nodal analysis question

Discussion in 'Homework Help' started by chrischivers1, Mar 23, 2015.

  1. chrischivers1

    Thread Starter New Member

    Mar 23, 2015
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    Hi,

    I am having a few issues with this question about Nodal Analysis (well part (b)).

    Can anyone help me out as i am very stuck.

    thank you.
     
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    In which part you have a problem ?? You do not know KCL ??
     
  3. chrischivers1

    Thread Starter New Member

    Mar 23, 2015
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    Overall.. I'm a mature student returning to it after 11 years.. Bit rusty..

    Any pointers?
    Thanks for taking a look btw
     
  4. chrischivers1

    Thread Starter New Member

    Mar 23, 2015
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    The circuit I have only has 1 current source.. How do I attack that as the one in the video has two.

     
  5. Jony130

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    All you have to to is to write a two KCL equation for two node.
    [​IMG]
    First a KCL for node A (Va) next write KCL for node B (Vb). But you need to write all component currents in terms of a its nodal voltages.
    For example R1 resistor current is IR1 = Va/R1. The voltage across the 20kΩ resistor is the difference between the voltage on the node on top (Va) of the resistor and the voltage on the node on the bottom of the resistor (Vground).
     
  6. chrischivers1

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    Mar 23, 2015
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    Thanks, I'll heed your advice and watch the video.
    It's not hard, I'm just rusty.
     
  7. Jony130

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    Are you managed to solve this circuit?
     
  8. chrischivers1

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    Mar 23, 2015
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    Just going thru it now.
    It's a similar layout to the video just doesn't have the 3rd variable
     
  9. Jony130

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    Exactly
     
  10. chrischivers1

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    Mar 23, 2015
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    How do you break it down for the current over each component tho? Surely when you know A and B you'd only be able to find the current of R1 and R2.. By ohms law.
    What about the others?
     
  11. Jony130

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    Hmm, what about R3 current?
    What is the voltage across R3?
    Also notice that the R4 and R5 are connected in series and when resistors are connected in series the current flowing in each is the same.
     
  12. chrischivers1

    Thread Starter New Member

    Mar 23, 2015
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    True. I am having a thick Monday damn.
    How would I get r3?
     
  13. chrischivers1

    Thread Starter New Member

    Mar 23, 2015
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    R2 even!!
     
  14. chrischivers1

    Thread Starter New Member

    Mar 23, 2015
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    A
    actually im being very thick as the currents are flowing away from the current source into all the resistors, I was getting hung up after the video as that has 2 sources
     
  15. Jony130

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  16. WBahn

    Moderator

    Mar 31, 2012
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    You seem to be stuck down in the weeds and not grasping the more general concepts. If it is just a matter of being rusty, then you should be able to go back to a basic Circuits I text and read through it very quickly to blow of the rust and come back up to speed on the various analysis techniques. I'm concerned that the way you are doing it now will only chip away at parts of the rust and take you much longer to end up at a place where the gears are still grinding. So go back and walk yourself through the whole thing to become a well-oiled machine. I really don't think it will take you that long -- a few hours, perhaps.
     
  17. chrischivers1

    Thread Starter New Member

    Mar 23, 2015
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    Cheers mate, yea it's just reawakening to all the concepts, I'll do some reading up and all being well it'll resink in.

    I'm just going to the gym so I'll resume once im home.
     
  18. chrischivers1

    Thread Starter New Member

    Mar 23, 2015
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    can somebody walk me through this question please? the attached PDF is where i am at so far.. had a family bereavement so my heads not been in it and the deadline is Thursday.

    asking for help isn't my thing but im up against it.

    thanks
     
    • Doc1.pdf
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  19. WBahn

    Moderator

    Mar 31, 2012
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    You use the same labels for different things. It appears that you are trying to color code them, but this isn't a good idea. First, what happens when you make a copy on a black and white copier or your photo gets rendered in grayscale? The point being that color information is too ephemeral to rely on. But even worse, you've got to combine these terms together at some point and it will become impossible to track them properly. Follow the rule that one label has one meaning.

    The work in red looks fine other than the fact that you fail to track your units properly.

    You last equation should be

    <br />
V_1 \( \frac{1}{20 \Omega} + \frac{1}{30 \Omega} \) - V_2 \( \frac{1}{30 \Omega} \) \; = \; 3 A<br />

    which can be reduced to

    <br />
V_1 \( \frac{3}{60 \Omega} + \frac{2}{60 \Omega} \) - V_2 \( \frac{1}{30 \Omega} \) \; = \; 3 A<br />
\,<br />
V_1 \( \frac{5}{60 \Omega} \) - V_2 \( \frac{1}{30 \Omega} \) \; = \; 3 A<br />
\,<br />
\( 60 \Omega \)\[ V_1 \( \frac{5}{60 \Omega} \) - V_2 \( \frac{1}{30 \Omega} \) \] \; = \( 60 \Omega \) \; \[ 3 A \]<br />
\,<br />
5 V_1 - 2 V_2 \; = 180 V<br />

    See how the units all work out?

    Also consider that if I can be bothered to track my units in as inconvenient a format as writing LaTex code into a forum post, why can't you be bothered to do it with pen and paper?

    Your work in red goes off the rails almost immediately due to sloppiness.

    Your first line is fine, but your second line is

    I_A = something = 0

    You are explicitly stating that I_A is equal to zero.

    Your final line isn't even an equation (it's just an expression), it is dimensionally inconsistent, and it doesn't follow from the expression in the line above it.

    Take the time to do it carefully and correctly.
     
  20. chrischivers1

    Thread Starter New Member

    Mar 23, 2015
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    Thank you for your detailed observations.
    Apologies for my sloppiness, the colours seemed a good call but now you mention it, it'll prove troublesome in the future
     
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