series-parallel CKT

Discussion in 'Homework Help' started by icanwalk, Nov 16, 2006.

  1. icanwalk

    Thread Starter New Member

    Nov 16, 2006
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    I'm trying to find the current in it. This is what I've got so far:

    Voltage applied=15v
    R1=2.5K(series)
    R2=15K(parallel)
    R3=10K(series)
    R4=5K(parallel)
    Rt=16.25k
    It=923micro amps......i think?

    I know the current's constant, but idk....i'm lost for some reason :confused:
     
  2. hgmjr

    Moderator

    Jan 28, 2005
    9,030
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    It would be clearer if you could post a diagram of your circuit.

    hgmjr
     
  3. icanwalk

    Thread Starter New Member

    Nov 16, 2006
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    I tryed to make it as discriptive so maybe you would be able to draw it out and plug the info in
     
  4. EEMajor

    Well-Known Member

    Aug 9, 2006
    67
    4
    I'm sorry, but I do not think your original post is clear at all. For example, you have R2=15K(parallel). Parallel to what? Is it in parallel before R1 or after R1, with R4 or without R4?

    Also, do you mean milliamps? Microamps just seems really small, Instinctively. (is it a little u shaped symbol, or a lower case m)?

    At least use MS Paint to make a primitive schematic, please.

    If you want us to help you, please help us understand what you need!!!!
     
  5. icanwalk

    Thread Starter New Member

    Nov 16, 2006
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    alright sorry about that. Here's this I made in pain ^_^ and yeah by the way, yeah it is microamps

    [​IMG]
     
  6. EEMajor

    Well-Known Member

    Aug 9, 2006
    67
    4
    Thank you so much for drawing it out, that helps TONS!

    Okay, you the circuit is a total of 16250 ohms, so using ohms law to find current, I=E/R you get I=15V/16250 ohms = .9mA or 923 microamps. Good job!

    To do this, first combine the parallel portion by using the formula
    1/((1/R2)+(1/R4)) which makes those two act the same as a single resistor of 3750 ohms. Then you can simply add all the values together to get the total resistance: 2500 + 3750 + 10000 = 16250 ohms total circuit resistance.

    Please ask if this doesn't make sense! I'm happy to help.

    But I think it does, because you got it!
     
  7. EEMajor

    Well-Known Member

    Aug 9, 2006
    67
    4
    I made a mistake in my original answer, but I fixed it there and will mention it here incase you already read that one before I corrected it. I had said it was 900microAmps, but you are correct, it is 923 microAmps. I punched it in wrong in the calculator! Opps.

    Here is the worked out answer on paper:
    [​IMG]
     
  8. EEMajor

    Well-Known Member

    Aug 9, 2006
    67
    4
    Man, how many more posts can I make?

    Here is a challange for you:

    1) Find the voltage on the parallel branch. (The voltage for R2 and R4, it will be the same voltage on both since they are parallel).

    2) Find the current in R2

    3) Find the current in R4.

    NOTE: The values of the current in R2 and R4 will add up to be the total current, 923 microAmps.

    If you can find these values, you have this type of circuit down!

    Post your answers and I will help if needed, but you can do it!
     
  9. icanwalk

    Thread Starter New Member

    Nov 16, 2006
    8
    0
    damn, you're awesome man. I can tell you really like your passion in order for you to take that time to do that. thans abunch. by the way, i'll post up the results for the challenge once i'm done =)
     
  10. Dave

    Retired Moderator

    Nov 17, 2003
    6,960
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    You will be able to verify your answer to EEMajor's challenge by looking at the appropriate weighting of R2 and R4:

    I(R2) = R4/(R2+R4)*I(tot)

    and

    I(R4) = R2/(R2+R4)*I(tot)

    You should notice that the bigger resistance in the parallel branch leads to a smaller current through that branch.

    Dave
     
  11. icanwalk

    Thread Starter New Member

    Nov 16, 2006
    8
    0
    alright, i guess i don't got these kind of circuits down cause i can't seem to get this. =/
     
  12. JoeJester

    AAC Fanatic!

    Apr 26, 2005
    3,280
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    Where are you having problems? did you follow EEMajor's explaination?
     
  13. icanwalk

    Thread Starter New Member

    Nov 16, 2006
    8
    0
    well i don't even know what the curren going through them is....i know it's not the same.....right?
     
  14. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    In the series portion ... the current is the same. In the parallel portion, it's different for each resistor ... but it can not exceed the resistance in the series portion.

    Look at what EEMajor did when he "combined" the two parallel resistors to be one "equvalent" resistor having the "same current" as the other two in the series circuit.

    He gave you the hint about the voltage being the same across each resistor, so using Ohms law, you should be able to manipulate the numbers to get the current through each resistor.
     
  15. EEMajor

    Well-Known Member

    Aug 9, 2006
    67
    4
    Hi again,
    I think JoeJester meant "current" instead of "resistance" in his first sentance in post #14.

    Like JoeJester said, look at "Step 2" in my drawing. (I'm glad you can read my awefull handwriting!)

    Calculate the voltage for the "combined" value of the parallel resistances of 3750 ohms. You know the resistance (3750 ohms) and you know the current (.000923 Amps) so you can find the voltage pretty easily.

    Now, you know the voltage for the parallel branch. Use this voltage, and the original resistances of R2 and R4 to calculate each current. Thats it! You got it!

    Good Luck!


    Thanks for jumping in JoeJester!
     
  16. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    Good catch EEMajor.

    It should have read ... but can not exceed the current in the series portion.
     
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