Solving a Parallel Resistor Circuit problem

Discussion in 'Homework Help' started by Gdrumm, Feb 20, 2009.

  1. Gdrumm

    Thread Starter Distinguished Member

    Aug 29, 2008
    684
    36
    A Newbie here.
    I'm taking a DC Circuits class, and I can't figure out how to solve some homework circuits.
    Here is one for which I have to solve for some missing values.

    It's a Simple Parallel Circuit
    Es (or Et) =?
    R1 = ?
    R2 = 82 K ohms
    R3= ?
    Between Es and R1, I have an "I" value of 1.37mA, with an arrow pointing down and flowing back toward Es
    Alongside R1 is shown another "I" value of 583 micro Amps, with an arrow pointing down and flowing back toward Es.
    Req (or Rt) = 25.55K ohms

    I have solved for Es, and I believe it to be 35V.

    The directional arrows are throwing me off, and I'm not good at algebra, so any help would be greatly appreciated.

    Thanks,
    Gary
     
  2. Gdrumm

    Thread Starter Distinguished Member

    Aug 29, 2008
    684
    36
    Here is a .bmp of the circuit, I'm solving for Es, R1, R3, IR2, and IR3.

    Thanks,
    Gary
     
  3. sstbrg

    Active Member

    Nov 29, 2008
    54
    0
    You can start with It = I1+I2+I3 and get an equation with R3, because It (total current), I1 (current through R1), Es(=It*Req) and R2 are known. You can calculate R3 from here.

    R1 is Es / I1, and you know both Es and I1 now.

    Arrows are showing the direction of the current. It flows from higher potential to the lower.
     
    Last edited: Feb 20, 2009
  4. slepkowski3

    New Member

    Feb 17, 2009
    4
    0
    You know that current flowing in and out of the voltage source is that same, right? Since you are given the equivalent resistance you know the voltage across each of the resistors. Then using the current through R1 you can find R1 using Ohm's law and using voltage loops come up with a system of equations to solve for the two remaining resistances. I guess I'm just restating a lot of what sstbrg just said.
     
  5. Shmurk

    Member

    Aug 13, 2007
    24
    0
    I hope I'm not spoiling too much but here is some help (or so I hope, I'm a newbie...) :

    • Es is easily found with R(EQ) and the current value going back to the battery.
    • It's a parallel circuit, therefore Es is the same in each branch.
    • Find R1 with Ohm's law.
    • The total current is a known value I = 1.37mA, let's say that I = I1 + I2 + I3.
    • Find I2 with Ohm's law.
    • Find I3 from I = I1 + I2 + I3.
    • And finally find R3 with Ohm's law.
     
  6. Gdrumm

    Thread Starter Distinguished Member

    Aug 29, 2008
    684
    36
    So, is I1 = 583 pico amps?
    Thats where I am getting confused.
    I don't recognize the placement of the two given amps inside the circuit.

    What is the 1.37mA? Is it the remaining amps after all has been done?

    Please forgive my ignorance, but my textbook doesn't help me.

    Where exactly do I start?

    It seems like I'm missing a value I should have?

    If I use 1.37mA x 25.55 K ohms, I get 35 volts.
    Does that sound right?

    If so, why is it right, what exactly does the 1.37mA represent? It's not IT.

    Thanks again
     
  7. Shmurk

    Member

    Aug 13, 2007
    24
    0
    I'm a newbie so it may be wrong but:
    I don't think I'll spoil anything by saying that Es is good, 35V :cool: I'm confused too though: according to the attachment you gave, you ALREADY know that I=1.37mA, and I1=583E-6 A (it's 583 micro-amps, not pico-amps). What do you really want now? You CAN get R1 with Es=R1 * I1, is this what you want?
    BTW, "micro" is 1E-6, "nano" is 1E-9, and "pico" is 1E-12.
    "IT" (1.37mA) seems to be the current flowing back "through" the battery, and before it is "split" in the different branches of the circuit (the 3 branches composed by R1, R2, and R3). I1 is the value of the current in the branch composed by R1.
    I don't think you miss something.
    It seems good to me.
    Again, from the picture attached previously, it's the "I(total)" of the circuit.
     
  8. mikeska

    New Member

    Feb 18, 2009
    7
    0
  9. Gdrumm

    Thread Starter Distinguished Member

    Aug 29, 2008
    684
    36
    Ok,
    Thanks for your help.
    I was able to solve it with the understanding that 1.37mA is indeed IT.
    Es=IxR gave me 35V
    R1=E/IR1= 60K ohms
    I2=E/R2 =427 pico Amps
    I3=I at battery return, minus I1 and I2, (it was already stated, I just didn't recognize it) = 1.37 mA
    R3=E/I3 = 97.2 K ohms
    and RT = 25.6 K ohms, derived with calculator, using 1/x key and etc...

    Thanks again, sorry for any confusion I may have caused.
    I just didn't understand the placement of the I1 and I3 values, and the arrows used.

    Gary
     
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