1. Not finding help here? Sign up for a free 30min 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!

Two EMF circuit help

  1. Dec 5, 2004 #1
    What must the emf E in the figure be in order for the current through the 7.00 - Ohm resistor to be 1.81 A? Each emf source has negligible internal resistance.

    I solved for I through the 3-ohm resistor using the outer loop: 24V - (1.81A)(7-Ohm) - (3 Ohm)(I) = 0 and got that I = 3.77. I don't know where to go from there, though. I tried using the two inner loops separately, as there are two unknowns (I through 2-Ohm resister and E), but they just cancelled out, so I don't know where to go now. Thanks a lot, all help is appreciated
    Josh
     

    Attached Files:

    • emf.jpg
      emf.jpg
      File size:
      9.1 KB
      Views:
      158
  2. jcsd
  3. Dec 5, 2004 #2
    Your in the right direction, use Kirchhoff's junction rule for current. (Sum of the currents entering any junction must equal the sum of the currents leaving that junction) This gives you one equation and two unknowns. Run the inner loop for the second equation.
     
  4. Dec 5, 2004 #3
    Clarification...?

    I tried using that, but I didn't see how to do it, because I don't know the current through the middle section w/the 2-ohm resistor. That being said, how do I make an equation? Would I do something like...man i don't know. I don't know what the current coming out of the 24V EMF is, either...maybe you can help me a bit further in the right direction? thanks
     
  5. Dec 5, 2004 #4
    ok, my fault. I did not see that the voltage was unknown in the center section. You now have three equations with three unknowns. You know the current through the third section. So your unknowns are the current through the middle section, current through the first section and the voltage from the middle bat. Use junction rule and two loops for the equations. (junc rule: i1 = i2 + i3)
     
  6. Dec 5, 2004 #5

    Gokul43201

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Let the junction at the middle of the top section be A.

    You have determined that the current entering A from the left is I = 3.77 A. This current must get split up into the two paths that leave A.

    The path that goes towards the 7 Ohm resistor carries 1.81 A (given). So the rest of the current must travel down the middle path, through the 2 Ohm resistor.

    With this you can find E, from either of the two inner loops.
     
  7. Dec 5, 2004 #6
    O I C, got it now. Thank you for the help guys
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?