from somewhere and there is no current between the resistor and the output
     
  11. Jan 23, 2017 #10

    NascentOxygen

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    You have an impressive 4-significant-figure intuition? Where did this value spring from?
     
  12. Jan 23, 2017 #11

    ehild

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    No, it is not sure that the output current is zero. You can check easily: in this case the voltage across R2, R3+R4, and R5 is Vout, and the sum of the currents through the resistors is equal to I1. Also,
    Vout + I1 R1 = Vin. Is it true?
     
  13. Jan 23, 2017 #12

    ehild

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    Vout was given as 3.367 V.
     
  14. Jan 23, 2017 #13

    NascentOxygen

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    Ah, that simplifies things!
     
  15. Jan 23, 2017 #14
    @ehild, you are absolutely correct. I retract everything I previously said. Thank you for correcting me.
    @itzernie, please take note of ehild's post.
     
  16. Jan 24, 2017 #15

    ehild

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    Check: The current through R6 might be zero with the data given!
     
  17. Jan 24, 2017 #16
    Well, I tried to check a while ago (when I first read your post) and it looked like the currents did not match. But now, since you brought it up, and since I have so little confidence in my abilities right at the moment . . .
    I1 = (6-3.367)/2k = 1.3165 mA
    I2 = 3.367/20k = 0.16835 mA
    I4 = 3.367/(2.2k + 4.7k) = 0.48797 mA
    I5 = 3.367/5.1k = 0.66020 mA
    I2 + I4 + I5 = 0.16835 + 0.48797 + 0.66020 = 1.31652 mA

    So it appears that the currents do match. However, due to my excessive fumbling through this problem, I have very little confidence that my calculations here are correct. If someone offered me a bet where they would pay me $100 if my calculations are correct, and I would have to pay them one donut if my calculations are wrong, I would probably have to consider that a bad risk for me.

    I think it is bedtime. Good night Mary Ellen. Good night John boy. Good night Jim Bob.
     
  18. Jan 24, 2017 #17

    ehild

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    Yes, they do :) Good night.
     
  19. Jan 24, 2017 #18
    Sorry for the late response! I it was 4am on my last post so I passed out lol. Anyway @ehild what you said in the way specific R's add up to Vout and how I1 is the result of all currents from the other resistors makes sense to me. Thanks ! @TomHart I will work on the problem asap and let you know if my answers are close enough to where you can take the donut risk.
     
  20. Jan 24, 2017 #19
    OK so I sat down and really went over what was talked about here. I fully understand 2 of the 3 statements that @ehild made.

    I'm just confused with how did you know the sum of the current on the resistors R2-R5 had to equal the I1 current? Is it because If you were to simplify all of those resistors together into one you would have a junction up top with 3 directions to go. Left is where I1 comes from, right has no current , and down is the combined resistor where Isum must go. Therefore by KCL we know at a junction the current sum must = 0 and therefore the Isum of R = I1 ?

    I can see why Vout + I1 R1 = Vin. - this alone can help find I1 and V1
    Also I see why R2 through R5 have the same Voltage as Vout.

    Using that statement alone helped me find I2,I4,V3,I5.
    V6 is in this case 0V.
    I believe that is all of the variables (and @TomHart, I basically computed the same numbers so you can make the bet hopefully)
     
  21. Jan 24, 2017 #20

    ehild

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    It would not be true if there was current through R6.
    You have a two-port circuit, with some input and output voltages and currents. Usually such circuit is part of a bigger one, so the current can flow out of it. You would be able to determine the currents with the information given, and the result is Iout = 0 if Vout=3.367 V. With an other output voltage this current would not be zero.
    Yes, assuming that the current through R6 is zero, the current through R1 is the same as the current through the combined resistors.

    It is correct.
    Can you write the equations for arbitrary Vout and Iout?
     
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