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Electrical engineering

Course: Electrical engineering > Unit 2

Lesson 3: DC circuit analysis
Science
Electrical engineering
Circuit analysis
DC circuit analysis
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Mesh current method

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

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 2, E 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, start text, I, end text and start text, I, I, end text, and one dashed loop, start text, I, I, I, end text, all the way around the outside.
If we trace the loops in the clockwise direction, the three loops in our circuit go through
Loop I:V1 - R1 - R3Loop II:R3 - R2 - V2Loop III: (dashed loop)V1 - R1 - R2 - V2\begin{aligned} &\text{Loop I:}\quad&\text{V1 - R1 - R3} \\ &\text{Loop II:}\quad&\text{R3 - R2 - V2} \\ &\text{Loop III: (dashed loop)}\quad&\text{V1 - R1 - R2 - V2} \end{aligned}
A mesh is a restricted kind of loop; a mesh is a loop that contains no other loops. In the circuit above, loops start text, I, end text and start text, I, I, end text 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, start subscript, start text, I, end text, end subscript and i, start subscript, start text, I, I, end text, end subscript flowing around meshes start text, I, end text and start text, I, I, end text, with the positive direction of the loop current indicated by the arrows.
It is clear that i, start subscript, start text, I, end text, end subscript is the current flowing in source start text, V, end text, 1 and resistor start text, R, end text, 1. Likewise i, start subscript, start text, I, I, end text, end subscript is the current flowing in resistor start text, R, end text, 2 and source start text, V, end text, 2. But what is going on with the current in start text, R, end text, 3?
Let's take a close-up look at start text, R, end text, 3 in the middle branch of the circuit. What is the element current flowing through start text, R, end text, 3?
The way the loop currents are drawn, it looks like both of them are going through start text, R, end text, 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 start text, R, end text, 3, we are using the principle of superposition when we say the two loop currents, i, start subscript, start text, I, end text, end subscript and i, start subscript, start text, I, I, end text, end subscript add up to the actual current in the resistor, i, start subscript, start text, R, end text, 3, end subscript.
plus, i, start subscript, start text, R, end text, 3, end subscript, equals, plus, i, start subscript, start text, I, end text, end subscript, minus, i, start subscript, start text, I, I, end text, end subscript
The two loop currents superimpose (add up) to form the actual element current in start text, R, end text, 3. The arrow for loop current i, start subscript, start text, I, end text, end subscript points in the same direction as branch current i, start subscript, start text, R, end text, 3, end subscript, giving it a plus sign in the superposition equation. The arrow for loop current i, start subscript, start text, I, I, end text, end subscript points in the opposite direction, so it gets a minus 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, equals, i, start text, R, end text
a, v, equals, a, i, start text, R, end text
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, start subscript, start text, R, x, end text, end subscript.
i, start subscript, start text, R, x, end text, end subscript, equals
start text, m, A, end text

problem 2
Find the element current i, start subscript, start text, R, y, end text, end subscript.
i, start subscript, start text, R, y, end text, end subscript, equals
start text, m, A, end text

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, start subscript, start text, I, end text, end subscript and i, start subscript, start text, I, I, end text, end subscript. 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 plus and minus) 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, start subscript, start text, I, I, end text, end subscript flows against the 1, start text, k, end text, \Omega current arrow. This is okay, it will work out fine in the end.
  • Then, label element voltages with the plus 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 times 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 start text, I, end text, step-by-step

We start in the lower left corner of the schematic, and travel clockwise around mesh start text, I, end text.
  • The first element we come to is the 5, start text, V, end text voltage source. We first encounter the orange minus 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 plus sign, as plus, 5, start text, V, end text.
start text, m, e, s, h, space, I, space, colon, end text, plus, 5, start text, V, end text, point, point, point
  • The second element we come to is the 2, start text, k, end text, \Omega resistor. The voltage across this resistor is 2, start text, k, end text, \Omega, dot, i, start subscript, start text, I, end text, end subscript. (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, start subscript, start text, I, end text, end subscript. The orange plus voltage sign tells us we will be experiencing a voltage drop going through this component, so this term goes into the equation with a minus sign, as minus, 2000, i, start subscript, start text, I, end text, end subscript.
start text, m, e, s, h, space, I, space, colon, end text, plus, 5, start text, V, end text, minus, 2000, i, start subscript, start text, I, end text, end subscript, point, point, point
  • The next component in the loop is the 1, start text, k, end text, \Omega resistor. It has two loop currents flowing through it, i, start subscript, start text, I, end text, end subscript and i, start subscript, start text, I, I, end text, end subscript. The net current in the resistor is left parenthesis, i, start subscript, start text, I, end text, end subscript, minus, i, start subscript, start text, I, I, end text, end subscript, right parenthesis. The voltage is therefore 1, start text, k, end text, \Omega, dot, left parenthesis, i, start subscript, start text, I, end text, end subscript, minus, i, start subscript, start text, I, I, end text, end subscript, right parenthesis. The orange plus 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 minus sign, as minus, 1, start text, k, end text, \Omega, dot, left parenthesis, i, start subscript, start text, I, end text, end subscript, minus, i, start subscript, start text, I, I, end text, end subscript, right parenthesis.
start text, m, e, s, h, space, I, space, colon, end text, plus, 5, start text, V, end text, minus, 2000, i, start subscript, start text, I, end text, end subscript, minus, 1000, left parenthesis, i, start subscript, start text, I, end text, end subscript, minus, i, start subscript, start text, I, I, end text, end subscript, right parenthesis, point, point, point
  • The trip around loop start text, I, end text is complete. All that's left is to set the sum of voltages around the loop to zero.
start text, m, e, s, h, space, I, space, colon, end text, plus, 5, start text, V, end text, minus, 2000, i, start subscript, start text, I, end text, end subscript, minus, 1000, left parenthesis, i, start subscript, start text, I, end text, end subscript, minus, i, start subscript, start text, I, I, end text, end subscript, right parenthesis, equals, 0
Summary of mesh start text, I, end text KVL terms:

The equation for mesh start text, I, I, end text, step-by-step

We start at the bottom of the 1, start text, k, end text, \Omega resistor, and travel clockwise around the mesh.
  • The first element is the 1, start text, k, end text, \Omega resistor, and it has two loop currents flowing through it. The net current in the resistor is left parenthesis, i, start subscript, start text, I, end text, end subscript, minus, i, start subscript, start text, I, I, end text, end subscript, right parenthesis. Since we are entering this resistor from the bottom, at its orange minus sign, we will experience a voltage rise going through it, so the term we include in the equation is plus, 1000, dot, left parenthesis, i, start subscript, start text, I, end text, end subscript, minus, i, start subscript, start text, I, I, end text, end subscript, right parenthesis.
start text, m, e, s, h, space, I, I, space, colon, end text, plus, 1000, left parenthesis, i, start subscript, start text, I, end text, end subscript, minus, i, start subscript, start text, I, I, end text, end subscript, right parenthesis, point, point, point
  • The next component is the 2, start text, k, end text, \Omega resistor at the top right of the schematic, with just i, start subscript, start text, I, I, end text, end subscript flowing through it. This is a voltage drop, so it enters the equation as minus, 2000, dot, i, start subscript, start text, I, I, end text, end subscript.
start text, m, e, s, h, space, I, I, space, colon, end text, plus, 1000, left parenthesis, i, start subscript, start text, I, end text, end subscript, minus, i, start subscript, start text, I, I, end text, end subscript, right parenthesis, minus, 2000, i, start subscript, start text, I, I, end text, end subscript, point, point, point
  • The last thing we come to is the 2, start text, V, end text 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 minus, 2, start text, V, end text.
start text, m, e, s, h, space, I, I, space, colon, end text, plus, 1000, left parenthesis, i, start subscript, start text, I, end text, end subscript, minus, i, start subscript, start text, I, I, end text, end subscript, right parenthesis, minus, 2000, i, start subscript, start text, I, I, end text, end subscript, minus, 2, start text, V, end text, point, point, point
  • And finish up by setting the sum around the loop to zero,
start text, m, e, s, h, space, I, I, space, colon, end text, plus, 1000, left parenthesis, i, start subscript, start text, I, end text, end subscript, minus, i, start subscript, start text, I, I, end text, end subscript, right parenthesis, minus, 2000, i, start subscript, start text, I, I, end text, end subscript, minus, 2, start text, V, end text, equals, 0
Summary of mesh start text, I, I, end text KVL terms:

Solve the system of mesh equations to find the currents

Our mesh equations copied from above:
start text, m, e, s, h, space, I, space, colon, end text, plus, 5, start text, V, end text, minus, 2000, i, start subscript, start text, I, end text, end subscript, minus, 1000, left parenthesis, i, start subscript, start text, I, end text, end subscript, minus, i, start subscript, start text, I, I, end text, end subscript, right parenthesis, equals, 0
start text, m, e, s, h, space, I, I, space, colon, end text, plus, 1000, left parenthesis, i, start subscript, start text, I, end text, end subscript, minus, i, start subscript, start text, I, I, end text, end subscript, right parenthesis, minus, 2000, i, start subscript, start text, I, I, end text, end subscript, minus, 2, start text, V, end text, equals, 0
To begin the solution process, multiply out terms and move constants to the right side,
start text, m, e, s, h, space, I, space, colon, end text, minus, 2000, i, start subscript, start text, I, end text, end subscript, minus, 1000, i, start subscript, start text, I, end text, end subscript, plus, 1000, i, start subscript, start text, I, I, end text, end subscript, equals, minus, 5
start text, m, e, s, h, space, I, I, space, colon, end text, plus, 1000, i, start subscript, start text, I, end text, end subscript, minus, 1000, i, start subscript, start text, I, I, end text, end subscript, minus, 2000, i, start subscript, start text, I, I, end text, end subscript, equals, plus, 2
Combine like terms to get a tidy system of equations,
start text, m, e, s, h, space, I, space, colon, end text, minus, 3000, i, start subscript, start text, I, end text, end subscript, plus, 1000, i, start subscript, start text, I, I, end text, end subscript, equals, minus, 5
start text, m, e, s, h, space, I, I, space, colon, end text, plus, 1000, i, start subscript, start text, I, end text, end subscript, minus, 3000, i, start subscript, start text, I, I, end text, end subscript, equals, plus, 2
Our strategy will be to eliminate i, start subscript, start text, I, end text, end subscript by multiplying the second equation by 3 and adding it to the first equation. Here's the multiplication of the mesh start text, I, I, end text equation,
mesh II :3×[+1000 iI−3000 iII=+2 ]=[+3000 iI−9000 iII=+6 ]\begin{aligned} \small\text{mesh II :}\qquad 3 \times [+1000 \, i_{\text I} - 3000\, i_{\text{II}} &= +2 \,] =\\ \\ \quad[+3000 \, i_{\text I} - 9000\, i_{\text{II}} &= +6 \,] \end{aligned}
Now add the two equations. The i, start subscript, start text, I, end text, end subscript terms cancel when we add, leaving us with just an i, start subscript, start text, I, I, end text, end subscript term,
mesh I :+[−3000 iI+1000 iII=−5 ]mesh II :+[+3000 iI−9000 iII=+6 ]sum : −8000 iII=+1iII=+1−8000iII=−0.125 mA\begin{aligned} \small\text{mesh I :}\qquad \phantom{+} [- 3000 \, i_{\text I} + 1000\, i_{\text{II}} &= -5 \,] \\ \\ \small\text{mesh II :}\qquad + [+3000 \, i_{\text I} - 9000\, i_{\text{II}} &= +6 \,] \\ \\ \small\text{sum :}\qquad\qquad\qquad\qquad\ - 8000\, i_{\text{II}} &= +1 \\ \\ i_{\text{II}} &= \dfrac{+1}{-8000} \\ \\ \\ i_{\text{II}} &= -0.125 \,\text{mA} \end{aligned}
Loop current i, start subscript, start text, I, I, end text, end subscript 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 start text, I, end text,
minus, 3000, i, start subscript, start text, I, end text, end subscript, plus, 1000, i, start subscript, start text, I, I, end text, end subscript, equals, minus, 5
minus, 3000, i, start subscript, start text, I, end text, end subscript, plus, 1000, dot, left parenthesis, minus, 0, point, 125, start text, m, A, end text, right parenthesis, equals, minus, 5
minus, 3000, i, start subscript, start text, I, end text, end subscript, equals, minus, 5, plus, 0, point, 125
i, start subscript, start text, I, end text, end subscript, equals, start fraction, minus, 4, point, 875, divided by, minus, 3000, end fraction
i, start subscript, start text, I, end text, end subscript, equals, plus, 1, point, 625, start text, m, A, end text
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,
i, start subscript, 2, start text, k, end text, \Omega, start text, space, l, e, f, t, end text, end subscript, equals, plus, i, start subscript, start text, I, end text, end subscript, equals, plus, 1, point, 625, start text, m, A, end text
i, start subscript, 2, start text, k, end text, \Omega, start text, space, r, i, g, h, t, end text, end subscript, equals, plus, i, start subscript, start text, I, I, end text, end subscript, equals, minus, 0, point, 125, start text, m, A, end text
The 1, start text, k, end text, \Omega resistor carries two loop currents, so we use superposition to find the element current,
i, start subscript, 1, start text, k, end text, \Omega, end subscript, equals, i, start subscript, start text, I, end text, end subscript, minus, i, start subscript, start text, I, I, end text, end subscript, equals, plus, 1, point, 625, start text, m, A, end text, minus, left parenthesis, minus, 0, point, 125, start text, m, A, end text, right parenthesis, equals, plus, 1, point, 75, start text, m, A, end text
And finally we get the voltage at the node between the three resistors using Ohm's Law on the 1, start text, k, end text, \Omega resistor,
v, start subscript, 1, start text, k, end text, \Omega, end subscript, equals, 1, start text, k, end text, \Omega, dot, 1, point, 75, start text, m, A, end text, equals, 1, point, 75, start text, V, end text
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, times, i, start subscript, l, o, o, p, end subscript.
    • If two loop currents flow in opposite directions in a resistor, the voltage goes in as R, times, left parenthesis, i, start subscript, l, o, o, p, 1, end subscript, minus, i, start subscript, l, o, o, p, 2, end subscript, right parenthesis. (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.

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  • hopper cool style avatar for user Gustavo Soares
    Posted 7 years ago. Direct link to Gustavo Soares's post “All the currents is defin...”
    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)
    Default Khan Academy avatar avatar for user
    • spunky sam orange style avatar for user Willy McAllister
      Posted 7 years ago. Direct link to Willy McAllister's post “You may define the curren...”
      Good Answer
      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!
      (11 votes)
  • blobby green style avatar for user babeyface01
    Posted 7 years ago. Direct link to babeyface01's post “Could you show an example...”
    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)
    Default Khan Academy avatar avatar for user
    • spunky sam orange style avatar for user Willy McAllister
      Posted 7 years ago. Direct link to Willy McAllister's post “When you begin studying c...”
      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)
  • blobby green style avatar for user tarikulislam.cse
    Posted 6 years ago. Direct link to tarikulislam.cse's post “−3000iI+1000⋅(-0.125mA)= ...”
    −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?
    (3 votes)
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  • aqualine ultimate style avatar for user Denis
    Posted 7 years ago. Direct link to Denis's post “What happens if we have a...”
    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)
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  • blobby green style avatar for user nima.jordan74
    Posted 6 years ago. Direct link to nima.jordan74's post “what if we could guess th...”
    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)
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    • spunky sam orange style avatar for user Willy McAllister
      Posted 6 years ago. Direct link to Willy McAllister's post “If you want to start off ...”
      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)
  • piceratops seedling style avatar for user anshuman.8
    Posted 7 years ago. Direct link to anshuman.8's post “How would one solve a mor...”
    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)
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    • purple pi purple style avatar for user APDahlen
      Posted 7 years ago. Direct link to APDahlen's post “Hello Anshuman, I'd like...”
      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)
  • blobby green style avatar for user Adel.mahfoud
    Posted 4 years ago. Direct link to Adel.mahfoud's post “How the current can be ne...”
    How the current can be negative?
    (1 vote)
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    • spunky sam orange style avatar for user Willy McAllister
      Posted 4 years ago. Direct link to Willy McAllister's post “At the start of the probl...”
      At the start of the problem you annotate your circuit with current arrows. The arrows establish what you will call a positive current direction. At this point you don't yet know the direction the actual currents are flowing (if you did the problem would already be solved).

      Then you dive into the circuit analysis and derive the actual current directions. The number you get for current will end up with either a positive or negative sign. If the sign comes out positive that means the actual current is flowing the same direction as the current arrow. If the sign comes out negative the current is flowing the other way, opposite the direction of the current arrow.
      (1 vote)
  • leafers ultimate style avatar for user Nahom
    Posted 3 years ago. Direct link to Nahom's post “I can't quite understand ...”
    I can't quite understand the purpose of the second voltage source(2V). Is it another battery? And if so, shouldn't the current that gets out of it travel counterclockwise instead of clockwise like i2??
    Thanks for the great lesson anyway!
    (1 vote)
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    • spunky sam orange style avatar for user Willy McAllister
      Posted 3 years ago. Direct link to Willy McAllister's post “The 2V source is just ano...”
      The 2V source is just another battery.

      The circuit analysis methods (Node Voltage, Mesh Current) are intended to be by-the-book algorithms (recipes) for solving any circuit. If you trust the steps you don't have to think about anything else. You don't have to worry that "If the 2V source points this way then I have to make sure its mesh current flows a certain direction."

      We started out curling both mesh currents in the clock-wise direction. When we got to the solution it turned out i2 had a negative value, indicating that the current flows out of the top of the 2V source, just like you would expect. There is no crime if you have a mesh current with a negative value.

      The reason we curl all the mesh currents in the same direction is to encourage correct writing of mesh equations, especially when two mesh currents flow through one resistor (the 1kohm resistor in this circuit). You get accustomed to writing that terms with a +meshcurrent and a -meshcurrent. This habit helps reduce careless sign errors. That feature is more important than trying to eliminate a negative sign from the final answer.
      (2 votes)
  • leaf green style avatar for user Biorni Greco
    Posted 5 years ago. Direct link to Biorni Greco's post “why should we write the l...”
    why should we write the loop III equation above,and then we use just two loops(I and II)?
    (1 vote)
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  • blobby green style avatar for user Colin080272
    Posted 5 years ago. Direct link to Colin080272's post “When calculating Mesh Cur...”
    When calculating Mesh Current, if a circuit has two voltage sources and four possible Mesh opening should the circuit be broken down into two equations each time?
    (1 vote)
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