If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

# Mesh current method (step 4)

We complete the solution of a circuit using the mesh current method. Fourth step (solve) out of four. Created by Willy McAllister.

## Want to join the conversation?

• When you talk about scaling the equations down (), wouldn't that affect the 5 Volts on the other side as well? Does it have to do with the currents being expected to be in mA? Thank you in advance. • The scaling up/down by 1000 happens on just the left side of both equations. Every term on the left side is current x resistance. If all currents are in mA, and all resistance is in kOhms, then 10^-3 from the mA cancels out the 10^+3 from the kOhms, and you are just left with the integer parts. In Ohm's Law, the units cancel like this: kOhms x mA = 1. And for big resistors: MegOhms x uA = 1.
• I'm a bit confused by the final results. I understand the algebra and where all the figures come from but-
How are the values for i_1 i_2 and i_3 related? Shouldn't KCL have something to say here?
Plus, is there some relationship between i_r2 i_r4 and I? In other words: what do 1.8mA, 1.6mA and 2mA have to do with one another?
Thanks! • I may be wrong, but at 3 : 40 when -i3 is factored out by -R4 to make +RI, this would make RI positive and then subtracted on the other side to make it -RI. On the video it's positive. • I'm 14 and I'm wondering whether this is university work or not.Thnaks • why don't you just show us solving it with matrices instead of making it longer and longer.. • You are correct that this problem could be solved with matrix methods. However, I don't assume the reader has taken a class in Linear Algebra. These systems of equations are fairly small, so solving by ordinary algebra is not too onerous, and is accessible to anyone who can solve simultaneous equations.

• Ugh.

So I use the alternative rule set for walking around the loop (add when encountering +, subtract when encountering -), which meant I was doing the inverse for all of the calculations. Everything went fine until you gave the hard values for the variables, then suddenly my calculations kind of got thrown out of whack. I got the inverse values for for i_1 and i_2 (which was expected, given that I had to multiply by -4 instead of 4 for the cancellation of i_2 to work at ), and I got the inverse for i_R2 (also expected), but i_R4 is coming out as 2.4mA instead of the expected -1.6mA. I noticed that everything checks out if I tweak the schematic slightly by inverting either i_3 or I, but I'm not sure which is allowed, if either.

So I guess my question is if using the alternative rule set when following these demonstrations is allowed, and if so, what aspects should I be changing in the given values for the math to still work? Sorry for being a pain, but the alternative rule set just works better for me and I don't want to go against what feels natural if I don't have to. • Your description suggests you have a sign error somewhere along the way.
The only thing you are doing different than the video is to change the sign of the terms in the KVL equations. The mesh currents still flow clockwise, right? When I did the problem I got these two equations,

3k i1 - 2k i2 = 5
-2k i1 + 8k i2 = -6

These are the same as the video's equations, with every sign flipped. So they should solve down to the same answer.

Where you got iR4 = 2.4mA, that points to a sign error in the second equation where you move the term +3k i3 = +3k (-2mA) over to the right side. It should end up -6 on the right.

Sign errors like this are the hardest part of manual circuit analysis. I have problems with this all the time. My habit is to triple check frequently as I go along, (and use a circuit simulator to help check my own calculations. http://spinningnumbers.org/a/circuit-sandbox.html)
• So at on the video why would multiply the equation by 4? So far that is my only problem that I have. • Could the combination of R1,R2,R3,R4,V and I be such that the resulting current flowing through R2 or/and R4 be zero? (for example if i1=i2 or/and i2=i3) • Kind of a fun question.

There is the degenerate case where you set V and I to zero. Then no current flows anywhere. That's not so interesting.

To get I_R2 = 0 or I_R4 = 0 you have to arrange for both of resistor terminals to be at the same voltage. Assume the bottom of both resistors is connected to ground (V = 0). Then you have to figure out how many ways you can get their upper node to be 0.

One way I can think of is to make V a negative value while I is still a positive value (current flowing up).

I bet you would have a lot of fun trying to figure this out using this circuit simulator: https://spinningnumbers.org/a/circuit-sandbox.html
(1 vote)
• I'm having trouble figuring out how the algebra was worked out starting at . He mentioned how I1 is mentioned twice. How does that turn into -(R1 + R2)I1 ?
(1 vote) • Why can't we use ohm's law on the resistors? 5 volts (from the source) over 3000 ohms = about 2.5 mA
(1 vote) • Where did you get 3000 ohms from?
This isn't a simple 1-resistor problem for Ohm's Law.
The current coming out of the 5V source splits after the 1k resistor. You don't know ahead of time how much current goes down through the 2k resistor. The ratio of the split is determined by everything else in the circuit.

This video sequence on Mesh Current is one way to work out how it all happens.