# Why does each mesh equation give different I3 value? What's the correct value and final answer?

The provided solution considers the third eqn as the correct one and proceeds with the same alone. Why is the answer achievable only thru the 3rd mesh and why not the first two? Why are they giving different values at all?

• What happens to the mesh current when a current source sits in a peripheral branch?
– Carl
Commented Apr 10 at 7:27
• Lumbini, I'll +1 your question now that you've added your thoughts. Best wishes! Commented Apr 10 at 8:58

You have three loops showing. (There are other loop arrangements that could be used to replace any of those three, just FYI.) Two of the loop currents are known because there are current sources that set them. So only the equation for $$\I_3\$$ is important.

Starting with $$\0\:\text{V}\$$ as the bottom left-hand corner and working clockwise around that loop:

$$0\:\text{V}-4\:\text{k}\Omega\cdot\left(I_3-I_2\right)-2\:\text{k}\Omega\cdot\left(I_3-I_1\right)-6\:\text{k}\Omega\cdot I_3+3\:\text{V}=0\:\text{V}$$

Given that $$\I_1=4\:\text{mA}\$$ and that $$\I_2=-2\:\text{mA}\$$ the above solves out as $$\I_3=250\:\mu\text{A}\$$. Just as you show at the bottom of your workpage.

Here's the Python/SymPy code:

solve(Eq(0-4e3*(I3-(-2e-3))-2e3*(I3-4e-3)-6e3*I3+3,0),I3)[0]
0.000250000000000000


Let's set that aside now, for a moment. You want to know about the other mesh equations and why you are getting different values for $$\I_3\$$.

The answer is that you need to assign a voltage difference across each of the current sources. Those two voltage differences are unknown. Allow me to annotate your schematic:

I've included the passive sign notations on every resistor and given the resistors names. (Please do that in the future.) You can also see that the sign interpretation for each resistor also depends upon the assumed direction of the current through each resistor for each loop.

Please review these choices with respect to the equation I set up for $$\I_3\$$ above.

Also note that I've added two new voltage differences, $$\V_1\$$ and $$\V_2\$$. ($$\V_3\$$ is assigned to the voltage supply in the $$\I_3\$$ loop.)

Note that I've also chosen to place the (+) sign for each current source's voltage at the end where the current arrow points.

So now let's do them. Each loop will start at the bottom left-hand corner and work clockwise.

\begin{align*} 0\:\text{V}+V_1-R_2\cdot\left(I_1-I_3\right)-R_1\cdot\left(I_1-I_2\right)&=0\:\text{V}\tag{loop 1} \\\\ 0\:\text{V}-V_2-R_1\cdot\left(I_2-I_1\right)-R_3\cdot\left(I_2-I_3\right)&=0\:\text{V}\tag{loop 2} \\\\ 0\:\text{V}-R_3\cdot\left(I_3-I_2\right)-R_2\cdot\left(I_3-I_1\right)-R_4\cdot I_3+V_3&=0\:\text{V}\tag{loop 3} \end{align*}

Now you have three equations and three unknowns.

But the unknowns are $$\I_3\$$, $$\V_1\$$, and $$\V_2\$$.

Let's use Python\SymPy (free and easy to get and use):

EQ1 = Eq( 0 + V1 - R2*(I1-I3) - R1*(I1-I2), 9 )
EQ2 = Eq( 0 - V2 - R1*(I2-I1) - R3*(I2-I3), 0 )
EQ3 = Eq( 0 - R3*(I3-I2) - R2*(I3-I1) - R4*I3 + V3, 0 )
for i,j in solve( [EQ1, EQ2, EQ3], [I3, V1, V2] ).items():
i, j.subs({ R1:4e3, R2:2e3, R3:4e3, R4:6e3, V3:3, I1:4e-3, I2:-2e-3 })
(I3, 0.000250000000000000)
(V1, 40.5000000000000)
(V2, 33.0000000000000)


And I think you can see that $$\I_3\$$ hasn't changed. It's still the same value as was computed in the earlier single equation.

Your only problem is in failing to recognize that the current sources will have their own voltage drops.

And also that you then, of course, will need three equations in three unknowns, so that you can solve not only for $$\I_3\$$ but then also for the voltage drops across the two current sources, as well: $$\V_1\$$ and $$\V_2\$$.

That's all there is, really. It's just that you missed seeing all the details. Once you gather all that up, correctly, nothing important changes. You still get the same $$\I_3\$$ value, as before.

You just get two additional unknowns that, earlier, you didn't need to worry about.

I sincerely hope that helps you see that everything is right about the process. The mesh approach does work well. You just need to keep an eye out for missing details.

• Ohk, I didn't know that current sources can also have voltage drops. I thought they hv 0 resistance and are treated like wires. Thankyou Commented Apr 10 at 9:18
• @LumbiniAshutoshTambat No problem. I'm just seriously glad it helped out. Best wishes! Commented Apr 10 at 9:23