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

Question. How Loop Equation 1 is written.

· We know that Voltage drop and rise in a loop is always equal to zero

· In loop 1 , we have two elements Voltage and Resistor

Analyzing the voltage across Three Elements for loop 1

Hence to Voltage Equation for loop 1 is

Voltage 1 (Across Battery )+ Voltage 2 (across Resistor)=0

=-10V+ (voltage across resistor=The value of resistor x current )=0

 

·Now if we observe diagram we will note that value of resistor is R3 which is 10 Ohm and the current through this resistor is I1-I2(as I2 is opposing I1).

· Hence the Voltage across R3= 10(I1-I2)

· So the total Equation for loop 1 becomes

Voltage 1 (Across Battery )+ Voltage 2 (across Resistor)=0

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

·-10 V + 10(I1-I2)=0

-10V+10I1-10I2=0

=10I1-10I2=10

Solving Loop 2

Question. How Loop Equation 2 is written.

· We know that Voltage drop and rise in a loop is always equal to zero

· In loop 2 , we have three elements Voltage and Resistor 1 and Resistor 3

· Hence to Voltage Equation for loop 1 is

Voltage 1 (Across Battery )+ Voltage 2 (across Resistor 1 ) +Voltage across resistor 3 =0

Analyzing the voltage across Three Elements for loop 2

From Above Expression Finalizing Equations

-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