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!

Current problem with emf and internal resistance

  1. Sep 22, 2012 #1
    1. The problem statement, all variables and given/known data

    Assuming each battery has internal resistance r = 1.0 Ω, determine the magnitudes and directions of the currents through each resistor shown in the following figure. The batteries have emfs of 1 = 9.0 V and 2 = 12.0 V and the resistors have values of R1 = 15 Ω, R2 = 20 Ω, and R3 = 30 Ω.

    Find Current through R1, R2, R3

    PLEASE SEE PICTURE:

    http://www.webassign.net/gianpse3/26-40.gif

    2. Relevant equations

    V=Emf-r*I
    V=I*R

    3. The attempt at a solution

    When two batteries are in a series, their total voltage is the algebraic sum of of their respective voltages.

    For R1: I used V=emf-r*I and set it equal to I*R. My first attempt I added the emf's and plugged into the equation and it didn't work. Then I tried adding (emf-r*I)+(emf-r*I) for both eq'ns and setting equal to I*R and was incorrect.

    I tried this for all three to no avail.

    Since the resistance is given for each, and I'm not adding them, the fact that they are parallel or in a series with each other was not taken into consideration. But I have been getting the feeling that maybe this should have been...

    Any help would be really nice
     
  2. jcsd
  3. Sep 22, 2012 #2

    gneill

    User Avatar

    Staff: Mentor

    The internal resistance of E1 is in series with R1, and the internal resistance of E2 is in series with R3. Why not "sneak" the internal resistances into the values of the corresponding resistors? In other words, add 1Ω to each of R1 and R2 and then proceed.

    There are several possible ways to proceed with analysis including using KVL and KCL equations, node voltage analysis, current mesh equations, or even superpostion. What have you learned about so far?
     
  4. Sep 22, 2012 #3
    Appreciate the reply:

    Do mean "sneak" one ohm in because V=E1-rI=RI? Since r=1, to solve that you add one I to both sides...I don't think that's what you meant, so could you expand?

    This is a undergraduate Physics 2 course. Are KVL eq'n's Kirchhoff's rules? The rest that you named are not familiar.

    I've just learned emf, internal resistance, parallel and series circuits, batteries, etc.
     
  5. Sep 22, 2012 #4

    gneill

    User Avatar

    Staff: Mentor

    I just meant that since those resistances are in series, you can combine them. This will not affect any current values in the circuit (since current is the same for components in series).
    Yes, KVL and KCL are Kirchhoff's Voltage and Current laws respectively. So it looks like you'll be writing KVL and KCL equations for the circuit.
    Okay, why not begin by labeling the currents and identifying where you'll be applying KVL and KCL?
     
  6. Sep 22, 2012 #5
    Alright, I've labeled some values points of reference, and am treating it basically like two closed loops to get equations:

    V=30I3-20I2=0
    V=15I1-20I2=0

    That's 2 eq'ns, three unknowns. I feel like combining R1 and R3 like you referenced earlier ^ might solve that. Am I in the right direction?
     
  7. Sep 22, 2012 #6

    gneill

    User Avatar

    Staff: Mentor

    When you write KVL for a loop, be sure to include any voltage sources (and pay attention to their polarity). Here you have two voltage sources, E1 and E2, one in each loop. It would be helpful to know what directions you picked to assign to the currents.

    Yes, you can add in the source resistances to R1 and R3 if you wish, or write separate terms for them in the equations --- algebraically they will end up added anyways.

    For a third equation you should pick a convenient node in the circuit for which you can write a KCL equation. This will then specify a relationship between the currents I1, I2, and I3.
     
  8. Sep 22, 2012 #7
    No rush on this, I'm in EST zone so this will be my last post for tonight...

    The third eq'n definitely makes sense, and there is an obvious node here to choose. Btw, I have currents running to the outside from each respective voltage source, then looping around through the middle...

    I went ahead and added your suggestions and solved for I1, and still came up wrong. Can you check my methods?

    -30I3+12-20I2=0
    -15I1+9-20I2=0
    I1+I3=I2 {The node I picked was the intersection of the two loops on the side opposite the batteries}

    Solving for I1:
    -30I3+12-20I2=0 gives I3=(15I1+3)/30

    I1+I3=I2 and ^ can be plugged into my second eq'n,

    -15I1+9-20[I1+{15I1+3}/30]=0 gives I1=11/25
     
  9. Sep 22, 2012 #8

    gneill

    User Avatar

    Staff: Mentor

    I don't see terms for the internal resistances (or alternatively, the internal resistances having been added to R1 and R3). Otherwise, your equation methodology looks fine.

    Your labeled circuit probably looks something like this:

    attachment.php?attachmentid=51132&stc=1&d=1348373336.gif
     

    Attached Files:

    • Fig1.gif
      Fig1.gif
      File size:
      1.6 KB
      Views:
      3,412
  10. Sep 23, 2012 #9
    Yes, that's exactly my model.
     
  11. Sep 23, 2012 #10

    Redbelly98

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Okay. Well, let's look at one of the equations you wrote:
    -30*I3 + 12 - 20*I2=0​
    You left out a term to account for the voltage drop across the r=1Ω resistor. (And you also left it out of your other KVL equation.)
     
  12. Sep 23, 2012 #11
    Redbelly, that's true, I added "1" to resitor's 1 and 3 and got the correct answer. Thanks!
     
  13. Sep 23, 2012 #12

    Redbelly98

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    You're welcome, glad it worked out :smile:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Current problem with emf and internal resistance
Loading...