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Chapter: Electrical and electronics - Circuit Theory

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Mesh Analysis

This is an alternative structured approach to solving the circuit and is based on calculating mesh currents.

MESH ANALYSIS:

This is an alternative structured approach to solving the circuit and is based on calculating mesh currents. A similar approach to the node situation is used. A set of equations (based on KVL for each mesh) is formed and the equations are solved for unknown values. As many equations are needed as unknown mesh currents exist.

 

Step 1: Identify the mesh currents

Step 2: Determine which mesh currents are known

 

Step 2: Write equation for each mesh using KVL and that includes the mesh currents Step 3: Solve the equations

 

Step 1:

The mesh currents are as shown in the diagram on the next page

 

Step 2:

Neither of the mesh currents is known


 

Step 3:

KVL can be applied to the left hand side loop. This states the voltages around the loop sum to zero.

 

When writing down the voltages across each resistor equations are the mesh currents.

 

I1R1 + (I1 - I2) R4 - V = 0

KVL can be applied to the right hand side loop. This states the voltages around the loop sum to

 

zero. When writing down the voltages across ea the equations are the mesh currents.

 

I2R2 + I2R3 + (I2 - I1) R4 = 0

Step 4:

Solving the equations we get

 


 

The individual branch currents can be obtained from the these mesh currents and the node voltages can also be calculated using this information. For example:


 

Problem 1:

 

Use mesh-current analysis to determine the current flowing in (a the 1Ω resistance of the d.c. circuit shown in


The mesh currents I1, I2 and I3 are shown in Figure

Using         Kirchhoff’s   voltage   law:

For loop 1, (3 + 5) I1 −I2 = 4   …………………………………………………………(1)

For loop 2, (4 + 1 + 6 + 5) I2 −(5) I1 −(1) I3 = 0…………………………………….(2)

For loop 3, (1 + 8) I3 −(1) I2 = 5− ……………………………………………………(3)

Thus

 

8I1 −5I2 −4 =0

 −5I1 + 16I2 −I3 =0

 −I2 + 9I3 + 5 =0




(a) Current in the 5 Ωresistance = I1 −I2

= 0.595 −0.152 

= 0.44A

(b) Current in the 1 Ωresistance = I2 −I3

= 0.152 −(−0.539) 

= 0.69A

 

Problem 2: For the a.c. network shown in Figure determine, using mesh-current analysis, (a) the mesh currents I1 and I2 (b) the current flowing in the capacitor, and (c) the active power delivered by the 1000◦V voltage source.


(a) For the first loop

(5−j4) I1 −(−j4I2) =1000◦……………………………………………………(1) For the second loop

(4+j3−j4)I2 −(−j4I1) =0   ………………………………………………………   (2)

 

Rewriting equations (1) and (2) gives:

 (5 −j4)I1 + j4I2 −100 =0

 

j4I1 + (4 −j) I2 + 0 =0 Thus, using determinants,


Thus total power dissipated = 579.97 + 436.81 = 1016.8W = 1020W

 

Problem 3: Calculate   current   through  -6Ω resistance   u


Case(1): Consider loop ABGH ; Apply KVL .

 


D3 = 6(220) +4(-80) +10(-24)

D3 = 760

 

I1 = D1/D = 260/284 = 0.915A

I2 = D2/D = -320/284 = -1.1267A I3 = D3/D = 760/284 = 2.676A

Current through   6Ω 2 +Iresistance3    =   I

= -1.1267+2.676 = 1.55A

 

Problem 4: Find the current through branch a-b using mesh analysis. 


 

Solution:                                 

                                      

Consider loops                                 

Loop HADE  - > 5I1+2I2+6(I2-I3) = 60

                             5I1+8I2-6I3 = 60 ------------ (1)

Loop ABCDA  - > 3I3+6(I3-I2) = -50

                                                                             3I3+6I3-6I2 = -50

                                                                             9I3-6I2 = -50---------- (2)

I2-I1 = 5A ------------------------- (3)

From (1), (2) & (3).


 

=     -1(-400+360)-(-250) +5(-30) 

=         40+250-150 

D3 =   140.

I3 = D3/D = 140/-81 = -1.7283

The current through branch ab is 1.7283A which is flowing from b to a.

 

 

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