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Homework Help: Using Kirchoff's Rules

  1. Feb 27, 2008 #1
    [SOLVED] Using Kirchoff's Rules

    1. The problem statement, all variables and given/known data

    http://img266.imageshack.us/img266/4120/kirchoffho1.gif [Broken]
    In the circuit, I_1 = 1.6 A. Find the value of: (a) the current I_2 and (b) the current I_4.

    2. Relevant equations

    Junction Rule, Loop Rule, V=IR

    3. The attempt at a solution

    Small loop, top left:

    2V + (2ohms)(I_2) - (4ohms)(1.6 A) = 0
    (2 ohms)(I_2) - (6.4) = -2
    2 ohms * I_2 = 4.4
    I_2 = 2.2 A

    That also means that the current in the entire circuit is 4.4 Amps.


    That is as far as I got. Assuming my answer for I_2 is correct, all I have to do is find I_4, which I am stumped on how to do. Should I combine the two resistors on the top left, since they're in series?

    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Feb 27, 2008 #2
    Small loop, top left:(moving anti-clockwise)

    [tex] 2+ 4*1.6-2i_2=0[/tex]
    [tex]i_2 =4.2 A[/tex]

    I don't think any two resistors in this circuit are in series :redface:
  4. Feb 27, 2008 #3
    So that means, I got something that *roughly* looks like the following:

    http://img297.imageshack.us/img297/7287/kirchoffeditck6.gif [Broken]
    (horrible MS paint editting by me)

    So that means total I will be 5.8 (1.6 + 4.2).

    Should I now do a loop rule for the whole circuit, calling the smaller loop Ir (where r is the total resistence of the right loop)? Then just solve for R using r and I?
    Last edited by a moderator: May 3, 2017
  5. Feb 27, 2008 #4
    No I don't think so. The resistances are neither in series nor parallel. You'll have to use KVL.
    The question doesn't ask you to find R. Only [tex]i_2[/tex] and [tex]i_4[/tex]
    Just use loop rule in lower big loop to find [tex]i_4[/tex] .
  6. Feb 27, 2008 #5
    Thanks. Got it!

    12 V - 4.2*2 - 3I_4 = 0

    I_4 = 1.20 A
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