# Mesh Current Analysis

Discussion in 'Homework Help' started by gajoline, Sep 10, 2013.

1. ### gajoline Thread Starter New Member

Aug 28, 2013
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As far as i know if we have a Current Source we asume its a voltage source hence the Vx.
After i do the analysis i don't know how to proceed to find the currents and specifically I2 which i think is the one that pass through 5kΩ resistor...

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2. ### WBahn Moderator

Mar 31, 2012
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What relationship between I1 and I2 is imposed by the current source?

3. ### gajoline Thread Starter New Member

Aug 28, 2013
10
0
I don't really know what you mean but the only thing this excercise gives is the circuit and the question to find the current(I) tha passes through 5kΩ resistor...
the other stuff drawn by me while i was trying to solve this

4. ### WBahn Moderator

Mar 31, 2012
22,852
6,818
In terms of I1 and I2, what is the current that is flowing downward through the middle vertical branch?

What does current have to be equal to?

5. ### gajoline Thread Starter New Member

Aug 28, 2013
10
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If that is a question to help me solve this then i would guess that I1 will go down the vertical branch and if i got it right i think the current equals I1-I2=2mA

6. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
522
Remember in this method (maxwells mesh method) the mesh currents are not the actual currents.

The actual currents are a composite of the appropriate mesh currents.

Does this help understand what WBahn is saying?

7. ### WBahn Moderator

Mar 31, 2012
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Very good.

Now, can you use that relationship to write both of the equations you got originally just in terms of I2 (the current you are interested in)?

After doing that, can you substitute the equation for Vx (the second equation) into the first equation to eliminate Vx?

What are you left with? Can you solve it for I2?

8. ### gajoline Thread Starter New Member

Aug 28, 2013
10
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Not really i think i am really confused with this circuit ...

9. ### WBahn Moderator

Mar 31, 2012
22,852
6,818
You are doing fine.

You got that, in terms of I1 and I2, the net current flowing down through the middle branch is I1-I2. You also got that this current has to be equal to the actual current that is imposed by the 2mA current source, hence

$
(1) \ \ I_1 \, - \, I_2 \, = \, 2mA
$

$
(2) \ \ 2V \, + \, V_x \, = \, 2k \Omega \cdot I_1
$

and

$
(3) \ \ -V_x \, = \, 9k \Omega \cdot I_2
$

So you have three equations and three unknowns (I1, I2, Vx}. You are only interested in one of them, I2, since that IS the current flowing in the 5kΩ resistor. So let's eliminate the other two unknowns from the equations.

What happens if you add equations (2) and (3) together? Do you know how to add equations?

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10. ### gajoline Thread Starter New Member

Aug 28, 2013
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But i think i should have turn mA into Amps before do the calculations...
2mA = 0.002A so we have 2 - 0.004 = 1,996 => 1,996 / 11 ≈ 0,18 (AGAIN)

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11. ### WBahn Moderator

Mar 31, 2012
22,852
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Very good.

I2 = -0.18.... what? Amps? mA? kA? Ducks? What?

You need to start using units properly.

$
(1) \ \ I_1 \, - \, I_2 \, = \, 2mA
\
(2) \ \ 2V \, + \, V_x \, = \, 2k \Omega \cdot I_1
\
(3) \ \ -V_x \, = \, 9k \Omega \cdot I_2
\
(4) \ \ 2V \, = \, 2k \Omega \cdot I_1 \, + \, 9k \Omega \cdot I_2
\
(5) \ \ I_1 \, = \, I_2 \, + \, 2mA
\
(6) \ \ 2V \, = \, 2k \Omega \cdot (I_2 \, + \, 2mA) \, + \, 9k \Omega \cdot I_2
\
(7) \ \ 2V \, = \, I_2 \cdot ( 2k \Omega \, + \, 9k \Omega) \, + \, 2mA \cdot 2k \Omega
\
(8) \ \ 2V \, = \, I_2 \cdot 11k \Omega \, + \, 4V
\
(9) \ \ I_2 \, = \, \frac{-2V}{11k \Omega} = -0.182mA
$

See how at no point was there any question about whether I should use 2 or 0.002. Tracking the units ensured that the results worked out correctly. If one of the resistances had been in MΩ or Ω, that would have been fine because 2mA*2000Ω is either 4000mV or 4V, which are equal to each other. But 4000 and 4 are just numbers and they are NOT equal to each other.

Get in the habit of tracking your units and checking them. Always. ALWAYS!

You WILL make mistakes (we all do) and most of your mistakes will mess up the units. Far better to catch the mistake right away and fix it then to waste hours of effort on a path that was doom in line three.

Last edited: Sep 10, 2013
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12. ### WBahn Moderator

Mar 31, 2012
22,852
6,818
The other thing you need to get in the habit of doing is always asking if the answer makes sense. In most cases in engineering, you can verify the correctness of an answer from the answer itself. This problem is no exception.

If I2 is -0.182mA, then what is Vx? It must be

Vx = -I2*9kΩ = 1.638V

If Vx is 1.638V, then what is I1? It must be

I1 = (2V+Vx)/2kΩ = 3.638V/2kΩ = 1.819mA

If I2 is -0.182mA and I1 is 1.819mA, what is the current down the middle branch? It must be

I1-I2 = 1.819mA - -0.182mA = 2.001mA

This agrees with the requirement imposed by the 2mA current source, so we have very high confidence that the answer is correct.

BTW: In writing this I started by copying over the result that I had originally written in the prior post, which was -182mA (I hadn't hit the period key hard enough) and that rather immediately resulted in a totally absurd value for Vx of 1638V. That slapped me in the face and I immediately saw and fixed the error. So see, it works!

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13. ### gajoline Thread Starter New Member

Aug 28, 2013
10
0
Thank you once again , i understood everything !!!