Circuit analysis

Discussion in 'Homework Help' started by pianolife, Jan 5, 2013.

  1. pianolife

    Thread Starter New Member

    Nov 15, 2012
    28
    0
    Dear members of All About Circuits, I'm asking for some help because I am a little bit confused about circuit analysis.
    In the attachment there is the circuit I need to analyse, and I need to find the Voltage V across the 12Ω resistor.

    I was trying to analyse it by using superposition. My idea was to get the total current through the circuit and then, multiplying it by 12Ω, I was looking for the voltage across 12Ω resistor. (Possibly I am already wrong).

    Steps I've done:
    1) Leave the 50V source, then [(20+12) || 40] + 10 = 27.7Ω. Then 50/27.7=1.8A
    2) Leave the 40V source. (40||10)+12+20 = 40. Then -40/40=-1
    3) Leave the 8A source. Here comes the problem. Doing (10||40+12)=20. Then I have a 8A source in parallel with 2 resistors of 20Ω each. What I thought is: current divider=4A per side.

    The actual problem is: if I sum all these current I get 4.8A, which multiplied by 12=57.6V.

    With a circuit simulator I get that the voltage drop is 48 (120V on the right side and 72 on the left). And 48/12=4, like the value I got from step 3.
    My silly idea is: shall I consider every single loop, and since the 12Ω resistor is "in the loop" with 4A, easily multiply that current times the resistance to find the voltage across that resistor?

    I'm sorry for this confusing message, I hope that someone can help me.

    Thanks in advance.

    All the best
     
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,957
    1,097
    You are doing good job so far.
    But here you find a total currant
    1) Leave the 50V source, then [(20+12) || 40] + 10 = 27.7Ω. Then 50/27.7=1.8A
    2) Leave the 40V source. (40||10)+12+20 = 40. Then -40/40=-1
    Not a current that is flow through 12R resistor. So all you need to do is to find a current that is flow through 12R resistor
     
    • aa.PNG
      aa.PNG
      File size:
      25.4 KB
      Views:
      17
  3. The Electrician

    AAC Fanatic!

    Oct 9, 2007
    2,281
    326
    You've calculated the current through the 10 ohm resistor, but that's not the same as the current through the 12 ohm resistor.
     
  4. pianolife

    Thread Starter New Member

    Nov 15, 2012
    28
    0
    Thank you so much to both of you!
    How can I find it?
    Looking at the sketch made by Jony130, can I find the current through the 12Ω resistor by doing first Itot - I1?
    Thank you very much!
     
  5. pianolife

    Thread Starter New Member

    Nov 15, 2012
    28
    0
    I was thinking about doing source transformation with the current source but another doubt comes out: which of the two resistors in parallel shall I use? I would use the 12Ω...

    Thank you very much
     
  6. ECC83

    New Member

    Jan 6, 2013
    13
    0
    A much easier way IMO is to do nodal analysis and leave the circuit as it is.

    If you consider the two nodes that the 12 ohm resistor is in between and call it node A and node B, consider the currents going into and out of nodes A and B. Since you're solving for two nodes you will get two simultaneous equations which is about a minute's work and you're done.

    What I did was for e.g. for node A:

    (50 - A)/10 = A/40 + (A - B)/12

    Similarly for node B, then solve.
     
  7. pianolife

    Thread Starter New Member

    Nov 15, 2012
    28
    0
    Thank you ECC83, I'll try this way as well. Just for double checking, could you please take a look at the picture I'm attaching? Do you mean thinking at nodes like this?
    Shall I consider the blue arrow as well?

    Thank you very much
     
  8. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,957
    1,097
  9. ECC83

    New Member

    Jan 6, 2013
    13
    0
    I'm not too sure why you've drawn so many arrows there. If you look at your labelled node A, consider the currents going into and out of that node.

    1 current path going into Node A, where A stands for the voltage to be solved.

    (50V - A)/10

    We have two current paths going out of Node A.

    A/40 and (A - B)/12

    Putting this all into an equation, remember currents going into a node must equal currents coming out, so remember which side they go:

    (50V - A)/10 = A/40 + (A - B)/12

    Do the same for Node B.

    I think for good practice you should try the basic methods given by Jony as well, the answers should all be the same in the end.
     
  10. pianolife

    Thread Starter New Member

    Nov 15, 2012
    28
    0
    Thank you Jony130!! So basically my work wasn't that bad, I only need to you a current divider formulae to find the current across 12Ω?

    Cheers
     
  11. WBahn

    Moderator

    Mar 31, 2012
    17,716
    4,788
    You are on the right track.

    Think of the following chain of reasoniing:

    I want to know V12 (the voltage across the 12Ω resistor). If I know the voltage on the nodes A and B (the nodes on either side of it) then I can trivially find V12. I can use KCL to write equations involving the voltages on those node, namely Va and Vb, giving me two equations and two unknowns. Hence I can solve them for Va and Vb and then use those results to find V12.
     
    ECC83 likes this.
  12. ECC83

    New Member

    Jan 6, 2013
    13
    0
    Va and Vb are much better labels for the unknown nodes than A and B, good point!
     
  13. pianolife

    Thread Starter New Member

    Nov 15, 2012
    28
    0
    Because I'm messy and I love arrows :D
    I just wanted to check if I got the loops right :)

    Thank you very much for your patience!
     
  14. WBahn

    Moderator

    Mar 31, 2012
    17,716
    4,788
    You can draw the loops whatever direction you want, you just have to be sure to be consistent with the resulting polarities of the terms in your equations. It is common practice to always use loops that go in the same direction in order to minimize the likelihood of sign errors.
     
  15. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,957
    1,097
    Yes, that all you need to do
     
    pianolife likes this.
Loading...