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: Need help with a circuit problem? (Picture attached)

  1. Dec 14, 2011 #1
    **PICTURE ATTACHED AT BOTTOM**

    1. The problem statement, all variables and given/known data
    The current in the 13.8-Ω resistor in the attached picture is .795 A. Find the current in the other resistors in the circuit.

    2. Relevant equations
    V=RI

    In a series circuit:
    V of Circuit = V1 + V2 + V3 +...Vn
    I of Circuit = I1 = I2 = I3 = ...In
    R of Circuit = R1 + R2 + R3 +... Rn

    In a parallel circuit:
    V of Circuit = V1 = V2 = V3 = ... Vn
    I of Circuit = I1 + I2 + I3 +... In
    1/R of Circuit = 1/(R1) + 1/(R2) + 1/(R3) +... 1/(Rn)

    3. The attempt at a solution

    For the 13.8-Ω resistor, I found that Voltage = 10.971 by using V=IR
    - Because V of Circuit = (V1 = V2 = V3 = ... Vn), the Voltage for the 17.2-Ω resistor is also 10.971
    --Based on this, the Current for the 17.2-Ω resistor is .638:
    V=IR
    I=V/R
    I=(10.971)/(17.2)
    I=.638

    After this, I proceeded to find the equivalent resistance for the entire circuit.

    To find this, I separated the circuits into groups.

    Group 1= the 15-Ω and 12.5-Ω resistors (they are in series).
    Group 2= the 13.8-Ω and 17.2-Ω resistors (they are parallel).
    Group 3= the 8.45-Ω and 4.11-Ω resistors (they are in series).

    When solved for, the equivalent resistance for each group is as follows:

    Group 1= 27.5-Ω
    Group 2= 7.66-Ω
    Group 3= 12.56-Ω

    Note that Groups 2 and 3 are now parallel. We will combine them into Group 4.

    Group 4 Resistance = 4.75-Ω

    Now we only have Groups 1 and 4 left in series. When combined, we will get the net resistance for the entire Circuit. This works out to be 32.56-Ω



    .....So that's everything I could find, which is decent, but doesn't come anywhere near answering the question. I know I must be missing something somewhere.

    How do I figure out the individual currents for each resistor with the information I currently have?
     

    Attached Files:

  2. jcsd
  3. Dec 14, 2011 #2

    gneill

    User Avatar

    Staff: Mentor

    You're doing so well using the voltage across the 13.8Ω resistor, why not continue? The same voltage must be across the series connected 8.45Ω + 4.11Ω resistors. So what's the current through them?

    What then is the total current for the subcircuit?

    attachment.php?attachmentid=41965&stc=1&d=1323922442.gif
     

    Attached Files:

    • Fig1.gif
      Fig1.gif
      File size:
      5.7 KB
      Views:
      1,044
  4. Dec 14, 2011 #3
    Ah, of course!

    Okay, so based on that...

    Current of the 8.45 Ω resistor should be (10.971)/(8.45) = 1.3 A

    and the 4.11 Ω will be (10.971)/(4.11) = 2.67 A

    So the total for the subcircuit will be

    .795 A
    +.638 A
    +1.3 A
    +2.67 A

    = 5.403 A

    ....
    Would that be correct?
     
  5. Dec 15, 2011 #4

    gneill

    User Avatar

    Staff: Mentor

    No, not quite. The 8.45 and 4.11 resistors are in series. How do you combine series resistances?
     
  6. Dec 15, 2011 #5
    So would it be

    (10.971)/(8.45+4.11) = .873 A

    And then the subcircuit would work out as

    .873 A
    + .795 A
    + .638 A

    = 2.3 A
     
  7. Dec 15, 2011 #6

    gneill

    User Avatar

    Staff: Mentor

    Yessir. That looks good. You should now be in a position to state the current through every resistor in the circuit.
     
  8. Dec 15, 2011 #7
    Alrighty, here are my final results. Everything seems to check out:

    13 Ω = .795 A
    17.2 Ω = .638 A
    8.41 Ω = .87 A
    4.11 Ω = .87 A
    15 Ω = 2.3 A
    12.56 Ω = 2.3 A
     
  9. Dec 15, 2011 #8

    gneill

    User Avatar

    Staff: Mentor

    Yup.
     
  10. Dec 15, 2011 #9
    thanks for all the help.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook