1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Finding the current in uneven parallel circuits

  1. Aug 7, 2007 #1
    1. The problem statement, all variables and given/known data
    This problem is regarding electric circuits. The diagram in the book is a parallel circuit. The voltage of a single battery is 1.5V and there are three resistors. The two resistors that are in parallel are 6 ohms and 12 ohms. The third resistor is 8 ohms and the resistance in the battery is negligible.:


    I apologize for the crude diagram.
    The question I got stuck is asking what the current flowing through the 6 ohm resistor is.

    2. Relevant equations

    The equivalent resistance (Req) of the two-resistor parallel combination is:
    1/Req= 1/6 + 1/12
    Req= 4ohms

    The total current flowing through the battery is:

    3. The attempt at a solution

    My prof actually wrote the answer of the question in my textbook, but I cannot understand how or why he did what he did.

    His answer was thus:

    I (of 6 ohms) = (12/18) x 0.125

    I thought I=V/R, so this does not make sense to me. Can someone please explain why he did this the way he did? (Classes are finished and I cannot get a hold of him before the exam).

    Any help would be greatly appreciated.



    Attached Files:

    Last edited: Aug 7, 2007
  2. jcsd
  3. Aug 7, 2007 #2


    User Avatar
    Homework Helper

    He used the current divider rule. When you have a current that divides into two parallel resistors R1 and R2, then the current through R1 = Itotal * R2/(R1+R2) and the current through R2 = Itotal*R1/(R1+R2). In other words the current through one resistor is the total current times the ratio of the other resistance to the total resistance.

    If you have a current that divides into more than two resistors in parallel for example R1, R2, R3, R4... then to get the current through one you can still use the current divider, but be careful... for example the current through R2 would be Itotal*(R1||R3||R4)/(R2 + R1||R3||R4). So what I did was combine R1, R3 and R4 into one resistance so that I could use the current divider rule.

    The current divider rules and voltage divider rules are derived from Kirchoff's laws. Check your textbook out for more examples. These rules are really useful to solve circuits.
  4. Aug 7, 2007 #3
    Cheers! Of course, it makes so much sense now :)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook