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# Mesh current method

The Mesh Current Method solves circuits by writing Kirchhoff's Voltage Law for currents flowing in the loops of a circuit.

## Introduction

The Mesh Current Method is another well-organized method for solving a circuit. (The other is the Node Voltage Method.) As with any circuit analysis challenge, we have to solve a system of $2E$ independent equations, where $E$ is the number of circuit elements. The Mesh Current Method efficiently manages the analysis task, resulting in a relatively small number of equations to solve.
The Mesh Current Method is based on Kirchhoff's Voltage Law (KVL).
The Loop Current Method is a small variation of the Mesh Current Method.

## Loops and meshes

The Mesh Current Method uses two special terms: loop and mesh.
A loop is any closed path around a circuit. To trace a loop, you start at any component terminal, and trace a path through connected elements until you get back to the starting point. A loop is allowed to go through an element just one time (so you don't get loops that look like a figure-8). In the circuit above, there are three loops, two solid loops, $\text{I}$ and $\text{II}$, and one dashed loop, $\text{III}$, all the way around the outside.
If we trace the loops in the clockwise direction, the three loops in our circuit go through
$\begin{array}{rlr}& \text{Loop I:}\phantom{\rule{1em}{0ex}}& \text{V1 - R1 - R3}\\ & \text{Loop II:}\phantom{\rule{1em}{0ex}}& \text{R3 - R2 - V2}\\ & \text{Loop III: (dashed loop)}\phantom{\rule{1em}{0ex}}& \text{V1 - R1 - R2 - V2}\end{array}$
A mesh is a restricted kind of loop; a mesh is a loop that contains no other loops. In the circuit above, loops $\text{I}$ and $\text{II}$ are meshes because there are no smaller loops inside. The dashed loop is not a mesh, because it contains two other loops.
In the Mesh Current Method, we use the meshes of a circuit to generate KVL equations.

## Loop current

We now define a new term, loop current. (You can also call them mesh currents.) Up to now when we've talked about current, it has usually been in the context of an element current (current flowing through an element). When we use the term loop current we are talking about an imagined current flowing around a loop. This is a bit of an odd idea, but stick with me. In the following circuit, let us define loop currents ${i}_{\text{I}}$ and ${i}_{\text{II}}$ flowing around meshes $\text{I}$ and $\text{II}$, with the positive direction of the loop current indicated by the arrows.
It is clear that ${i}_{\text{I}}$ is the current flowing in source $\text{V}1$ and resistor $\text{R}1$. Likewise ${i}_{\text{II}}$ is the current flowing in resistor $\text{R}2$ and source $\text{V}2$. But what is going on with the current in $\text{R}3$?
Let's take a close-up look at $\text{R}3$ in the middle branch of the circuit. What is the element current flowing through $\text{R}3$?
The way the loop currents are drawn, it looks like both of them are going through $\text{R}3$, but in opposite directions. Can this be true? Yes it can, because we can use a very important concept called the principle of superposition.

## The principle of superposition

Superposition is a fancy word for add. In the case of $\text{R}3$, we are using the principle of superposition when we say the two loop currents, ${i}_{\text{I}}$ and ${i}_{\text{II}}$ add up to the actual current in the resistor, ${i}_{\text{R}3}$.
$+{i}_{\text{R}3}=+{i}_{\text{I}}-{i}_{\text{II}}$
The two loop currents superimpose (add up) to form the actual element current in $\text{R}3$. The arrow for loop current ${i}_{\text{I}}$ points in the same direction as branch current ${i}_{\text{R}3}$, giving it a $+$ sign in the superposition equation. The arrow for loop current ${i}_{\text{II}}$ points in the opposite direction, so it gets a $-$ sign in the equation.

## Linearity

The reason we get to use superposition with ideal resistors is because an ideal resistor is linear. Linearity for an ideal resistor means if we multiply the voltage by a constant $a$, then the current is multiplied by the same constant $a$.
$v=i\phantom{\rule{0.167em}{0ex}}\text{R}$
$a\phantom{\rule{0.167em}{0ex}}v=a\phantom{\rule{0.167em}{0ex}}i\phantom{\rule{0.167em}{0ex}}\text{R}$
For a real-world resistor there is a limit to how large $a$ can be before the resistor burns up. But for our ideal resistor, it works for any $a$, so an ideal resistor is linear.
Linearity means we get to use the principle of superposition. Superposition means it makes sense to have multiple loop currents flowing in one circuit element. Multiple currents means we can use loop currents as our independent variables. And that means we can use the Loop Current Method to solve circuits!
If you want to learn more about the concept of linearity, check out the main linearity article.

## Loop current practice

Problem 1
Find the element current ${i}_{\text{Rx}}$.
${i}_{\text{Rx}}=\phantom{\rule{0.167em}{0ex}}$
$\text{mA}$

problem 2
Find the element current ${i}_{\text{Ry}}$.
${i}_{\text{Ry}}=\phantom{\rule{0.167em}{0ex}}$
$\text{mA}$

The Mesh Current Method described next works on planar circuits (circuits that can be drawn flat, without crossing wires). Most of the circuits you will come across will be planar. If the circuit is non-planar (can only be drawn with crossing wires), there is a small modification to the method, and it gets the name Loop Current Method. Most of the process is the same for both methods, so let's learn the Mesh Current Method for planar circuits first. We'll talk more about the Loop Current Method in the main Loop Current Method article.

## Mesh current method

The Mesh Current Method is based on loop currents flowing around meshes. The analysis is performed with this sequence of steps:
• Identify the meshes, (the open windows of the circuit).
• Assign a current variable to each mesh, using a consistent direction (clockwise or counterclockwise).
• Write Kirchhoff's Voltage Law equations around each mesh.
• Solve the resulting system of equations for all mesh currents.
• Solve for other element currents and voltages you want using Ohm's Law.
Here is the circuit we will analyze to demonstrate the Mesh Current Method,

### Identify the meshes

Our circuit has two meshes. We identify two loop currents and call them ${i}_{\text{I}}$ and ${i}_{\text{II}}$. These are going to be our independent variables. Important: the loop current directions are the same, both flow in a clockwise direction.
By defining a loop current in every mesh, you will have enough independent equations to solve the circuit.

### Write Kirchhoff's Voltage Law (KVL) around each mesh

#### Important Mesh Current skill - scribble on the schematic!

To get ready to write KVL equations, we mark up the schematic with element voltages (orange $+$ and $-$) and element currents (green arrows), using the sign convention for passive components. I also add extra blue arrows to the loops, so I always know which direction the loop current is flowing.
• Label the element currents first (before assigning the voltages). It is a good idea draw currents so they point in the same direction as the nearest loop current. You can't always do this; we see an example where ${i}_{\text{II}}$ flows against the $1\phantom{\rule{0.167em}{0ex}}\text{k}\mathrm{\Omega }$ current arrow. This is okay, it will work out fine in the end.
• Then, label element voltages with the $+$ sign near the incoming current arrow (the passive sign convention).
Now we write an equation for each mesh using Kirchhoff's Voltage Law, (add up the voltages around a mesh and set it equal to zero). While writing a mesh equation, here is how to include the voltage terms:
• When you come to a voltage source, it enters into the equation as a voltage value.
• When you come to a resistor, express the voltage as resistance $×$ loop current. This is equivalent to doing Ohm's Law in your head.
• If two loop currents flow through a component, include their difference in the Ohm's Law expression.

#### The equation for mesh $\text{I}$‍ , step-by-step

We start in the lower left corner of the schematic, and travel clockwise around mesh $\text{I}$.
• The first element we come to is the $5\phantom{\rule{0.167em}{0ex}}\text{V}$ voltage source. We first encounter the orange $-$ sign of the voltage source. That means we will experience a voltage rise going through the source. Because it is a rise, it goes into the equation with a $+$ sign, as $+5\phantom{\rule{0.167em}{0ex}}\text{V}$.
$\text{mesh I :}\phantom{\rule{1em}{0ex}}+5\phantom{\rule{0.167em}{0ex}}\text{V}\phantom{\rule{0.167em}{0ex}}\text{…}$
• The second element we come to is the $2\phantom{\rule{0.167em}{0ex}}\text{k}\mathrm{\Omega }$ resistor. The voltage across this resistor is $2\phantom{\rule{0.167em}{0ex}}\text{k}\mathrm{\Omega }\cdot {i}_{\text{I}}$. (This is what it means to "do Ohm's Law in your head.") This resistor's current arrow is in the same direction as loop current ${i}_{\text{I}}$. The orange $+$ voltage sign tells us we will be experiencing a voltage drop going through this component, so this term goes into the equation with a $-$ sign, as $-2000\phantom{\rule{0.167em}{0ex}}{i}_{\text{I}}$.
$\text{mesh I :}\phantom{\rule{1em}{0ex}}+5\phantom{\rule{0.167em}{0ex}}\text{V}-2000\phantom{\rule{0.167em}{0ex}}{i}_{\text{I}}\phantom{\rule{0.167em}{0ex}}\text{…}$
• The next component in the loop is the $1\phantom{\rule{0.167em}{0ex}}\text{k}\mathrm{\Omega }$ resistor. It has two loop currents flowing through it, ${i}_{\text{I}}$ and ${i}_{\text{II}}$. The net current in the resistor is $\left({i}_{\text{I}}-{i}_{\text{II}}\right)$. The voltage is therefore $1\phantom{\rule{0.167em}{0ex}}\text{k}\mathrm{\Omega }\cdot \left({i}_{\text{I}}-{i}_{\text{II}}\right)$. The orange $+$ voltage sign as we enter the component tells us we will be experiencing a voltage drop going through the component, so this term goes into the equation with a $-$ sign, as $-1\phantom{\rule{0.167em}{0ex}}\text{k}\mathrm{\Omega }\cdot \left({i}_{\text{I}}-{i}_{\text{II}}\right)$.
$\text{mesh I :}\phantom{\rule{1em}{0ex}}+5\phantom{\rule{0.167em}{0ex}}\text{V}-2000\phantom{\rule{0.167em}{0ex}}{i}_{\text{I}}-1000\phantom{\rule{0.167em}{0ex}}\left({i}_{\text{I}}-{i}_{\text{II}}\right)\phantom{\rule{0.167em}{0ex}}\text{…}$
• The trip around loop $\text{I}$ is complete. All that's left is to set the sum of voltages around the loop to zero.
$\text{mesh I :}\phantom{\rule{1em}{0ex}}+5\phantom{\rule{0.167em}{0ex}}\text{V}-2000\phantom{\rule{0.167em}{0ex}}{i}_{\text{I}}-1000\phantom{\rule{0.167em}{0ex}}\left({i}_{\text{I}}-{i}_{\text{II}}\right)=0$
Summary of mesh $\text{I}$ KVL terms:

#### The equation for mesh $\text{II}$‍ , step-by-step

We start at the bottom of the $1\phantom{\rule{0.167em}{0ex}}\text{k}\mathrm{\Omega }$ resistor, and travel clockwise around the mesh.
• The first element is the $1\phantom{\rule{0.167em}{0ex}}\text{k}\mathrm{\Omega }$ resistor, and it has two loop currents flowing through it. The net current in the resistor is $\left({i}_{\text{I}}-{i}_{\text{II}}\right)$. Since we are entering this resistor from the bottom, at its orange $-$ sign, we will experience a voltage rise going through it, so the term we include in the equation is $+1000\cdot \left({i}_{\text{I}}-{i}_{\text{II}}\right)$.
$\text{mesh II :}\phantom{\rule{1em}{0ex}}+1000\phantom{\rule{0.167em}{0ex}}\left({i}_{\text{I}}-{i}_{\text{II}}\right)\phantom{\rule{0.167em}{0ex}}\text{…}$
• The next component is the $2\phantom{\rule{0.167em}{0ex}}\text{k}\mathrm{\Omega }$ resistor at the top right of the schematic, with just ${i}_{\text{II}}$ flowing through it. This is a voltage drop, so it enters the equation as $-2000\cdot {i}_{\text{II}}$.
$\text{mesh II :}\phantom{\rule{1em}{0ex}}+1000\phantom{\rule{0.167em}{0ex}}\left({i}_{\text{I}}-{i}_{\text{II}}\right)-2000\phantom{\rule{0.167em}{0ex}}{i}_{\text{II}}\phantom{\rule{0.167em}{0ex}}\text{…}$
• The last thing we come to is the $2\phantom{\rule{0.167em}{0ex}}\text{V}$ source. We deal with sources as a special case, just using their voltage value. We see a voltage drop going through this source, so it enters the equation as $-2\phantom{\rule{0.167em}{0ex}}\text{V}$.
$\text{mesh II :}\phantom{\rule{1em}{0ex}}+1000\phantom{\rule{0.167em}{0ex}}\left({i}_{\text{I}}-{i}_{\text{II}}\right)-2000\phantom{\rule{0.167em}{0ex}}{i}_{\text{II}}-2\phantom{\rule{0.167em}{0ex}}\text{V}\phantom{\rule{0.167em}{0ex}}\text{…}$
• And finish up by setting the sum around the loop to zero,
$\text{mesh II :}\phantom{\rule{1em}{0ex}}+1000\phantom{\rule{0.167em}{0ex}}\left({i}_{\text{I}}-{i}_{\text{II}}\right)-2000\phantom{\rule{0.167em}{0ex}}{i}_{\text{II}}-2\phantom{\rule{0.167em}{0ex}}\text{V}=0$
Summary of mesh $\text{II}$ KVL terms:

## Solve the system of mesh equations to find the currents

Our mesh equations copied from above:
$\text{mesh I :}\phantom{\rule{1em}{0ex}}+5\phantom{\rule{0.167em}{0ex}}\text{V}-2000\phantom{\rule{0.167em}{0ex}}{i}_{\text{I}}-1000\phantom{\rule{0.167em}{0ex}}\left({i}_{\text{I}}-{i}_{\text{II}}\right)=0$
$\text{mesh II :}\phantom{\rule{1em}{0ex}}+1000\phantom{\rule{0.167em}{0ex}}\left({i}_{\text{I}}-{i}_{\text{II}}\right)-2000\phantom{\rule{0.167em}{0ex}}{i}_{\text{II}}-2\phantom{\rule{0.167em}{0ex}}\text{V}=0$
To begin the solution process, multiply out terms and move constants to the right side,
$\text{mesh I :}\phantom{\rule{1em}{0ex}}-2000\phantom{\rule{0.167em}{0ex}}{i}_{\text{I}}-1000\phantom{\rule{0.167em}{0ex}}{i}_{\text{I}}+1000\phantom{\rule{0.167em}{0ex}}{i}_{\text{II}}=-5$
$\text{mesh II :}\phantom{\rule{1em}{0ex}}+1000\phantom{\rule{0.167em}{0ex}}{i}_{\text{I}}-1000\phantom{\rule{0.167em}{0ex}}{i}_{\text{II}}-2000\phantom{\rule{0.167em}{0ex}}{i}_{\text{II}}=+2$
Combine like terms to get a tidy system of equations,
$\text{mesh I :}\phantom{\rule{1em}{0ex}}-3000\phantom{\rule{0.167em}{0ex}}{i}_{\text{I}}+1000\phantom{\rule{0.167em}{0ex}}{i}_{\text{II}}=-5$
$\text{mesh II :}\phantom{\rule{1em}{0ex}}+1000\phantom{\rule{0.167em}{0ex}}{i}_{\text{I}}-3000\phantom{\rule{0.167em}{0ex}}{i}_{\text{II}}=+2$
Our strategy will be to eliminate ${i}_{\text{I}}$ by multiplying the second equation by $3$ and adding it to the first equation. Here's the multiplication of the mesh $\text{II}$ equation,
$\begin{array}{rl}\text{mesh II :}\phantom{\rule{2em}{0ex}}3×\left[+1000\phantom{\rule{0.167em}{0ex}}{i}_{\text{I}}-3000\phantom{\rule{0.167em}{0ex}}{i}_{\text{II}}& =+2\phantom{\rule{0.167em}{0ex}}\right]=\\ \\ \phantom{\rule{1em}{0ex}}\left[+3000\phantom{\rule{0.167em}{0ex}}{i}_{\text{I}}-9000\phantom{\rule{0.167em}{0ex}}{i}_{\text{II}}& =+6\phantom{\rule{0.167em}{0ex}}\right]\end{array}$
Now add the two equations. The ${i}_{\text{I}}$ terms cancel when we add, leaving us with just an ${i}_{\text{II}}$ term,
Loop current ${i}_{\text{II}}$ has a negative sign. That means this current flows in the opposite direction of its blue arrow.
Now we know one of the loop currents. Plug this value into either loop equation to get the other current. Let's use the equation for loop $\text{I}$,
$-3000\phantom{\rule{0.167em}{0ex}}{i}_{\text{I}}+1000\phantom{\rule{0.167em}{0ex}}{i}_{\text{II}}=-5$
$-3000\phantom{\rule{0.167em}{0ex}}{i}_{\text{I}}+1000\cdot \left(-0.125\phantom{\rule{0.167em}{0ex}}\text{mA}\right)=-5$
$-3000\phantom{\rule{0.167em}{0ex}}{i}_{\text{I}}=-5+0.125$
${i}_{\text{I}}=\frac{-4.875}{-3000}$
${i}_{\text{I}}=+1.625\phantom{\rule{0.167em}{0ex}}\text{mA}$
The two loop currents are now solved. Now we are ready to find element currents and voltages.

### Solve for other element currents and voltages

For any element carrying only one loop current, we immediately know its element current, it's the same as the loop current,
The $1\phantom{\rule{0.167em}{0ex}}\text{k}\mathrm{\Omega }$ resistor carries two loop currents, so we use superposition to find the element current,
${i}_{1\text{k}\mathrm{\Omega }}={i}_{\text{I}}-{i}_{\text{II}}=+1.625\phantom{\rule{0.167em}{0ex}}\text{mA}-\left(-0.125\phantom{\rule{0.167em}{0ex}}\text{mA}\right)=+1.75\phantom{\rule{0.167em}{0ex}}\text{mA}$
And finally we get the voltage at the node between the three resistors using Ohm's Law on the $1\phantom{\rule{0.167em}{0ex}}\text{k}\mathrm{\Omega }$ resistor,
${v}_{1\text{k}\mathrm{\Omega }}=1\phantom{\rule{0.167em}{0ex}}\text{k}\mathrm{\Omega }\cdot 1.75\phantom{\rule{0.167em}{0ex}}\text{mA}=1.75\phantom{\rule{0.167em}{0ex}}\text{V}$
All done! We analyzed a circuit using the Mesh Current Method.

## Choosing a method

Now we have two efficient methods for analyzing circuits, Node Voltage Method and Mesh Current Method. Which is the best one to use in a given situation? To choose between the two methods, count the number of meshes in the circuit and compare that to the number of nodes. Which number is smaller, meshes or nodes? It is usually best to choose the method that generates fewer simultaneous equations. If the meshes and nodes are the same, or nearly the same, you can choose the method you understand the best.

## Summary

The Mesh Current Method is an alternative to the Node Voltage Method for solving a circuit.
The steps in the Mesh Current Method are,
• Identify the meshes.
• Assign a current variable to each mesh, using a consistent direction (clockwise or counterclockwise).
• Write Kirchhoff's Voltage Law around each mesh.
• Voltage sources go in as voltages.
• Resistor voltages go in as $R×{i}_{loop}$.
• If two loop currents flow in opposite directions in a resistor, the voltage goes in as $R×\left({i}_{loop1}-{i}_{loop2}\right)$. (It's a plus instead of a minus if they move in the same direction.)
• Set the sum of voltages equal to zero. (If this is confusing, check out the KVL article.)
• Solve the resulting system of equations for all loop currents.
• Solve for any element currents and voltages you want using Ohm's Law.
If the circuit is non-planar, or there is a current source shared between two meshes, it's best to use the Loop Current Method.

## Want to join the conversation?

• All the currents is defined going around in the same direction (clockwise). It must be defined all clockwise? I mean, can I define one of them, let's say i1, counter-clockwise?
(5 votes)
• You may define the current directions as you wish. The best practice is to choose one direction for all loops, either clockwise or counterclockwise. This advice is intended to reduce the chance of making a sign mistake as you write the loop equations. Components sharing two loop currents will always have the pattern of (i1 - i2) because the loop currents will flow in opposite directions. However, you are perfectly free to assign directions as you wish, but stay alert for current direction!
(12 votes)
• Could you show an example of where there is a loop that only consists of resistors please. I'm struggling on how to set the polarities
(3 votes)
• When you begin studying circuit analysis there is often a strong desire to set the problem up (label the voltage polarities) so they are "right". It is important to let go of this desire. Instead, realize that, at the beginning of the problem, the voltage assignments have only one job: to help you create the independent mesh equations. There is no "right" and "wrong" way to label the different resistor voltages. You can do it any way you want. The critical skill is the following step, where you write the mesh equations using the polarity assignments you just made. If you do that step right, all the voltage signs will come out right at the end of the problem.

One way to test your understanding is to solve a simple circuit twice with different voltage polarity assignments. During the process you will see how each polarity assignment generates a slightly different mesh equation. In the end you should get the same answer both ways. After all, it's the same circuit both times.
(5 votes)
• −3000iI+1000⋅(-0.125mA)= -5
-5 + 0.125−3000iI=−5+0.125
I counld't understand this. why you did'nt multiply 1000 with 0.125?
(4 votes)
• Yeah I'm also annoyed when the equations are sloppy about units. I think Danielle is right. The 1000 gets multiplied by the -0.125mA to cancel out the m part.
(1 vote)
• I can not understand the relation between linearity and superposition, if the element is non linear, then we can not use the principle of superposition?
(2 votes)
• Linearity is a property of certain functions. Examples of linear functions are the i-v equations of resistors, capacitors, and inductors.

Superposition is an analysis method makes use of (exploits) the linearity property of R, L, and C.

Non-linear elements (like diodes) do not have the linearity property (duh, that's why they are called non-linear. That means you can't apply superposition to a circuit with diodes.

But we love superposition so much there's a really nice trick. Sometimes the circuits we build have non-linear devices, like transistors. If we want to use transistors to build a linear amplifier what do we do? The trick is to restrict the analog signal levels so they only use a small portion of the operating range of the non-linear transistors. Over that small range the operating characteristics look pretty much linear. Example: A diode i-v curve has a steep upward portion at about +.6v. If I restrict the voltage range to 0.6-0.65 it kind of looks like a straight line (kind of looks like a resistor). If I'm careful I can use the superposition theory on this circuit.

When you work with transistors in analog circuits you will learn to create "small-signal models". That's another way to say "operate over a small range so the device seems pretty much linear".

If you push a linear amplifier with too big of a signal it sounds bad, which we call distortion. That's the indication that you've pushed the circuit too far and the non-linear characteristic is exposed.
(3 votes)
• What happens if we have a common voltage source between the meshes?
Something like this:
https://imgur.com/1MBmBfG
I'm not exactly sure if my sollution is right.
(2 votes)
• what if we could guess the direction of one of the currents? should we continue with our guess or let it go and use the clockwise direction for all?
(1 vote)
• If you want to start off with a guess for one of your mesh currents, you can do that. I recommend you start with that mesh, and then use the same direction for all the rest. Your guess (assuming it is correct) will earn you a + sign for that mesh current.

The Mesh Current Method is set up as a robotic recipe for solving circuits, with the emphasis on consistency and getting the signs correct. It does not need a good guess to get the right answer. This won't take too long: solve your circuit two ways, with your guess, and with all clockwise. Notice how you get a few sign changes in the set of equations that emerge. And notice if it is easier or harder to write the terms where two mesh currents flow in one element.
(3 votes)
• How would one solve a more complicated loop such as the following :
https://s3.postimg.io/gfw569zjn/IMG_20160824_023012_1.jpg
I know I could start with using Ohms Law and hence start writing KVL but I don't get how to assign polarity to the obtained V.
(1 vote)
• Hello Anshuman,

I'd like to help but the drawing is unclear. From what I can tell you will need three loop equations. It looks like you have three voltage sources. I can't make out if the thing in the middle is a meter or a constant current source...

Please leave a comment below.

Regards,

APD
(3 votes)
• why should we write the loop III equation above,and then we use just two loops(I and II)?
(1 vote)
• Loop III is never written out as an equation in this article. It was mentioned at the beginning to distinguish meshes from loops.
(2 votes)
• Sir can i know what is the relationship between mesh current and branch current?
(1 vote)
• The relationship between mesh and branch (also called element) current is described above. Search for the phrase "Let's take a close-up look at R3 in the middle branch of the circuit." In the illustration you see the element current is the combination of the two mesh currents that flow through R3.

Element currents seem more "real" than mesh currents. Mesh current is a very clever computational trick at the heart of the Mesh Current Method for solving circuits. I actually enjoy using the Mesh Current Method because these currents seem so weird, and yet they always work.
(2 votes)
• Doesn't current flow from the positive terminal to the negative terminal? So, shouldn't the currents be going in the same direction through the middle resistor?
(1 vote)
• You are citing the Sign Convention for Passive Components. In the circuit above there are two mesh currents iI and iII. The oddball one is iII because it is flowing UP through the middle resistor.

The Sign Convention is not saying you can't have that; what its telling you is you better be careful when applying Ohm's Law to that resistor and iII. Because iII is flowing "backwards" through the resistor (coming in the - voltage terminal), the Ohm's Law equation has to include a negative sign in the product of (iII R).
(2 votes)