Mesh Analysis

Mesh analysis is a method used to determine the values of a circuit . Lets supposed we have a circuit as per given below

For mesh analysis it is important that we determine the loops in circuit. all loops should be in one direction

Now determine the loop direction in terms of current

In this circuit we have two loops and we have made the same directions

One thing that is to be notice here is that current is opposing each other at one branch

Note that current presented by blue and red color is opposing each other .

Solving Circuits with Mesh Analysis

Step 1

Step 2

Here we have two loops

1. Loop 1

2. Loop 2

Step 3

Lets say the current in loop 1 is I1

And current in loop 2 is I2

Solving Loop 1

-10 V + 10(I1-I2)=0

-10V+10I1-10I2=0

=10I1-10I2=10

Solving Loop 2

-5V+10(I2)+10(I2-I1)=0

-5V+10I2+10I2-10I1=0

-5v-10I1+0=0

-10I1=5

Computing Values

I1=-0.5 AMPERE

I2=-1.5 AMPERE

Now solving Circuit

Voltage across R1= R1xI2=10*-1.5=-15 V

Voltage across R2= R2(I1-I2)=10(-0.5+1.5)=10V

Example 2

First, label each mesh with a mesh current.

Now write KVL equations for each loop.

KVL for i1:

-18V + 5(i1-i2) + 4(i1-i3) + 1(i1) = 0

10i1 – 5i2 – 4 i3 -18V = 0

KVL for i2:

-5i1 + 10i2 – 3i3 – 12 = 0

KVL for i3:

-4i1 -3i2 +9i3 = 0

Now solve the three equations in three unknowns:

i1 = 7.02A
i2 = 6.28A
i3 = 5.21A

The current through the 5 ohm resistor is just i1 – i2, or 0.74A. The current through the 18V is i1, or 7.02A. All the branch currents are shown below from a Pspice simulation:

Solve for the current through the 5 ohm resistor and the current through the 4V source using Mesh-Current Analysis

First, define a mesh current around each mesh

 

 

 

Now we are ready to write the KVL equations.

KVL loop 1:

-V5 + 5(i1-i2) = 0

KVL loop 2:

5(i2-i1)+10(i2-i4)+2(i2-i3) = 0

-5i1 + 17i2 – 2i3 = 0

KVL loop 3:

2(i3-i2) + V2 = 0

KVL loop 4:

10i4 + 4V + 10(i4-i2) = 0

-10i2 + 20i4 = -4

We have two extra unknowns, so we need two more equations. These can be found by examining the loops with the sources:

i1= 5A

i3 = -2A

We can then solve for the remaining unknowns:

i2 = 1.58A

i4 = 0.59A

We can then find the current through the 5 ohm resistor as:

I = i1 – i2 = 3.42A

The current through the 4V source is simply i4, or 0.59A.