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\$I\$ \$2\$ gets different value by different equations , Using mesh analysis . Please explain why \$I\$ \$2\$ has two different values using mesh analysis .

My approach : Schematic

\$-12I1+2I2=0\$

\$-2I2+2I1-20I3=0\$

\$I3-I2=2I\$

\$I2-I1=I\$

\$I1=-5A\$

Solving this I find \$I2=-30A\$ also \$I2=-105/31A\$ . Please tell me which value of \$I2\$ is correct and why .

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    \$\begingroup\$ This is clearly a homework problem, so we expect you to show all of your work and ask a specific question. \$\endgroup\$ – Elliot Alderson Nov 7 '19 at 20:53
  • \$\begingroup\$ Hint, due to the arrangement of the CCCS and the control branch, the CCCS can be replaced with a resistor. \$\endgroup\$ – The Photon Nov 7 '19 at 21:05
  • \$\begingroup\$ @ElliotAlderson Reframed the question , now help please !! \$\endgroup\$ – Nilabja Saha Nov 8 '19 at 1:28
  • \$\begingroup\$ @ThePhoton The question has been reframed , now look . \$\endgroup\$ – Nilabja Saha Nov 8 '19 at 1:32
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    \$\begingroup\$ Your equation for the I1 mesh is not valid because there is no term for the voltage across the 5A current source. You can't assume that the voltage across a current source is zero. \$\endgroup\$ – Elliot Alderson Nov 8 '19 at 2:35
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Well, we have the following circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

Using KCL, we can write:

$$ \begin{cases} \text{I}_2+\text{I}_4=\text{I}_1\\ \\ \text{n}\cdot\text{I}_2+\text{I}_3=\text{I}_4\\ \\ \text{n}\cdot\text{I}_2+\text{I}_3+\text{I}_5=0\\ \\ \text{I}_1+\text{I}_5=\text{I}_2 \end{cases}\tag1 $$

Now, using KVL, we can write:

$$ \begin{cases} \text{I}_1=\frac{\text{V}_1-\text{V}_2}{\text{R}_1}\\ \\ \text{I}_2=\frac{\text{V}_2-\text{V}_4}{\text{R}_2}=\frac{\text{V}_2-0}{\text{R}_2}=\frac{\text{V}_2}{\text{R}_2}\\ \\ \text{I}_3=\frac{\text{V}_2-\text{V}_3}{\text{R}_3}=\frac{\text{V}_2-0}{\text{R}_3}=\frac{\text{V}_2}{\text{R}_3} \end{cases}\tag2 $$

Now, we can solve for \$\text{I}_3\$:

$$\text{I}_3=\frac{\text{I}_1\text{R}_2}{\text{R}_2+\text{R}_3\left(1+\text{n}\right)}\tag3$$

Using your values we get:

$$\text{I}_3=\frac{-5\cdot2}{2+20\left(1+2\right)}=-\frac{5}{31}\tag4$$


Solving all the other values:

enter image description here

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  • \$\begingroup\$ Correct !! Thanks !! \$\endgroup\$ – Nilabja Saha Nov 23 '19 at 14:32
  • \$\begingroup\$ @NilabjaSaha You're welcome! You can accept my answer but pressing the green V sign. \$\endgroup\$ – Jan Nov 23 '19 at 14:36
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You need to analyze the question using a supermesh, or the equivalent of just combining the equations for mesh 2 and 3. It will essentially look like this:

\$ 2(I2-I1)+20(I3)=0 \$.

You can then easily derive equations to reduce the number of unknown variables from two to one to solve it. This can be done by analyzing the current across the CCCS in terms of I2 and I3, and the current across the 2 Ohm resistor in terms of I1 and I2. I1 is, of course, a known mesh current and can just be plugged in.

The fact that you derived two values of current for the same mesh should be a red flag that your analysis wasn't correct and needed rethinking.

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  • \$\begingroup\$ Ok , Used KVL in this way and got the right answer !! Thanks !! \$\endgroup\$ – Nilabja Saha Nov 23 '19 at 14:34

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