# Mesh law in a resistor circuit with current and voltage sources

• Engineering
• PinguZaide
In summary: The idea of the supermesh is to write a loop (KVL) equation for a loop that doesn't include the current source, yet includes the components of the loops that it borders. This avoids introducing the current source potential difference as an unknown, but still allows you to "touch" all the components of the circuit. For the circuit under discussion, the following path in green would satisfy that requirement: Note that you still write the loop equation using the original mesh currents. The constraint equation relating the individual loop current to the current source is the only additional equation you need, so in this case, iC - iA = 2.
PinguZaide
Hello everyone. Hope somebody can help.

## Homework Statement

We are given the following circuit.
http://puu.sh/70uaQ.png
We need to find the currents i1, i2 and i3 by explicitly using the mesh law.

## Homework Equations

There are no relevant equations.

## The Attempt at a Solution

So, I draw these 4 meshes on the circuit.

Let's call iA the current in mesh A, iB the current in mesh B, and so on.
Let's also call v_I0 the voltage at I0 and v_I1 the voltage at I1. For each current source, we assume the current enters at + and leaves at -.

Considering we already know the current in mesh D, there is no need to build an equation for that one. We only need to build the equations for meshes A, B and C.

We know that iD = 4A and that iC - iA = 2A, so that iC = 2 + iA.

I ended up with those equations.

Mesh A
12 = 2 * (iA) + 2 * (iA - iB) - v_I0

Mesh B
-6 = 2 * (iB - 4) + 2 * (iB - iA) + 2 * iB

Mesh C
24 = 1 * iC + v_I0 + 1 * (iC - 4)

After some algebra, and by substituing iC = 2 + iA, we end up with 3 equations and 3 unknowns in iA, iB and v_I0.

I end up finding iA = 6.875A, iB = 2.625A and v_I0 = 10.25V, so iC = 8.875.

Therefore, the currents that we want to know are :

i1 = iA = 6.875A
i2 = iA - iB = 4.25A
i3 = iD = 4A

The problem I have is that I used an online tool called DoCircuits to simulate the circuit, and by the result of the run, I would be suppose to have :
i1 = -4.375A, i2 = 5.25A and i3 = 4A
(Link to the circuit : http://www.docircuits.com/my-circuit/7691/2)

I therefore think there must be something I did wrong...

Thanks to anyone who helps me. :)

Last edited by a moderator:
The voltage across 2(iB+4), not 2(iB-4).

.Scott said:
The voltage across 2(iB+4), not 2(iB-4).
Can I ask why ? The way I've put the currents (all clockwise), iB meets iD at R5 in opposite directions, no ? Is it wrong to put the current iD in that direction ? If I take iB+4, would that mean that I also have to take iC+4 and not iC-4 at R3 ?

And it does not change the fact that I would still have a positive current for iA, when I'm getting a negative one with the sim... Unless I just put the ammeter in the wrong direction too.

PinguZaide, have you been introduced to the concept of the supermesh yet? If so, you can avoid having to introduce a potential difference variable for the current supply.

Also take note that the current dource in loop D flows contrary to your defined loop current direction, so iD should be negative.

1 person
.Scott said:
The voltage across 2(iB+4), not 2(iB-4).

gneill said:
PinguZaide, have you been introduced to the concept of the supermesh yet? If so, you can avoid having to introduce a potential difference variable for the current supply.

Also take note that the current dource in loop D flows contrary to your defined loop current direction, so iD should be negative.
So I tried by switching the direction of the current so that iD is not negative, and found out I made an algebra mistake somewhere, and now it works. :)

And no I have not been introduced to the concept of supermesh, but I heard of it. Is it simply using a mesh that runs around two or more "independant" mesh? I know my solution now works, but I'm curious as to where I should have put that "supermesh"?

Thanks for your input guys. :)

PinguZaide said:
And no I have not been introduced to the concept of supermesh, but I heard of it. Is it simply using a mesh that runs around two or more "independant" mesh? I know my solution now works, but I'm curious as to where I should have put that "supermesh"?
The idea of the supermesh is to write a loop (KVL) equation for a loop that doesn't include the current source, yet includes the components of the loops that it borders. This avoids introducing the current source potential difference as an unknown, but still allows you to "touch" all the components of the circuit. For the circuit under discussion, the following path in green would satisfy that requirement:

Note that you still write the loop equation using the original mesh currents. The constraint equation relating the individual loop current to the current source is the only additional equation you need, so in this case, iC - iA = 2.

#### Attachments

• Fig2.gif
20.4 KB · Views: 695
1 person
I see. I will remember that for further circuit analysis that I will need to do. Thanks a lot, gneill.

## 1. What is the concept of mesh law in a resistor circuit?

Mesh law, also known as Kirchhoff's voltage law, states that the algebraic sum of voltages in a closed loop in a circuit must equal zero. It is used to analyze the voltage drops and currents in a circuit with multiple loops.

## 2. How does a current source affect mesh law in a resistor circuit?

A current source does not affect mesh law, as it is a type of ideal source that maintains a constant current regardless of the voltage across it. Therefore, the voltage drop across a current source can be considered zero and does not need to be included in the mesh law calculations.

## 3. Can mesh law be applied to circuits with voltage sources?

Yes, mesh law can be applied to circuits with voltage sources. In this case, the voltage source is considered to be a part of the loop and its polarity (positive or negative) must be taken into account when applying the law.

## 4. What is the difference between mesh law and nodal analysis?

Mesh law and nodal analysis are two methods used to analyze circuits. While mesh law focuses on the voltage drops in a closed loop, nodal analysis focuses on the currents at different nodes in the circuit. They both follow the principles of Kirchhoff's laws and can be used together to solve complex circuits.

## 5. Are there any limitations to using mesh law in resistor circuits?

Mesh law can only be applied to circuits with ideal voltage and current sources. It also assumes that the resistors are linear and have a constant resistance. Additionally, it can become more complex and difficult to apply in circuits with a large number of loops.

• Engineering and Comp Sci Homework Help
Replies
7
Views
2K
• Engineering and Comp Sci Homework Help
Replies
8
Views
1K
• Engineering and Comp Sci Homework Help
Replies
5
Views
1K
• Introductory Physics Homework Help
Replies
9
Views
1K
• Engineering and Comp Sci Homework Help
Replies
7
Views
2K
• Engineering and Comp Sci Homework Help
Replies
4
Views
6K
• Engineering and Comp Sci Homework Help
Replies
8
Views
2K
• Engineering and Comp Sci Homework Help
Replies
26
Views
2K
• Engineering and Comp Sci Homework Help
Replies
9
Views
1K
• Engineering and Comp Sci Homework Help
Replies
62
Views
10K