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Given

My solution

The answers are V1= 3.043 , V2= -6.956 , V3= 0.6522. I've already searched how to use mesh analysis to find voltage, there are a lot with this example but they all used nodal analysis to find V1 , V2 & V3. I'm not sure what I'm doing wrong. I have 4 variables but only 3 equations.

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3 Answers 3

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I find the following mesh equations, recognizing that \$i=-i_2\$ given your handwritten labeling of the loop currents:

$$\begin{align*} 0\:\text{V} + 10\:\text{V} -i_1\,6\:\Omega - \left(-i_2\right)5\:\Omega &= 0\:\text{V} \\\\ 0\:\text{V} -i_2\,2\:\Omega -10\:\text{V} -\left(i_2-i_3\right)\,4\:\Omega &= 0\:\text{V} \\\\ 0\:\text{V}-\left(i_3-i_2\right)\,4\:\Omega + \left(-i_2\right)5\:\Omega - i_3\,3\:\Omega&=0\:\text{V} \end{align*}$$

You need to check your own work against the above.

The solution, using SymPy (freely available), is:

eq1 = Eq( 0 + 10 - i1*6 - (-i2)*5, 0 )
eq2 = Eq( 0 - i2*2 - 10 - (i2-i3)*4, 0 )
eq3 = Eq(0 - (i3-i2)*4 + (-i2)*5 - i3*3, 0 )
ans = solve( [ eq1, eq2, eq3 ], [ i1, i2, i3 ] )

For \$v_1\$:

-ans[i2]*2.n()
3.04347826086957

For \$v_2\$:

(ans[i2]-ans[i3])*4.n()
-6.95652173913043

And for \$v_3\$:

ans[i3]*3.n()
0.652173913043478

LTspice finds:

enter image description here

Everything matches up.

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  • \$\begingroup\$ If you think getting a little dirty is not too much price to pay, you can use the voltage source with 0 Rser=<R> as a resistor, to avoid an extra node for the CCxS. Then you can use the symbol for the resistor to cover the dirty deed. Example. Optionally modify it to have a small arrow for the direction of the current. Not sure but I think it may cause problems in subcircuits. But then this whole thing would be pointless. \$\endgroup\$ Commented Sep 7, 2022 at 19:51
  • \$\begingroup\$ @aconcernedcitizen Hmm. Thanks. I was just pasting things up quickly in LTspice and didn't want to think beyond old habits at the moment. (Costs me time to break a habit.) I suppose I should develop new habits when the mood strikes, though. Thanks! \$\endgroup\$
    – jonk
    Commented Sep 7, 2022 at 20:49
  • \$\begingroup\$ No need to spend too many CPU cycles on it. For my part I find the extra voltage sources as too much information but then, it's the same with having R1 1 as a value for the CCVS. Call me Whimshmael. \$\endgroup\$ Commented Sep 8, 2022 at 7:16
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The last equation you need is one that relates the value of \$i\$ (the current that controls the dependent voltage source) to one of your mesh currents. Once you have that, you can replace \$5i\$ in the first equation with an expression that is only in terms of the mesh currents.

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Simply note that \$ i = - i_2 \$, so the term \$ -5i\$ in the last equation can be rewritten as \$ 5i_2 \$.

Now you have just 3 variables in your 3 equation system.

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