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!

Series circuit problem

  1. May 11, 2015 #1
    1. The problem statement, all variables and given/known data
    Find the current for the next circuit. Consider the values given:
    V1= 2 V, V2=3V, V3= 5V, R5= 4Ω Y R4= 2Ω

    2. Relevant equations
    RT=R1+R2+R3+Rn
    I=V/RT
    P=VI=V^2/R
    3. The attempt at a solution
    My attempt of solutions is to sum the first two voltages and divide it between the first resistance. My background context is that I only know very few about this topic.
     

    Attached Files:

  2. jcsd
  3. May 11, 2015 #2

    gneill

    User Avatar

    Staff: Mentor

    I suggest that you review Kirchhoff's laws. KVL would be very useful here.
     
  4. May 12, 2015 #3
    According to the Kirchoff's law of voltages:
    RT=6 ohms
    VT=?
    -2+3-2I+5-4I=0
    6+-6I=0
    6I=6
    I=6/6=1A
    is the result right? I have some doubts in the signs of the voltages.
     
    Last edited: May 12, 2015
  5. May 12, 2015 #4

    gneill

    User Avatar

    Staff: Mentor

    Definitely sign problems. Pay attention to the polarities of the voltage sources as you do your "KVL walk" around the loop in the direction of the current.
     
  6. May 12, 2015 #5
    VT=2-3+5=4V
    it's correct?
     
  7. May 12, 2015 #6

    gneill

    User Avatar

    Staff: Mentor

    No. Check the polarities of the voltage sources as you go around the loop. Suppose you "walk" clockwise around the loop starting at the bottom of V1. Is there a potential rise or drop as you pass through each source? Do they have different polarity orientations with respect to your "walk"?
     
  8. May 12, 2015 #7
    The first one is=-2
    The second has the same polarity, and it has a rise of potential
    For the third, it has the same polarity and it has a rise of potential.
    that's correct?
     
  9. May 12, 2015 #8

    gneill

    User Avatar

    Staff: Mentor

    Why is the first one -2? If all the polarities are the same, shouldn't they all have the same sign? Are you summing potential rises or potential drops? (Some prefer to sum drops as positive, rises as negative, which is perfectly fine as long as one is consistent about it)
     
  10. May 12, 2015 #9
    Yes, I didn´t see the sign
    So it would be VT=2+3+5=10V
     
  11. May 12, 2015 #10

    gneill

    User Avatar

    Staff: Mentor

    Okay, you should now be able to write out KVL for the loop, including the drops associated with the resistances.
     
  12. May 12, 2015 #11
    10V-(6 ohms*I)=0
    10V=6 ohms*I
    I=10V/6 Ohms
    I=1.66 A
     
  13. May 12, 2015 #12

    gneill

    User Avatar

    Staff: Mentor

    Yup. That's better!
     
  14. May 12, 2015 #13
    Ok, thank you so much for the help.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted