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Solving for elements on a circuit using KCL and KVL

  1. Feb 6, 2013 #1
    1. The problem statement, all variables and given/known data
    calculate the voltages and currents for each element in the circuit? the resistances are as follows R1 = 1 KΩ R2 = 2.4 KΩ R3 = 1.2 KΩ R4 = 1 KΩ R5 = 1.2 KΩ. Vs= 6V


    2. Relevant equations
    Is - I1 = 0
    I1 - I2 - I3 = 0
    I3 - I4 = 0
    I2 + I4 - I5 = 0

    -Vs +V1 +V2 +V5 = 0
    -V2 +V3 +V4 = 0

    3. The attempt at a solution
    Ok so I havent finished this problem yet as I got confused. What I did is this Is=I1 and therefore Vs=IsR1, which I then plug in values for R1=1 K ohm and Vs= 6 volts and solve for Is. I get Is=6mA and I1=6mA so if I1=6mA then the potential drop accross the resistor is V1=6 volts. Then Vs-V1=0 volts which then gives me 0 amps accros I2. The I1-I2=I3 so I3=6mA because I3=I4 then I4=6mA and finally I2+I4-I5=0 using algebra and plugging in I get I5 also equals 6 mA. I know this must be incorrect what am I dooing wrong?
     

    Attached Files:

  2. jcsd
  3. Feb 6, 2013 #2

    gneill

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    Staff: Mentor

    Vs is not across R1 alone, it "sees" the entire resistor network. So your statement (in blue highlight) is not correct. This issue will invalidate the rest of your deductions that depended on it.

    So far you've identified some currents which have several names (Like I1 and Is which are two names for the same current) and written KCL for the junctions. That's fine, but in order to proceed further you'll want to identify some loops and write KVL equations for them.
     
  4. Feb 6, 2013 #3
    Ok so can I use the two KVL equations I mentioned above -Vs+V1+V2+V5=0 and -V2+V3+V4=0 with the first equation defining the first loop which I am calling the square area on the left and the second equation being the right side square.
     
  5. Feb 6, 2013 #4
    I am also considering the current in each loop to be clockwise
     
  6. Feb 6, 2013 #5

    gneill

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    Staff: Mentor

    Yes, fine. Now to link your KVL and KCL equations, replace the individual resistor voltages with the appropriate individual current x resistance expressions. Solve.
     
  7. Feb 6, 2013 #6
    Ugh ok here is what I have so far i1R1+i2R2+i5R5=6. Which by my equations becomes (i2+i3)R1+i2R2+(i2+i4)R5=6. I then plug in R values which gives i2+i3+2.4i2+1.2i2+1.2i4=6. Again using more equations I get it down to 4.6i2+2.2i3=6, then solve for i2=1.3043-.4782i3. This is ok with me except that I have two unknown variables still and cant seem to get rid of it. I could keep solving more of these but I would end up with 2 in each one. What am I doing wrong?
     
  8. Feb 7, 2013 #7
    Ok I think I got after looking at it for a while I figured out i2R2=i3R3+i4R4 which I then applied equations to which gave (1.3043-.4782i3)R2=i3R3+i3R4 then plugging in my R values I get 3.34768i3=3.13032 which gives i3=.93507mA. Then I solve for i2=1.3043-.4782(.93507)=.85714mA. i4=i3=.93507mA so i4=.93507. Then I solve for i1 by using i1=i2+i3=.85714+.93507=1.792mA and because i1=is is=1.792mA as well. i5=i2+i4 which gives i5=.85714+.93507=1.792mA. I then solved for V2=2.057 Volts, V1=1.792 volts, V3=1.122 volts, V4=.9349 volts, and finally V5=2.151 volts. Ok so did i do it right?
     
  9. Feb 7, 2013 #8
    Oh hey can I ask one more question the assignment also says Compute the percentage error in the two measurements and provide a brief explanation for the error. With measured values of Vs=6 volts, V1=1.81 volts, V2=2.1 volts, V3=1.15 volts, V4=.953 volts, V5=1.987 volts, and then currents are ,i1=1.35mA,i2=.877mA,i3=.967mA,i4=.967mA,i5=1.8544mA, and is=1.35mA. So comparing those to my (theoretical calculated values from above). I would sum the measured values then subtract these from the sum of the theoretical values then divide these by the theoretical values and multiply by 100. I would then proceed to use this for current and resistance. Would this work for finding the percent error for all three categories and then I can give my explanation of why this is?
     
  10. Feb 7, 2013 #9
    I forgot to give the R values but I hopefully tou can see what I was getting at. I also thought can I just sum the current,voltage, and resistance all together for the measured values and theoretical values. An the use the equation ((measured values)-(theoretical values))/((Theoretical Values))*100=percent error or should I do them all seperate and get 3 seperate percent error as mentioned above.
     
  11. Feb 7, 2013 #10

    gneill

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    Staff: Mentor

    It looks to me like you've found correct values for all the variables (although your "block of text" presentation style makes it a chore to follow).

    Note that you could make things simpler for yourself by recognizing that serially connected components must all have the same current. A glance at the circuit diagram will let you pick out portions of the circuit where serial components will allow you to pick a single "representative" current for that section, and write down which currents are equal to it:

    attachment.php?attachmentid=55460&stc=1&d=1360242340.gif

    This will help avoid a lot of KCL equation substitutions and manipulations as you write and solve the KVL equations. In the figure, note that ##I_s = I_1 = I_5## , so choose one of them, say I1, to represent all of them. Then when you write your KVL equations you're dealing with just three current variables, and you'll need only one KCL equation (from either junction b or junction d) to relate them.
     

    Attached Files:

  12. Feb 7, 2013 #11

    gneill

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    Staff: Mentor

    It sounds like they're looking for the percentage error in each individual measurement, not sums or averages of errors.

    As for commenting on the errors, consider that you can work out resistance values from the measured values. How do these calculated resistance values compare to the theoretical ones? How would a change in one resistor value affect voltage and current measurements throughout the circuit?
     
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