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### Course: Electrical engineering > Unit 2

Lesson 3: DC circuit analysis- Circuit analysis overview
- Kirchhoff's current law
- Kirchhoff's voltage law
- Kirchhoff's laws
- Labeling voltages
- Application of the fundamental laws (setup)
- Application of the fundamental laws (solve)
- Application of the fundamental laws
- Node voltage method (steps 1 to 4)
- Node voltage method (step 5)
- Node voltage method
- Mesh current method (steps 1 to 3)
- Mesh current method (step 4)
- Mesh current method
- Loop current method
- Number of required equations
- Linearity
- Superposition

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# Loop current method

The Loop Current Method is closely related to the Mesh Current Method. Use it for two special cases: non-planar circuits, and when a current source is shared between two meshes.

## Introduction

The Loop Current Method is a small variation on the Mesh Current Method. It accounts for two special cases that are bothersome for the Mesh method. In this article we describe the special cases and show how to deal with them using the Loop method.

The Loop Current Method, just like the Mesh Current Method, is based on Kirchhoff's Voltage Law (KVL).

### What we're building to

The two special cases are a non-planar circuit (one that can't be drawn without crossing wires) and a circuit with a current source shared between two meshes.

To analyze circuits like this, you include equations for some non-mesh loops. Make sure every loop includes a circuit element that is not part of any other loop. The steps in the Loop Current Method are otherwise the same as the Mesh Current Method.

## Special case: non-planar circuit

The Mesh Current Method defines equations based on meshes. This works for circuits that are

*planar*.### Planar vs. non-planar

- A circuit is
*planar*if it can be drawn on a flat surface without crossing wires. All the schematics you have seen up to now are planar. The schematic below on the left is planar. For planar circuits, we use the Mesh Current Method and write the equations based on*meshes*. This always works for planar circuits. - A
*non-planar*circuit is shown below on the right. It has has to be drawn with at least one crossing wire, meaning it cannot be drawn flat. Since there is no way to redraw the circuit to avoid a crossing wire, the circuit on the right is non-planar.

When faced with a non-planar circuit, we

*must*use the Loop Current Method (described below).## Another special case: current source shared by two meshes

A second special case comes up when you have a current source shared between two meshes. This is another time when you may want to include a non-mesh loop in the system of equations.

Both mesh ${\text{I}}$ and mesh ${\text{II}}$ go through the current source. It is possible (but irksome) to write and solve mesh equations for this configuration. (Try it and see what it's like. It is quite awkward to figure out the voltage at the node above the current source.)

This is a time when you might want to use a loop. You can drop one of the meshes and replace it with the loop that goes around both meshes, as shown here for loop ${\text{III}}$ .

You then solve the system of equations exactly the same as the Mesh Current Method.

You may see loop ${\text{III}}$ referred to as a

*supermesh*.## Selecting loops

We can make a small adjustment to the Mesh Current Method to help us with the two special cases: We allow loops to participate in the equation-building step (not just meshes). This isn't a big deal. When selecting which loops to include:

- Make sure every element is included in a loop or mesh. Every element needs to have a chance to influence the solution.
- Make sure at least one element in each loop is not part of any other loop or mesh. This assures the loop equations are independent.

These rules generate just the right number of independent equations to solve the circuit.

## Loop Current Method

The Loop Current Method is a small variation on the Mesh Current Method. The changes are highlighted in this list of steps.

- Identify the meshes, (the open windows of the circuit)
**and loops (other closed paths)**. - Assign a current variable to each mesh
**or loop**, using a consistent direction (clockwise or counterclockwise). - Write Kirchhoff's Voltage Law equations around each mesh
**and loop**. - Solve the resulting system of equations for all mesh
**and 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 by two meshes, it is beneficial to switch to the Loop method. Just make sure every loop includes a circuit element that is not part of any other loop.

## Want to join the conversation?

- Could you show how equation would look like with 1 mesh and 1 loop?

For very last pircutre let's say that our battery is 5V, R1 = 1k Ohms, R2 = 3k Ohms, R3 = 4k Ohms and current source = 2mA.

Equation for mesh current = V -R1*i1 - Is(?)

Equation for loop current = V - R1*i2 - R2*i2 - R3*i2

i1 is current in the mesh and i2 is current in the loop.

And if we write resistors as R * I for KVL, how do we write Is? Is*Rtotal?(5 votes)- Willy

You are not teaching Mesh Analysis correctly, you cannot create a variable V2 for an ideal current source as there is no relationship between the current through the source and the voltage developed across it. The correct way to solve this circuit is using a Super-Mesh (not a Super-Node) and write the KCL equation that defines the Super-Mesh.

For this problem let i1 be the loop current in the left loop and i2 the loop current in the right loop. Then applying KCL at the top middle node gives:

-i1 +Is +i2 = 0

Choose i1 or i2 to be the unknown for the Super-Mesh, choose i1 (it is arbitrary) and then solve the Super-Mesh equation for i2 => i2 = i1 - Is

Write KVL for the Super-Mesh:

5V - i1*R1 - i2*R2 - i2*R3 = 0

Substitute the equation for i2:

5V - i1*R1 - (i1 - Is)*R2 - (i1 - Is)*R3 = 0

Solve 1 equation for one unknown. This is the proper way to solve circuits using Mesh Analysis.

Creating a variable V2 is wrong. You are creating a Hybrid method combining Mesh and Branch Analysis and hence no longer doing Mesh Analysis. This will also most likely lead to more mistakes.(1 vote)

- Could we still use KVL and KCL for a circuit with motors in it, in addition the resistors? Should we consider their counter-electromotive force in the same way as a battery, then? And how would we know whether to add or subtract its voltage?

Let's say we had a circuit just like the one in the second image, but with a motor instead of a current source. In that case, would the method of analysis have to be different?

An example would be very useful, thank you!!(2 votes)- It is important to note that KVL and KCL, also known as conservation of charge and current, is always true. Superposition though, is not true in non-linear components I believe.(1 vote)

- The loop current method and node voltage method are both pretty universal. So how do we determine when to utilize which one?(2 votes)
- Great question. The first thing I do is compare the number of nodes to the number of meshes. If one of them is smaller, that method will produce the fewest independent equations to solve. It the numbers come out close, I choose the method I like best (that's usually the one I'm less likely to make a mistake.)(5 votes)

- When analyzing a non planar circuit is there a method to determine what would be the best spanning tree that we can select.(2 votes)
- I'm not aware of a way to pick a "best" set of loops for a non-planar circuit. Any set that gives you a system of independent equations is "good" enough to solve the circuit.

On a practical note, you might attack a small circuit with a by-hand analysis, in which case any system of equations you develop will not be much worse than the "best" (if there is one). Any circuit complex enough to merit searching for a "best" spanning tree would be complicated enough that you would be unwise to attack by hand. You would instead resort to simulation tools that sort themselves out according to their own internal logic.(3 votes)

- So we pick loops that account for every element and that each have at least one unique element, right? So for your non-planar example, we could pick three loops. Loop 1 would go around the square, loop 2 would go through a diagonal and a wire outside of the square, and loop 3 would go through the other diagonal and the other wire outside of the square. All elements would be included in exactly one loop. But would 3 equations really be enough to solve a circuit with a voltage source and 9 resistors?(2 votes)
- Have you tried to solve it this way? Give the resistors some simple values and see if you can do it. Visit spinningnumbers dot org/circuit-sandbox/index.html and build the circuit in the simulator. Does your answer match the simulator?(3 votes)

- may i know how to determine the voltage of the current source in order to write KVL equation for mesh I ?(3 votes)
- You don't put a current source as a voltage source in the equation. You would claim the current source once as one of your I values, which would be the give value in A or mA.(1 vote)

- can you draw non-planar example's meshes and loops please?...(2 votes)
- Can KVL and KCL be applied to non-conservative electric fields i.e. the field produced by a varying magnetic field? Also, can these rules be applied to circuits including inductors?(2 votes)
- Hello Vishnu,

For non-conservative fields you will need Faraday's Law. This video will explain:

https://www.youtube.com/watch?v=nGQbA2jwkWI

This is a subtle distinction. Most electrical engineer types simplify the circuit by lumping these time varying magnetic field and calling them inductors or transformers.

Yes, KVL and KCL apply to AC circuits with reactive components such as inductors and capacitors. Ref:

https://www.khanacademy.org/science/electrical-engineering/ee-circuit-analysis-topic/ee-ac-analysis/v/ee-ac-analysis-intro1

Regards,

APD(1 vote)

- If you wanted to solve the current source problem without using a supermesh then how would you do the KVL equation through the current source?

The two equations I have are:

1) V_bat - R_1*i_1 - ? = 0

2) -R_2*i_2 - R_3*i_2 + ? = 0

I'm unsure of what to put for the "?".(1 vote)- The "?" term is the voltage across the current source. Make up a name for it, like VS or anything else.

When you insert the made-up voltage name and add equations 1) and 2) the made-up voltage vanishes. What is left is a single KVL equation going all the way around the outside. You've basically derived something like the supermesh equation from first principles.

But this leaves you with 1 equation in 2 unknowns (i_1 and i_2). You need one more equation to solve the circuit. Do you think you can create a KCL equation at the node above the current source? Does that give you the second equation you need?

As you can see from this, the shared current source is a trouble maker. You end up falling back on your wits to come up with a system of equations. That's okay. You have the wits to succeed.(2 votes)

- If a loop had an independent current source in it would that be the current for the loop? For example if loop 1 had a 2mA current source would the I equal 2ma(1 vote)
- You are correct. If you are using the Mesh Current method and you have a current source participating in one of the meshes, that current source defines that mesh current.

You have to be careful with circuits that have current sources. If you have a current source participating in two meshes then the Mesh Current Method is not the best choice for solving the circuit.

If you are doing the Loop Method the same answer holds. If one of your loops passes through a current source, that defines the loop current.(2 votes)