Finding the voltage of Vs

Discussion in 'General Electronics Chat' started by Majorwiree, Oct 28, 2017.

  1. Majorwiree

    Thread Starter New Member

    Oct 13, 2017
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    How can i find the voltage of Vs using only ohm's law and kirchhoff, i started by finding the current in the 2K resistor which is 1mA, but i dont know what to do next, all help is welcome . THANKS!
     
  2. kubeek

    Expert

    Sep 20, 2005
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    Try finding out the voltage across the 6k resistor.
     
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  3. Majorwiree

    Thread Starter New Member

    Oct 13, 2017
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    I actually im trying but can't find any way to find the voltage of 6K, only if it has the same voltage as the 2K but they aren't in parallel so its not possible right?
     
    Last edited: Oct 28, 2017
  4. kubeek

    Expert

    Sep 20, 2005
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    You know the voltage at the 2k resistor. Between the 6k and the 2k there is clearly a voltage source, wonder what that could do with the voltage at the 6k resistor.
     
  5. Majorwiree

    Thread Starter New Member

    Oct 13, 2017
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    Using KVL -4+Vr2+2=0<=>Vr2 = 2, is that correct? If so i calculated the current going through the resistor wich is 0.333(3)mA, which is lower than the 1mA , whats going on is Vs stealing current?
     
  6. kubeek

    Expert

    Sep 20, 2005
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    That is not correct, you got the minus signs in wrong places. Just look at it with some common sense, if on the right side you got 2V, the voltage source is 4V, and its plus sign is on the left side, how many volts have to be on the left side?
     
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  7. Majorwiree

    Thread Starter New Member

    Oct 13, 2017
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    This is really hurting my brain right now... i've been taught to use the concept that if the current flows from + to - on the source it means it takes the negative sign -Vs+Vr...=0, by what your saying and to make the KVL true the sum of the voltages must be 0... so it would mean the 6k resistor must be -6 to nulify the sum of 4+2?...but on the right side the 2k voltage goes from + to - thats the opposite side the current is flowing from the source so it should be -4+Vr2-2=0 ending in Vr2 = 6
    Im getting super confused with so many stuff interacting and signs and grounds...
     
  8. kubeek

    Expert

    Sep 20, 2005
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    Depends how you draw the current loop for the KVL. If you go from bottom left through top to bottom right, then you have -I*R2 -4V -2V =0
    If you go the other way, then 2V + 4V -I*R2 =0. You should not use symbols like V(r2) because it is ambiguous which way it is oriented and will get you confused, use current times resistance to get the voltage across a resistor.
     
  9. Majorwiree

    Thread Starter New Member

    Oct 13, 2017
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    How should i proced next after knowing V(r2)=6V, found the current wich is 1mA and tried to use KCL to find the current from I(r1) to find the voltage of R1 and use KVL, but not sure if my maths are right I(r2)=4/(6*10^3)=0.666mA only using the voltage from 4v source, and then subtracting 1mA-0.666mA=0.334 and for the voltage 0.334*1.5
     
  10. kubeek

    Expert

    Sep 20, 2005
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    R3 has 2k ohms and 2V across it, that is 1mA. R2 has 6k ohms and 6V across it, both currents are flowing towards ground.
    That means that the sum of those two currents has to be flowing through R1, which makes the drop across R1 xV and which brings the Vs to yV.
     
  11. Majorwiree

    Thread Starter New Member

    Oct 13, 2017
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    i feel like killing myself after this 2 hours on the same stupid exercise, thanks for your patience.....
     
  12. kubeek

    Expert

    Sep 20, 2005
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    1,062
    No problem, you could also draw it up in ltspice for example and measure the circuit, then correct your method of applying kirchhoff equations so that it works.
     
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