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How to calculate the total resistance in this circuit?

  1. Nov 13, 2015 #1
    1. The problem statement, all variables and given/known data
    How to calculate the value of total resistance RT in the shown circuit?

    2. Relevant equations


    3. The attempt at a solution
    I used Kirchhoff` first law: i1+i2=I where i1, i2 and I are the current passing through R1, R2 and the total current in the circuit, respectively.
    similarly, i4+i5=I,
    i1+i3=i4
    i3+i5=i2
    Then I used Kirchhoff` second law,
    20 i1+10 i4=V0
    10 i2+20 i5=V0.
     

    Attached Files:

  2. jcsd
  3. Nov 13, 2015 #2

    phinds

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    Given that you have not specified any points from which to take the Rth, you have an incomplete problem statement and as such it has no answer.
     
  4. Nov 13, 2015 #3
    I edited the figure after puting labels of currents and points of intersection.
     

    Attached Files:

  5. Nov 13, 2015 #4
    I tried this as well,
    considering the contour that includes i1 and i4
    $$ 20 i_1 + 10 i_4=V_0 $$
    considering the contour that includes i2 and i5
    $$ 10 i_2 + 20 i_5=V_0$$
    multiplying the second equation by 2 and collect term $$(i_1 + i_2)$$.
    $$ 20(i_1 + i_2) + 10 i_4 +i_5=3V_0 $$
    divide on $$ I=i_1+i_2 $$ and consider $$ R_T=\frac{V_0}{I} $$
    $$ 3R_T=20+\frac{10 i_4}{I}+\frac{40 i_5}{I} $$
     
  6. Nov 13, 2015 #5

    phinds

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    And you STILL haven't specified any points on which to base an Rth.
     
  7. Nov 13, 2015 #6
    What do you mean by specific point to base an Rth?
     
  8. Nov 13, 2015 #7

    phinds

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    Do you understand what a Thevenin Equivalent circuit IS? When your question said find Rt, and there is no Rt specified in your diagram, I assumed you meant Rth, the Thevenin Equivalent resistance. Did you mean something else?
     
  9. Nov 13, 2015 #8
    I meant the total resistance of the circuit.
     
  10. Nov 13, 2015 #9

    phinds

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    Based on what ???

    If you mean what is seen by the power supply, you have to say so. If that is what you mean then you would profit by looking up delta-Y transforms.
     
    Last edited: Nov 13, 2015
  11. Nov 13, 2015 #10
    That is great thank you.
    So transforming Δ circuit to Y circuit as in the diagram,
    Transforming the Δ component on the right side of the circuit to Y-component yields,
    $$R_1=\frac{R_a R_c}{R_a+R_b+R_c}$$
    $$R_2=\frac{R_b R_c}{R_a+R_b+R_c}$$
    $$R_3=\frac{R_a R_b}{R_a+R_b+R_c}$$
    so, $$R_1=\frac{10}{4}$$
    $$R_2=5$$
    $$R_3=5$$
    simplifying the circuit now yields,
    the left side of the circuit is a parallel circuit while the right side is series one.
    the resistance of the left side is $$R=\frac{(20+\frac{10}{4})(15)}{(20+\frac{10}{4})+(15)}=9$$
    summing to the right side of the circuit with a R=5 yields 9+5=14.
     

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  12. Nov 13, 2015 #11

    phinds

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    I didn't check your calculations, but that's definitely the right way to do it.
     
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