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Ohm's Law and the Principles of DC Circuits

  1. Jun 10, 2010 #1
    1. The problem statement, all variables and given/known data

    R1 = 9.29 V= _____ I = _______
    R2 = 9.19 V= _____ I = _______
    R3 = 8.45 V= _____ I = _______
    R4 = 4.47 V= _____ I = _______
    R5 = 9.12 V= _____ I = _______
    R6 = 4.60 V= 9.15 v I = 1.99 a

    1/(r1, r2, r3) = (1/R1+R2) + (1/R3) = 5.7986 equivalent for R1, R2, R3
    1/(r4, r5) = (1/R4) + (1/R5) = 2.9997

    Picture is attached. How do I got about finding these terms:


    2. Relevant equations

    1/(r1, r2 ...) = (1/r1) + (1/r2) ...

    V = IR


    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Jun 10, 2010 #2

    tiny-tim

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    hi tigerwoods99! :smile:
    Yes, those are the correct formulas (I haven't checked the actual numbers). :smile:

    I don't understand what you're asking :confused:

    you have the equivalent resistances, so find the currents, and then find the voltages. :wink:
     
  4. Jun 10, 2010 #3
    I am not sure how to get from equivalent resistances to currents...
     
  5. Jun 10, 2010 #4

    tiny-tim

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    First add all the equivalent resistances to get the total equivalent resistance for the whole circuit.

    That will give you the current, I6, in the undivided part of the circuit.

    Then write down the Kirchhoff's rules equations for the other currents, I12 I3 and I45 (one equation for each junction, and one for each loop) …

    what do you get? :smile:
     
  6. Jun 10, 2010 #5
    Simplify the circuit...
    R4||R5 = 3.00 ohms
    (R1 + R2)||R3 = 5.80 ohms

    Then work back through the circuit using the voltage divider...
    V1/VT=R1/RT
    This gives R4||R5 = 5.97 V (They both have this voltage as they are in parallel)

    To get the currents use the current division equation I1 = IT X R2/RT.

    You then repeat this process for the (R1 + R2)||R3 part of the circuit.
     
  7. Jun 10, 2010 #6
    we havn't learned about capacitors, all we go is like:

    V = IR
    Vt = v1 + v2...
    It = i1 + 12..
    (1/r(x)) = (1/...) + (1/...)

    I got I6 done:

    R1 = 9.29 V= _____ I = _______
    R2 = 9.19 V= _____ I = _______
    R3 = 8.45 V= _____ I = _______
    R4 = 4.47 V= _____ I = _______
    R5 = 9.12 V= _____ I = _______
    R6 = 4.60 V= 9.15 v I = 1.99 a
     
  8. Jun 10, 2010 #7
    How did you get R4||R5 = 5.97 volts?
     
  9. Jun 10, 2010 #8
    Please HElp!!
     
  10. Jun 11, 2010 #9
    Here's how to do it using only V=IR and series and parallel resistors.

    Combine R4 and R5 in parallel
    1/Rt=1/R1 + 1/R2
    This gives Rt=3.00 ohms

    Now add R1 + R2 in series = 18.48 ohms
    Then combine that with R3 in parallel to give 5.80 ohms.

    [PLAIN]http://www.xphysics.co.uk/x/E1.jpg [Broken]

    This is a series circuit so current will stay constant.

    Finding the voltage across the 3 ohm resistor V=IR =1.99 x 3 = 5.97 V

    Now expand that part back out...

    [PLAIN]http://www.xphysics.co.uk/x/E2.jpg [Broken]

    Clearly V4=5.97 V and also V5 = 5.97 V.

    The currents are I = V/R = 5.97/4.47 = 1.34 A for I4.
    And similarly I5 = 0.65 A

    Now go to the top part of the circuit and do the same process.

    [PLAIN]http://www.xphysics.co.uk/x/E3.jpg [Broken]

    V3=11.54 V, I3=1.37 A

    Then for the little series part...

    [PLAIN]http://www.xphysics.co.uk/x/E4.jpg [Broken]

    V1=5.76 V, V2=5.74 V, I1=I2=0.62 A

    Note that my current direction is "electron flow" and I'm using European resistor symbols.

    There are lots of different ways of doing this sort of problem but this is a simple approach even if a bit long winded.

    Hope this helps!
     
    Last edited by a moderator: May 4, 2017
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