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Solving circuits with superposition (2 problems)

  1. Mar 27, 2014 #1
    1. The problem statement, all variables and given/known data
    Screenshots have everything.


    2. Relevant equations
    ##V = iR##
    Voltage/current division
    Mesh analysis
    Adding resistors in parallel/series
    Kirchhoff's Laws

    3. The attempt at a solution

    First problem

    I started with turning off the current source. You are left with a simple series circuit with a dependent source because the part with the current source becomes an open circuit. I used Kirchhoff's voltage rule around the loop and got: $$-12 + i_{1} + 3i_{1} + 2i_{1} = 0$$

    Which yields ##i_{1} = 2##

    I then turned off the voltage source and activated the current source. I chose to use mesh analysis because it seemed to be the quickest way. The current you are left with is a resistor and a dependent source, another resistor, and your current source in parallel. The mesh current on the left is ##i_{x}## and on the right is ##i_{y}##. Using KVL for the supermesh around the perimeter: $$i_{x} +3i_{y}+2i_{x} = 0$$
    and also
    $$i_{y} - i_{x} = 6$$

    Which solves ##i_{x}## as -3 amperes. So, adding the first current and the second current gives -1 Amperes?

    Second question

    Again, I turned off the current source first. This leaves you with 2 resistors in series (parallel with the V source) and another 30 Ohm resistor in parallel. Since they all shared one node, the voltage should stay the same, right?

    If that's true, you can use Ohm's law to find the current over the strand on the right will be $$\frac {200}{100} = 2$$

    This next part is where I have a bit of a trouble. Turn off voltage source, turn on current source. If you look at the second picture, and replace the voltage source with a short circuit, does this mean that the 30 Ohm resistor can be ignored, because the current will always go through the short circuit instead of the path with the resistor?

    If this is true, you can just use current division, like so $$3 \frac {60}{40+60} = 1.8$$

    And you know this will be in the opposite direction from the previous current value, so ##2 - 1.8 = .2## so the current ##I_{0}## will be .2 Amperes. Is this correct? Thanks!
     

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    Last edited: Mar 27, 2014
  2. jcsd
  3. Mar 27, 2014 #2
    No one?
     
  4. Mar 27, 2014 #3

    SammyS

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    (It's too soon to be bumping your own thread ! Please review the Forum Rules.)

    attachment.php?attachmentid=68045&d=1395919681.png

    You should get ##-12 + i_{1} + 3i_{1} + v_{2i_{1}} = 0\ .##

    But you would also need ##\ i_1 = -2i_1 \ ## by KCL.


    (It's easy enough to check your answer to this one by using Kirchhoff's Current Law directly.)
     
    Last edited: Mar 27, 2014
  5. Mar 27, 2014 #4

    SammyS

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    attachment.php?attachmentid=68046&d=1395919681.png

    The second one looks good.
     
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