Unknown voltage value

Discussion in 'Homework Help' started by emagdnim47, Mar 5, 2013.

  1. emagdnim47

    Thread Starter New Member

    Jul 16, 2011
    21
    0
    Hi,

    Couple questions. I'm having a hard time figuring out how you'd get the
    applied voltage to these circuit's. My guess is to find the unknown
    voltage first before getting the individual branch current values.

    Once I got the RT. I used the Ohm's law E = I x R

    I ended getting 4.2 V for the circuit for problem number 1), does this even
    come close? Once you have the applied voltage we can use the I1 = E/R
    forumla and get individual values....



    I believe I'm pretty close to solving problem 1), but problem 2) has been
    insane. staring at it for over twenty hours now.



    Your help is GREATLY needed! Thanks in advance








    1) A parallel circuit contains the following resistor values:



    R1= 360 Ω, R2= 470 Ω, R3=300 Ω, R4= 270 Ω, IT= 0.05 A



    Find the following missing values:



    RT= Ω

    I1= A

    I2= A

    I3= A

    I4= A







    2) A parallel circuit contains the following resistor values:



    R1= 270k Ω, R2= 360k Ω, R3=430k Ω, R4= 100k Ω, IT= 0.006 A



    Find the following missing values:



    RT= Ω

    I1= A

    I2= A

    I3= A

    I4= A
     
  2. DerStrom8

    Well-Known Member

    Feb 20, 2011
    2,428
    1,328
    Well I'll just help you out with #2. It's actually fairly simple once you get the hang of it.

    Now, when it says "a parallel circuit contains the following resistor values...." I assume it means each individual resistor is in parallel with each other one. Therefore, you know that the voltage across them must be the same. The first thing you'll need to do is find the total resistance. The formula for parallel resistors is RT = 1/(1/R1 + 1/R2 + 1/R3... + 1/Rn) where Rn is the final resistor value. That formula will give you the total resistance. Since you're given the total current, you can now use Ohm's law (V=IR) to determine the total voltage.

    That takes care of the hard part. Since it's a parallel circuit, the voltage remains the same across each element. Therefore, to find the individual currents, all you have to do is use Ohm's law for each resistor, since you're given the voltage across it and the resistance value. Just divide the total voltage you found in the first part by the resistance of each element to get the current through that resistor.

    You might have to re-read this once or twice just to make sure you get it, but like I said it's still quite simple. I know you can do it. You just have to take it piece by piece.

    Hope this helps!
    Best wishes,
    Matt
     
  3. emagdnim47

    Thread Starter New Member

    Jul 16, 2011
    21
    0
    Hello Matt:)

    I think I figured out the problem 1). I hope someone can check my math, I don't know if any of this is right. Before moving on to problem 2) I need verification my math is on the right track.

    I've attached my work using Math type software, saved it in a pdf format.

    Thank you so much for your guidance here!!

    Alex
     
  4. emagdnim47

    Thread Starter New Member

    Jul 16, 2011
    21
    0
    For the second problem, I'm totally stuck. PLEASE PLEASE HELP!!

    Alex
     
  5. DerStrom8

    Well-Known Member

    Feb 20, 2011
    2,428
    1,328
    I'm afraid I don't have a pdf reader on this machine, but I practically walked you through how to do #2. Reread my first response.
     
  6. WBahn

    Moderator

    Mar 31, 2012
    17,743
    4,795
    Your answer of 4.2V for #1 is correct. As a rule, though, you should report values with 3 sig figs (and use 4 or 5 for intermediate results).

    But you don't need someone to check your math. If you come up with an answer of 4.2V then you can easily determine the individual currents. Add them up and see if you get 50mA.
     
  7. emagdnim47

    Thread Starter New Member

    Jul 16, 2011
    21
    0
    Thanks a ton Matt. The walk through was helpful indeed. I believe I'm on the right track for problem 1 now.

    Cheers!

    Alex
     
  8. emagdnim47

    Thread Starter New Member

    Jul 16, 2011
    21
    0

    Oh thank goodness!! Sig figs okay, i'll have to brush up on that, clearly a weak point of mine. I was really worried the applied voltage was off in some way, if that was off the individual resistor current values would be off as well.

    Fascinating, it added up to 0.05 A exactly!!!

    0.0116A + 0.0089A + 0.014A + 0.0155A = 0.05A

    I'm still having one heck of an issue w/problem two. I know the some steps need to be taken like problem one but somehow I'm lost:eek:
     
    Last edited: Mar 5, 2013
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