mesh analysis

We already have a background in solving circuits using the mesh analysis. We are now ready to put things more complicated using the complex number as what we had tackled previously.

Mesh analysis

Mesh is a loop which does not contain any other loop within it.

                In solving with mesh analysis, our basis is the Kirchhoff's voltage law or (KVL). This will give us a set of equations that we will solve to find the mesh currents. Once we find the mesh currents we can use them to calculate any other currents or voltages. Remember that in using KVL in every mesh we have to make all resistors to be positive. And for the voltages that are passed by the loops, there are two signs shown, so you have to get the first sign. You may wonder why I solve that way.Iit is because I want to lessen the negative signs so that I may not be confused in distributing signs.

Remember again that we have to consider that there are some elements that contain two meshes but we don't need to explain it further for i know that you already taken circuits 1.

(See problem below to see how it works).

Example Problem:

For the circuit above our task is to get the current across the 4 Ω resistor.

Loop 1, loop 2 and loop 3 are what we called the meshes. We will follow this loop to get the equations from the circuit and to solve for the current which is Io = - I2 or I2 = -Io.

Negative because the direction of I2 is opposite to the direction of our Io.

We will start,

              @ Mesh 1

           

                  8 I1 + j10 I3 - j2 I1 + j2  I2 = 0

                  I1 (8+j8) + j2   I2 - j10 I3 = 0   ------> equation 1

             @ Mesh 2

             

                  -j2 I2 + j2 I1 - j2 I2 + j2 I1 + 4 I2 + 20∠90 = 0

                    I2 (j4-4) -  - j2 I1  - j2 I3  =  20∠90 ----------> equation 2

                 

             

              @ Mesh 3

                   I3  =  5∠0 A            ------------->  equation 3

             

                   They have the same direction so current is positive.

              We already have I3 so we will substitute  equation 3 to  equation 1 and  equation 2 since the two equation contain I3.

          Our new equations are:

                 I1 (8+j8) + j2   I2  = j50   ------> equation 1

                  I2 (j4-4)  - j2 I1   =  j30 ----------> equation 2

Using matrix,

              (8+j8)      j2     =  j50

               - j2       (j4-4)   =  j30 

   

            Δ =  (8+j8)(j4-4) - ( - j2)( j2)

            Δ =  - 68

The required value is the Io so we have to get the I2. we will solve only I2. We don't need I1.

            Δ2 =  (8+j8)(j30 ) - ( - j2 ) (j50)

Yields,

           Δ2 =  -340 + j240

So,

              I2 =  (Δ2 / Δ)

              I2 =  ( -340 + j240)  /   (- 68)

              

              I2 =   6.12∠-35.22

                       Io =   6.12∠-35.22



              I've learned that In mesh analysis, from previous discussion, its basis is still the Kirchhoff's Voltage Law but the current, voltages and impedance are expressed in complex numbers and the voltage source and current source is expressed in phasor or time domain.

            As I have said from nodal analysis, in solving mesh analysis, it has the same steps in solving circuits. The only difference is it gets more complicated because there are sometimes you will confused yourself because it makes longer equation to write, additional variables and has two terms that is assigned in one element. In super mesh has the same step from previous discussion of mesh analysis we only need patience and apply the basic to more complicated problems.