Step (2): Apply KVL to super mesh and to other meshes
Applying KVL to this super mesh (combination of meshes 1 and 2 ) we get
R1.I1 + R3 ( I2 – I3) = V..............(1)
Applying KVL to mesh 3, we get
R3 ( I3 – I2) + R4.I3 = 0............(2)
Step (3): Make the relation between mesh currents with current source to get third equation.
Third equation is nothing but he relation between I , I1 and I2 which is
I1 - I2 = I............(3)
Step(4) : Solve the above equations to get the mesh currents.
Malla Reddy College of Engineering and Technology (MRCET)
Department of EEE ( 2017-18 )
Electrical Circuits EEE Example(1) : Determine the current in the 5 Ω resistor shown in the figure below.
Solution: Step(1): Here the current source exists between mesh(2) and mesh(3).Hence, super mesh is the
combination of mesh(2) and mesh(3) .Applying KVL to the super mesh ( combination of mesh 2
and mesh 3 after removing the branch with the current source of 2 A and resistance of 3 Ω ) we
get :
10( I2– I1) + 2.I2 + I3 + 5( I3 – I1) = 0
-15.I1 +12 I2 + 6.I3 = 0...................(1)
Step (2): Applying KVL first to the normal mesh 1 we get :
10( I1 – I2) + 5( I1 – I3) = 50
15.I1 –10. I2 – 5.I3 = 50....................(2)
Step (3): We can get the third equation from the relation between the current source of 2 A , and
currents I2 & I3 as :
I2 - I3 = 2 A...................(3)
Step (4): Solving the above three equations for I1, I2 and I3 we get I1 = 19.99 A I2 = 17.33 A
and I3 = 15.33 A
The current in the 5 Ω resistance = I1 - I3 = 19.99 - 15.33 = 4.66 A
Example(2): Write down the mesh equations for the circuit shown in the figure below and find
out the values of the currents I1, I2 and I3
Malla Reddy College of Engineering and Technology (MRCET)
Department of EEE ( 2017-18 )