Superposition Question Help

Discussion in 'Homework Help' started by Ronny2412, Sep 3, 2014.

  1. Ronny2412

    Thread Starter New Member

    Sep 2, 2014
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    Hello!

    So, this is a question I received for my Homework & I decided to tackle it using Superposition Theorem.

    1)
    Firstly they have asked to find out Voltage at B with respect to GND.
    In this case, 1st I ignore V2
    Considering V1, I calculate the Voltage @ B wrt GND which turns out to be I*(R1+R2)
    To Calculate I: I take V1/(R1+R2+R3+R4)

    I then repeat the above step ignoring V1.

    Finally Vb wrt GND= V1st step-V2nd step.

    PLEASE LET ME KNOW IF THIS IS CORRECT

    2)
    Secondly they ask me to find out Voltage between A&B
    I ignore V2.
    Considering V1:
    I used Voltage divider rule to get Vab= V1*(R2/R).

    I then ignore V1 and repeat the step.

    Finally Vab=V1st step-V2nd step.


    PLEASE LET ME KNOW IF THIS IS CORRECT.


    Thank you
     
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    I * (R1 + R2) is equal to a voltage drop across R1 + R2, but this is not Vb voltage.
    In this step V1step = I * (R? + R?)

    OK
    Why you put a "minus" sign ?

    Again why "minus" sign. And are your sure that you need all this steps ?
    In part A you already find the current, so Vab = I * R2
     
  3. Ronny2412

    Thread Starter New Member

    Sep 2, 2014
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    Dear Jony,

    Thank you so much for your time.
    Can you please explain to me where I am going wrong when I calculate Vb wrt GND?
     
  4. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    In step Vb1 you replace V2 with a short circuit. So Vb = I*(R3 +R4) or Vb = V1 - I*(R1+R2).
     
  5. anhnha

    Active Member

    Apr 19, 2012
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    It is OK to use superposition method here but honestly I don't like that approach much!
    You just need to find the current I = (V2 - V1)/(R1 + R2 + R3 + R4) and then everything is simple to calculate.
     
  6. Ronny2412

    Thread Starter New Member

    Sep 2, 2014
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    I meant we short circuit V2. .
     
  7. Ronny2412

    Thread Starter New Member

    Sep 2, 2014
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    Yes. I got that. I calculated Vb wrt ground across R3 & R4. I used Voltage Divider Rule to arrive to the formula:

    Vb= V1* (R3+R4)/(R1+R2+R3+R4).

    Is this correct?

    Also, we did this by shorting V2. Shouldn't we then Short V1 & do the whole process again with V2?
     
  8. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    Here are the steps: Step 1 Set V1=0V, and solve.

    69a.jpg

    Step 2: Set V2=0V and solve.

    69b.jpg

    Now add V(b) in step 1 to V(b) in step 2 to get 4.2+3.6 = 7.8V

    Here is the full solution, all at once, with V1 and V2 as specified:

    69c.jpg

    Note that V(b) = 7.8V, as expected.
     
  9. Ronny2412

    Thread Starter New Member

    Sep 2, 2014
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    yeap! I calculated Vb using V2 over R1+R2 only!
    and Finally I do Vb2-Vb1. .That is correct right?
     
  10. Ronny2412

    Thread Starter New Member

    Sep 2, 2014
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    Oh yes. I get it. Since both voltages are in the same direction, we add Vb2 & Vb1. OK. Thanks all of you :)
     
  11. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    No, this is incorrect.
    Vb1 + Vb2
     
  12. Ronny2412

    Thread Starter New Member

    Sep 2, 2014
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    Thanks all of you :). .I have understood the concept perfectly.
     
  13. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    I appologize. I was mixing Node-Voltage and Superpositon. My mistake. I finished doing Node-Voltage and I got Vb=7.8 volts like Mike showed.
     
  14. Ronny2412

    Thread Starter New Member

    Sep 2, 2014
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    Guys, to find out Vab. .I need to do Vab1 (shorting V2) - Vab2 (Shorting V1) right?
     
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