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R and I of DC circuit

  1. Apr 29, 2014 #1
    1. The problem statement, all variables and given/known data
    http://postimg.org/image/8oweasluj/
    Each rectangle in the figures represents a resistor with resistance=2.9kΩ.

    a) What is the net resistance of the circuit of Fig 1 connected to the battery? Each resistance has R=2.9 kΩ.

    b) What is the current flowing into the junction c and the branch of ca from the battery?

    c) If we replace the branch of bc with a 12V battery as Fig 2 shows and Vb>Vc, what is the current flowing into the branch of CA?

    3. The attempt at a solution

    a) Is the circuit in Fig 1 equal to that of Fig 3 as below?
    http://postimg.org/image/t5cwrllzt/

    Thank you.
     

    Attached Files:

  2. jcsd
  3. Apr 29, 2014 #2
    Yes, you can take it and solve it
     
  4. Apr 29, 2014 #3
    Thank you so much!! :D
    I know part a then, just apply the parallel and series equations.

    But for part b, does it mean that first I have to calculate the total current =It,
    then apply Ic=It*(Rt-Rc/Rt), where Rt represents the total resistance?
     
  5. Apr 29, 2014 #4
    No, just apply ohm's law
     
  6. Apr 29, 2014 #5
    V=ir. R means effective resistance
    "i"means total current through the cell.
    And then divide current into the branches of circuit in corresponding ratios.
     
  7. Apr 29, 2014 #6
    I see! I understand now :D Thank you so much!!
    How about part c then?
    Should I use Kirchhoff's Rule? But can Kirchhoff's Rule be applied if there's more than two loops?
     
  8. Apr 29, 2014 #7
    And in part b, the current flowing into junction c and branch ca would be the same then?
    (Since the ratio of resistance are the same?)
     
  9. Apr 29, 2014 #8
    For part b, can I rather use V=IR to find out the current through branches directly after finding out the total current?
     
  10. Apr 29, 2014 #9
    Kirchoff law can be applied separately to any number of loops and solve then separately to find currents
     
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