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# Mesh current method (steps 1 to 3)

We solve a circuit by writing Kirchhoff's Voltage Law in terms of "mesh currents." This video covers the first three steps out of four. Created by Willy McAllister.

## Want to join the conversation?

• Is it possible to eliminate current source from the circuit by converting it to voltage source or vice-versa ?
• At , why you define iR2 as (i1 - i2) and not the other way around?You are writing KVL for the second mesh. Thanks.
• Hello Asal,

Yes, we could also have written -R2(i2-i1) and substituted this into the video's equation.

We could have simplified things even more and written:

0 = R2(i2-i1) + R3(i2) + R4(i2-i3)

It all depends on how you look at things and what you define as a voltage rise and a voltage drop. Personally I try to make the math as simple as possible. It is so easy to miss a sign...

Regards,

APD
• I don't understand why i3 only equals to -I, why isn't that -I-i3=0?
• Hello! One question: why is i3 = -I and we don't include R4(i3-i2) also?
• Mesh current i3 has to be equal to -I because it is the only current flowing through the current source. When you look at R4, however, the resistor current is the combination of (superposition of) two mesh currents, i3 and i2. You nearly got the expression for the voltage across R4 correct. It equals R4(i2 - i3).

Mesh currents are kind of strange little animals. Sometimes they exist in a wire or component by themselves, as i3 does in the right side of the circuit. Another example is i2 flowing in R3. Other times mesh currents exist only in combination with other mesh currents, as the current in R4 or R2. I really like this weirdness. Mesh Current is my favorite analysis method, just for the entertainment value.
• At , why don't we account for the current coming in from the current source, I, since they're on the same node?
I'm sure you're right, but I don't understand why we wouldn't have our 2nd equation as:
R2(i1-i2) - R3(i2+I) - R4(i2-i3) = 0
^
• You are anticipating the path of I (the current from the current source on the right) before letting the math do its job. I does not necessarily flow where you think it does. When it reaches the junction where R3 and R4 connect, it has the option of splitting between those two resistors. So not all of it (necessarily) flows through R3.

The Mesh Current method is based on mesh currents (i1, i2, and i3, but not big I ). When you write the mesh equations you do it in terms of mesh currents. That means you won't mention element currents like I when you set the equations up.

The Mesh Current method is kind of fun because you are asked to put your faith in these made-up mesh currents as you set up the problem. Everything works out in the end.
• @ Why don't you include R4 into your i3 calculation?

Sal says: i3 = -I
why not: i3 = -I + R4(i2-i3)
• i3 is completely defined by the current source. You don't need any other information.

Take a look at the last term in your second equation. The units of R4(i2-i3) is what? That expression is in volts. It is the voltage across R4. The element current through R4 is the combination of two mesh currents, (i2-i3).
• Instead of having i3 going clockwise and equaling -(current soruce), could you have made i3 go counter-clockwise and equal +(current source)?
• Hello Emily,

Yes but...

The direction of the loops is completely arbitrary. If you changed i3 you would need to go back to and change your R4 calculations.

For your own sanity may I recommend consistency. Make all loops in one direction this way you can quickly use a robotic process to generate the equations.

Regards,

APD
• Can we use thevenin' s law to analyze this circuit?
• Hello Varney,

Absolutely 💯. The circuit may be simplified by converting Thevenin circuits to their equivalent Norton circuits and vise versa.

Do whatever makes the work easy - or whatever your instructor is looking for...

Regards,

APD
• sir,
how do we know that there is voltage rise or voltage drop... as the teacher says that there is always voltage drop accross the resistor but how can we judge it by + and - signs?
(1 vote)
• The voltage drops in the direction the current passes through the resistor. The current goes from high potential to low.
• Hello,

This question is for the working done in .

I am quite connfused as to how the second equation written for Mesh 2 is not R2(i2-i1) - R3i2 - R4(i2-i3). Particularly, why the voltage across R2 changes from -R2(i1-i2) in M1 to R2(i1-i2) in M2? This is what I did after assuming that going from negative to positive terminal gives a positive voltage:

I wrote the equation for M1 as:
5 - R1i1 -R2i1 + R2i2 = 0
5 - R1i1 -R2(i1-i2) = 0

M2:
R2i2 - R2i1 - R3i2 -R4i2 + R4i3 = 0
R2(i2-i1) - R3i2 - R4(i2-i3) = 0
-R2(i1-i2) - R3i2 - R4(i2-i3) = 0

I attempted to take the voltage with respect to each current loop. This way I thought that the voltages will remain same (and they are the same), but your working is different? Is it because that you take voltages with respect to the mesh instead of the loop? As you can see, I am quite confused. So, please help me in understanding the real process.

(1 vote)
• We are talking about the sign of the first voltage term. The video has -R2(i1-i2) for the first mesh, but uses +R2(i1-i2) for the second mesh. You are asking why we don't use the same expression, -R2(i1-i2) (or equivalently +R2(i2-i1)), for both meshes.

The reason is how you encounter R2's voltage polarity signs as you travel around the two different KVL arcs.

Assuming you start your KVL trip in the lower left corner of each mesh... For mesh 1 when you get to R2 you 'enter' it through its + voltage sign and exit through its - sign. In that direction of travel, that voltage is a drop and the overall term gets a - sign.

When you do the trip for the second mesh the first thing you encounter is the - sign of R2's voltage polarity. That means as you go through R2 you are getting a voltage rise. That's why this term gets a + sign in the second mesh equation.

A point to emphasize... When you have a resistor with two mesh currents, like R2 (and R4), be careful with which mesh current goes first in the subtraction, (i1-i2) is correct, whereas (i2-i1) is not. The first mesh current in the subtraction is always the one that flows into the + voltage terminal of the resistor. That is true for both mesh 1 and mesh 2.

Now that you have the order of the subtraction always right, the + and - signs in the KVL equation depend on the order of travel as discussed above.