Help for 3 DC Source Analysis

Discussion in 'Homework Help' started by LymanT, Feb 24, 2015.

  1. LymanT

    Thread Starter New Member

    Sep 25, 2014
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    Hey guys,
    Any help would be appreciated with this problem I am trying to solve. Lately I have been reviewing a lot of material I've forgotten over the years. With this problem I can't seem to get the right numbers for certain measurements. I was wondering if you guys could lead me in the right direction or provide some math arithmetic to help me solve all voltages and current branches. I keep getting lost in Mesh and tried superposition but I'm messing up somewhere.

    Thanks!
     
  2. crutschow

    Expert

    Mar 14, 2008
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    Sounds like this should be in the homework section.
     
  3. LymanT

    Thread Starter New Member

    Sep 25, 2014
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    Crap.. Could I get a mod to move this to the HW section. Thanks
     
  4. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    Call the voltage at junction of R4, R5, and V2 = V4
    Call the voltage at junction of R1, R2, and R3 = V3

    IR4 = IR5 + IR1 eq1 by KCL.
    IR1 +IR2 = IR3 eq2 by KCL.

    IR4 = E/R = (10 - V4)/2000 by Ohm's Law.
    IR5 = V4/2000
    IR1 = ((V4+5)-V3)/1000
    IR3 = V3/2000
    IR2 = (15-V3)/2000

    Subst into eq 1 and multiply each term by 2000:
    10-V4 = V4 + 2*((V4 + 5) - 2*V3
    2*V3 = 4*V4
    V3 = 2*V4 is equ 3,

    Subst into eq 2 and multiply each term by 2000:
    2*((V4+5) -V3) + (15-V3) = V3
    2*V4 +10 - 2*V3 +15 -V3 = V3
    2*V4 +25 = 4*V3 is equ 4,

    Subst equ 3 into equ 4:
    2*V4 + 25 = 8*V4
    V4=25/6 = 4.16667

    Therefore V3 = 2 * 25/6 = 25/3 = 8.33333

    16.gif
     
    Last edited: Feb 24, 2015
  5. LymanT

    Thread Starter New Member

    Sep 25, 2014
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    @MikeML Thanks bud. I see how you did the math throughout the problem by following, but where I am a little lost is how you developed the

    IR4 = E/R = (10 - V4)/2000 by Ohm's Law.
    IR5 = V4/2000
    IR1 = ((V4+5)-V3)/1000
    IR3 = V3/2000
    IR2 = (15-V3)/2000

    equations...
    I understand the I=V/R, but how IR5 = V4/2000(Rt for branch), or how the V4 comes into play in the Ohm's Law equations. I guess I'm a little lost as to how you wrote the equations for each branch that included 5V, 15V , and 10V. Do polarities determine whether Vs are +/- in the equation. For instance, for IR4 = (10-V4)/Rt, how does this differ from IR1 = ((V4+5)-V3)/Rt in regards to adding or subtracting the Vs...
     
  6. WBahn

    Moderator

    Mar 31, 2012
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    Moved from General Electronics to Homework Help.
     
  7. WBahn

    Moderator

    Mar 31, 2012
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    We can't tell where you are getting lost unless you show us your efforts. So please make the best attempt you can (leveraging what MikeML showed you) so that we can see what you are doing right and what you are doing wrong. I've got a pretty good guess where you are going astray, but I don't like to guess if I can avoid it.
     
  8. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    Let's dissect that one:
    IR4 = E/R = (V at one end of R4 - V at other end of R4)/R4 = (V1-V4)/R4 (refer to the LTSpice schematic to see where V1 is) .

    Now, look at the V1 voltage source. Its bottom end is tied to 0V (gnd, or common), so the voltage at node 1 = V1 = 10V. We know that V1 = 10 (look at the polarity of the voltage source). We don't know what V4 is yet, but the voltage across R4 is V1-V4, which by substitution is 10-V4.

    Ok?
     
  9. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    IR5 = (V4-V5)/1000.
    IR6 = (V5-0)/1000 = V5/1000

    By KCL on node 5, IR5 = IR6
    (V4-V5)/1000 = V5/1000, multiply by 1000
    V4 - V5 = V5
    V4 = 2*V5
    or V5=V4/2

    so finally, IR6 = IR5 = V5/1000 = (V4/2)/1000 = V4/2000.

    I just did that one by inspection by initially adding R5 (1000) to R6 (1000), so IR5 = V4/(R5+R6) = V4/(1000+1000) = V4/2000
     
  10. LymanT

    Thread Starter New Member

    Sep 25, 2014
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    Well, when I initially saw this problem I thought I would use Mesh because from what I remember we used mesh to solve for problems with loops. So the way MikeML is showing me seems to be new to me or badly forgotten, therefore it is taking me time to fully comprehend. Can this problem be solved using Mesh? I am not fresh on Mesh as well so I had to do some book digging to help get me started. When I attempt Mesh, this is what I am getting...:

    From right to left:
    Loop A: 15v - 2kI1 - 2k(I2-I1)
    Loop B: 2k(I2-I1) - 1kI2 - 5v -2k(I3-I2) - 10v
    Loop C: 10v + 2k(I3-I2) - 1kI3 - 1kI3

    This is one of many attempts... and am getting different equations so I am wondering what I am missing here. I went counter clockwise with this analysis with current and loops. I stopped here because when I am simplifying further, I am getting myself lost and am wondering if Mesh is even the right way.
    @MikeML I don't know why I am getting confused on your approach. Maybe I never learned this way but it has been years, close to a decade since I solved problems such as this. Thanks for your insight. I am slowly making my way through your arithmetic. Is this the simplest way to look at this problem? The reason I am asking is because the problem ask for multiple measurements. With your schematice, 4 and 3 junctions are A and B, with a 3rd junction on the other end of the R3 labeled as C. So the questions ask for Vac, Vbc, Vab, etc...This is why I want to work out the problem so that I get all the currents and voltage drops label so I can work from that. I thought Mesh would be easiest but even that is getting confusing.
     
  11. LymanT

    Thread Starter New Member

    Sep 25, 2014
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    Ok...I see where I went wrong. With the analysis I just provided, I got the loops switch around. I needed (I1-I2) instead of (I2-I1), and (I2-I3) instead of (I3-I2). What a difference that makes. Initially I wanted to use this but had thought since I was working counter clockwise, these values would need to be switched as well. Trial and error I guess. Once fixed, I got a whole other equation. So I did finally get the loop currents for A, B, and C. Working with that I was able to work out the problem as needed. Wow, I've been working 3 days on this.. However, I would still like to understand the way MikeML approached. Thanks for your help, it got my brain moving.
     
  12. WBahn

    Moderator

    Mar 31, 2012
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    Your problems start right from the beginning.

    Your Loop A equation is apparently summing up the voltage gains going counterclockwise around the loop. That's fine. Your second term requires that the current I1 is going counterclockwise. That's fine. But the next term requires that I2 is going downward in R3 (and hence clockwise in Loop B, which is fine as long as you are consistent with that choice) and that I1 is going clockwise in Loop A, which contradicts the choice made in the second term. So this equation (and it's not an equation because it's not set equal to anything) is wrong.

    You have a similar problem with the next to last term in your Loop B expression and with the second term in your Loop C expression.
     
  13. WBahn

    Moderator

    Mar 31, 2012
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    It's not trial and error -- it's being systematic and organized.

    Starting off with a well-annotated diagram can help prevent many of these kinds of mistakes as well as communicate what you are doing to others instead of making them guess and reverse engineering your work.

    mesh1.png

    Now it is easy to see that, when doing the right hand mesh, that the current in R3 in the direction that you are going (counterclockwise) is (I1-I2).

    Also, since I1 is going the same direction (counterclockwise) through all of the components in Loop A, then sign of all of the terms involving the voltage drops due to I1 in Loop A have to have the same sign. Use that as a sanity check.

    The method that MikeML is using is basically the application of KCL at the two unknown essential nodes (i.e., where more than two branches come together).

    Again, start with a well-annotated diagram.

    node2.png

    Applying KCL at Nodes A and B, respectively, we have

    <br />
I_1 + I_2 + I_3 = 0<br />
I_4 + I_5 + I_6 = 0<br />

    Now we just need to write these currents in terms of the two node voltages Va and Vb. First, we'll write them in terms of the most convenient node voltages, whatever they happen to be, exploiting the fact that if we can find the current through ANY element that is in series with the current we want, that will work since all elements in series have the same current.

    <br />
I_1 = \frac{\(V_A-V_G\)}{\(R_5+R_6\)}<br />
I_2 = \frac{\(V_A-V_C\)}{\(R_4\)}<br />
I_3 = \frac{\(V_E-V_B\)}{\(R_1\)}<br />
I_4 = \frac{\(V_B-V_E\)}{\(R_1\)}<br />
I_5 = \frac{\(V_B-V_G\)}{\(R_3\)}<br />
I_6 = \frac{\(V_B-V_D\)}{\(R_2\)}<br />

    Now we eliminate all of the node voltages except two unknown ones because we know that, by definition (i.e., because we chose it as our ground reference) we have

    <br />
V_G = 0<br />

    And by inspection we have

    <br />
V_C = 10V<br />
V_D = 15V<br />
V_E = V_A + 5V<br />
     
  14. LymanT

    Thread Starter New Member

    Sep 25, 2014
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    Ok, that puts a lot into perspective. Thank you. Given all the values, how is the power dissipation calculated with multiple voltage sources? I've seen simple circuits with one source and 2 resistors. Given P=IV, I can't seem to get a correct answer.

    I1 = 3.33mA
    I2 = -.833uA
    I3= 2.08mA

    Vt = ?
     
  15. WBahn

    Moderator

    Mar 31, 2012
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    You find the total current in each resistor and/or the voltage across each resistor and calculate the power dissipated in each resistor and sum them up. Alternatively you can find the total current in each voltage source and calculate the power supplied by each source and sum them up.

    What you cannot do us try to apply superposition since superposition only holds for properties that are linear and power is not linear.

    Again, show your work. I can't tell why you can't get the correct answer unless you show my how you are getting whatever wrong answer you are getting.
     
  16. LymanT

    Thread Starter New Member

    Sep 25, 2014
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    Initially I wanted to just take 30V(6.25mA) but that didn't seem right. What I got was 187mW or so and was not what I should have gotten. This is when I knew I was missing some important factors. By alternatively finding the total current for voltage sources, do you mean shorting 2 of the 3 sources and solving each for total current? Then adding each of the 3 separate analysis to get total power dissipation? But then you mention I can't use superposition, so I am a bit confused.
     
  17. WBahn

    Moderator

    Mar 31, 2012
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    What does P=IV mean?

    It means that the power dissipated by THAT current flowing through THAT voltage is equal to THAT power. You can't just grab a voltage from over here and a current from over there and throw them at an equation that has an I and a V and expect to get anything meaningful.

    Let's take R4. The total current flowing downward in R4 is (I2-I3). Calculate that current and then use I²R to find the power in R4.

    You cannot find the power in R4 by taking I2 and I3 and plugging them into the I²R formula separately. Let's say that I have a current in a 10Ω resistor that is equal to I0 = 1A + 2A = 3A. The total power is

    P = (3A)²(10Ω) = 90W

    But if I try to use superposition I get

    P = (1A)²(10Ω) + (2A)²(10Ω) = 10W + 40W = 50W.

    The reason is simple. (a+b)² ≠ (a²+b²).

    Do it right:

    P = (1A+2A)²(10Ω) = (1A)²(10Ω) + [2(1A)(2A)(10Ω)] + (2A)²(10Ω) = 10W + 40W + 50W = 90W

    If you apply superposition, you miss the term in square brackets, called the crossproduct term,
     
  18. nDever

    Active Member

    Jan 13, 2011
    154
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    It looks like most here are solving your circuit using the node voltage method, which is another way besides mesh current to do network analysis.
     
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