Confused on superposition

Discussion in 'Homework Help' started by electronicstech07, Nov 1, 2008.

  1. electronicstech07

    Thread Starter Member

    Nov 16, 2007
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    There's are some circuit theorems that's got me confused. I'll make this post the first of many problems that I am confused on. I understand that superposition theorem states that I can find current produced by each source independently in a multiple source circuit by shorting all voltage sources leaving one. If it is a current source I can open one at a time. I then add the current algebraically.

    The problem I am working is asking me to find I through the right most branch. I am not sure what I am doing wrong as my answer does not match what is on the back of the textbook. I have attached my work. It is a bit sloppy, hopefully someone can point out what I am doing wrong. Thanks!
     
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  2. mik3

    Senior Member

    Feb 4, 2008
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  3. vvkannan

    Active Member

    Aug 9, 2008
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    For calculating current due to Vs2 (killing vs1) you have considered (R12+R3) and R4 as series connections.Add R4 to the equivalent resistance between the parallel combination of (R12+R3) and R5 and find the total current.
     
  4. hgmjr

    Moderator

    Jan 28, 2005
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    Perhaps it might help you to rearrange the circuit as I have shown (see attached).

    You can see that in both superposition cases, the equivalent resistance seen by each of the voltage sources is the same. That means that once you calculate the equivalent resistance correctly for one case, you can use the same value of equivalent resistance to calculate the current for the second case.

    hgmjr
     
  5. electronicstech07

    Thread Starter Member

    Nov 16, 2007
    18
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    You're right about that. I didn't pay enough attention to how the resistors were connected.

    So Itotal due by Vs1=1.04 mA

    I entering pt. B = .568 mA

    Using the current divider formula, Ir5 = .189 mA

    It due by Vs2 = 3V/1.908k ohms = 1.57 mA

    I entering pt. B = 1.57 mA

    Using the current divider formula, Ir5 = r12eq + r3/r12eq + r3 + r5 * 1.57 mA

    = .666k + 1k/.666K + 1k + 2k * 1.57 mA = 1.666k/3.666k * 1.57mA = .713 mA

    With the current both entering r5 in the same direction, then .189 mA + .713 mA = .902 mA

    Please let me know how I did and thanks a lot!
     
  6. vvkannan

    Active Member

    Aug 9, 2008
    138
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    yes correct
     
  7. electronicstech07

    Thread Starter Member

    Nov 16, 2007
    18
    0
    Here is a problem in which I am asked to find the Thevenin equivalent as seen by RL. I am able to find Vth by opening terminal AB.

    Using voltage divider formula, Vth = r2/r1+r2+r3 * 25 V = 75/225 ohms * 25 V = 8.333 V

    But when finding Rth by shorting Vs, here is where I need some clarification. Looking from r1, can r1 be considered in // with r2 in series with r3?

    Or looking from r3, can r3 be considered in // with r1 in series with r2?

    The reason I ask is because I am only getting the right answer for Rth when I consider r2 to be in // with r1 in series with r3.

    When calculating, Rth would then be 75 * 150 / 225 ohms = 50 ohms + (r4) 25 ohms = 75 ohms.

    Help is appreciated!
     
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  8. electronicstech07

    Thread Starter Member

    Nov 16, 2007
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    There is another problem that is asking me to find the voltage across r4 using Thevenin's theorem. The thing is I am looking at the circuit and am unsure how to simplify the circuit to equivalent resistance once I open r4. The way the resistors are connected is confusing me. How can I rearrange the circuit so it is easier to look at and simplify it?

    Another problem involves finding Irl using Norton's theorem, but my answer is not matching with the book's answer of .1 mA. I attached my work and would appreciate it if someone can help me.
     
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  9. vvkannan

    Active Member

    Aug 9, 2008
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    Since you have to find thevenin equivalent as seen from AB ,you look from R L.

    You can calculate the equivalent resistance looking from any terminal by assuming flow of current from one terminal to the other
    Assume current flows from A to B.

    Now looking from R l you cannot consider R1 parallel with series combination of R2 and R3 because at the junction of R2 and R3 ,current would flow towards B and not through R3.
    At the junction of R2 ,R1,R4 the current flowing from A has two paths to reach B,through R1-R3-B and R2-B.
    Thus R1 series with R3 and R2 are parallel paths
     
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