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: Req Calculation. Very basic

  1. Sep 26, 2008 #1
    1. The problem statement, all variables and given/known data
    Calculate i2, given that r1=12, r2=6, r3=12, r4=6.
    There is a 20V source.

    The first step here is calculate Req, I believe.

    http://media.newschoolers.com/uploads/cache/images/1222468444-632813-600x325-1222468266Untitled2423423.jpg [Broken]

    2. Relevant equations
    Req=(r1 ll r2) + (r3 ll r4)

    note: ' ll ' means parallel.

    3. The attempt at a solution

    1/Req = 1/(6+12)+1/(6+12)


    The answer says that Req should be 8. Is my formula for Req correct?
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Sep 26, 2008 #2


    User Avatar
    Homework Helper

    I think you have the right idea but you're getting the math wrong.

    For two resistors in parallel, the equivalent resistance is [tex]\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}[/tex]

    This is NOT the same as [tex]\frac{1}{R_{eq}} = \frac{1}{(R_1+R_2)}[/tex]! (which is what you have done.)

    So take it one step at a time, and solve for the Req of R1 and R2, then for R3 and R4, then the Req of the circuit.
  4. Sep 26, 2008 #3

    The Electrician

    User Avatar
    Gold Member

    Imagine that you remove the wire carrying I2. Assume that the negative side of the voltage source is taken to be a reference (ground). What is the voltage at the junction of R1 and R3? What is the voltage at the junction of R2 and R4? I think that if you calculate those two voltages, you will see that it doesn't take much more calculation to determine I2.
  5. Sep 26, 2008 #4
    Thanks a lot! I got the answer and have a better understanding of circuits!

    By the way thanks for not giving me the answer, haha.

    For those others trying the question the answer is: i2=0.833A
  6. Sep 26, 2008 #5

    The Electrician

    User Avatar
    Gold Member

    Be careful. It looks to me like if you remove the wire carrying I2, the voltage at the junction of R1 and R3 will be 10 volts (with respect to the negative end of the voltage source), and the voltage at the junction of R2 and R4 will also be 10 volts. Do you agree?

    And, if that's the case, what will be the value of I2?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook