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Thevinin' s circuit ANALYSIS

  1. Sep 25, 2004 #1
    Va = -10 V, Vb = 10 V, I (through R2)= 2mA

    R4 = 2*R2 and R3 = 4*R5

    Find the resistances.

    Now, we have just started talking about Thevinin' s method and not yet discused Norton' s method. We have also finished Mesh current, Nodal and Branch methods for circuit analysis.

    I need help on this cause I would say Thevinin' s method is the only one. However, that ground symbol is too confusing for me. IF I pretend it is not there, will that give me the same resutl if it is there?

    Also, how do I approach this question?

    Anything would be really useful,
    James
     

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  2. jcsd
  3. Sep 25, 2004 #2

    ehild

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    You do not need Thevenin's method, and the ground symbol is a big help to solve the problem. It means that the node where R2, R3, R4, and R5 are connected together is at zero potential. The potential at the other end of both R2 and R4 is UA=-10 V, so the voltage across R2 is 10 V, and you know the current through it, just apply Ohm's law to get R2. I hope you will find the way from here...
     
  4. Sep 25, 2004 #3
    " You do not need Thevenin's method, and the ground symbol is a big help to solve the problem. It means that the node where R2, R3, R4, and R5 are connected together is at zero potential. The potential at the other end of both R2 and R4 is UA=-10 V, so the voltage across R2 is 10 V, and you know the current through it, just apply Ohm's law to get R2."

    What do you mean by "the node where R2, R3, R4, and R5 are connected together is at zero potential."? Do you mean the node between R4 and R5 ?

    Also, UA= -10 V but how do you know that it corresponds to the voltage across R2 and not across R4?

    Could I use the mesh current method for each loop?
     
  5. Sep 25, 2004 #4

    ehild

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    I mean the node C, which is common point of four resistors. The node between R4 and R5 here is the same node as it is between R2 and R4. The ends of a "wire" are at the same potential in a circuit diagram.

    UA refers to the ground, and the ground is the zero level of the potential. The voltage UAC = UA-UC , but C is connected to the ground, so the voltage
    UAC=-10 V. R2 and R4 are connected parallel, as they have to terminals in common, one is connected to A, the other is connected to C. Therefore the voltage across both of them is the same, 10 V.

    Notice, that R3 and R5 are also connected parallel and the resultants, R24 and R35 are connected in series together with R1.

    Anyway, you can use also the mesh method if you like.
     
    Last edited: Jun 29, 2010
  6. Sep 26, 2004 #5
    I am going to use your method since it makes sense. So I would have
    (R24 +R35 +R1)*I = V where V is given as 24 V. Wouldn' t this be I1? But then what about UA and Ub? And how do I get the currents through the four resistors in the loops?
     
  7. Sep 27, 2004 #6

    ehild

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    I1 is that what you wrote.

    UA and UB are the voltages with respect to the ground (C). UBC = 10 V, UAC = -10 V, that is UCA = 10 V. You know that the voltages add up, so UBA=UBC+UCA=20 V. From this you see that the voltage across R! must be
    4 V.

    The voltage across both R2 and R4 is 10 V. The current through R2 is I2 = 2 mA, so R2 = 10 V / 2 mA = 5 kohm. R4 = 2*R2 = 10 kohm.
    R2 and R4 are connected in parallel, so the voltage is the same across both resistors, 10 V. The current through R4 is I4=10 V/ 10 kohm = 1 mA. But I2 and I4 flow upward. The nodal low (Kirchhoff's first law) states that the sum of the incoming currents at a node is the same as the sum of the outflowing currents. I3+I5=I2+I4=3 mA, and I1=I2+I4=3 mA.
    The voltage across both R3 and R5 UBC = 10 V.
    The current through the parallel resultant R35 is 3 mA, so R35 = 10/3 kohm. R3 = 4* R5 so the parallel resultant of R3 and R5 is
    R35=0.8*R5 = 10/3 kohm. So R5=25/6 kohm and R3 = 100/6 kohm.
    The voltage across R1 is 4 V, the current is 3 mA, R1=4/3 kohm.

    Check if I haven't made mistakes.

    ehild
     
    Last edited: Jun 29, 2010
  8. Sep 27, 2004 #7
    Thank you . I took a quick glance at it and it seems so easy now. I will try it a bit later on my own. Your explanations were very useful. Thanks, James. :smile:
     
  9. Sep 29, 2004 #8
    Obtain the Thevenin equivalent circuit voltage and resistance for the circuit
    below, considering as output terminals

    1) AC
    2) AB

    Ok now this is a Thevenin circuit question. Rather than making up a new thread I just thought this might be more appropriate. What do I do with the two batteries? If there was one, I could do it PROBABLY but I am lost. If I see how one is done, then I should be able to get something for the second. Anything would really help me.

    James
     

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    Last edited: Sep 29, 2004
  10. Sep 30, 2004 #9
    Please guys anything you can offer would be great
     
  11. Sep 30, 2004 #10
    I am confused by the prescencce of the second battery. Maybe someone could explain its prescence and effect to me.
     
  12. Oct 1, 2004 #11
    Guys, sorry that I uploaded the wrong picture. This is the correct circuit for the Thevenin analysis...sorry forthe confusion. Any feedback would be great.

    Obtain the Thevenin equivalent circuit voltage and resistance for the circuit
    below, considering as output terminals

    1) AC
    2) AB
     

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