1. ECE 221 Electric Circuit Analysis I Chapter 9 Mesh-Current Method
Herbert G. Mayer, PSU
Status 11/11/2014
For use at Changchun University of Technology CCUT

2. Syllabus Definition Circuit for Mesh-Current
First Solve Via Substitution
Example 4.4
Example 4.5
Conclusion

3. Definition
The Mesh-Current Method is similar to the substitution method exercised earlier in order to compute voltages in a circuit
It expresses voltages as a function of currents along a mesh
So why have another method?
The mesh-current method is simpler due to a smaller number of equations needed
Method is applicable only to planar circuits
Mesh-Currents are not equivalent to branch currents
They are fictitious, and therefore not always measurable with an instrument (Amp meter)

4. Definition
Mesh-Current is the current associated with any one complete path through a mesh; naturally obeying Kirchhoff’s law
Valid only in a mesh, i.e. a loop with no interior loops enclosed
Valid only in the perimeter of a mesh
Has a defined direction, indicated with arrow, and to be used consistently across mesh
If a basic element is affected by multiple mesh-currents, they all have to be accounted for in computing currents
Here an example:

5. Circuit for Mesh-Current

6. First Solve Via Substitution
Number of essential nodes is 2
Number of essential branches is 3
To compute the currents i1, i2, and i3, using substitution, there is just one independent KCL equation
So we need 2 more independent voltage equations for a solution by substitution
When complete, we compare the result with the Mesh-Current Method

7. First Solve Via Substitution
(1) KCL: i2 + i3 = i1
(2) KVL: R1*i1 + R3*i3 - v1 = 0
(3) KVL: R2*i2 - R3*i3 + v2 = 0
Solve (3) for i3 and substitute into (2) and (3):
(2)’ v1 = i1*(R1 + R3) - i2*R3
(3)’ v2 = -i2*(R2 + R3) + i1*R3
Solve Via Substitution

8. Then Solve Via Mesh-Current
(1) R1*ia + R3*(ia – ib) – v1 = 0
(2) R2*ib + v2 + R3*(ib – ia) = 0
(1)” v1 = ia*(R1 + R3) – ib*R3
(2)” v2 = -ib*(R2 + R3) + ia*R3

9. Substitution vs. Mesh-Current
i1 == ia // == stands for “identical to”
i2 == ib
i3 = ia – ib

10. Conclusion
Mesh-Current Method is simpler than Substitution

11. Example 4.4 via Mesh-Current

12. Example 4.4 via Mesh-Current
To find the power associated with each voltage source:
First find the currents, there are 2 interesting currents, the ones through the constant voltage sources
Their voltage is known, hence their power can be computed
There are b = 7 branches with unknown currents
And n = 5 nodes
Hence we need b-(n-1) = 7-(5-1) = 3 equations

13. Example 4.4 via Mesh-Current
(1) 2*ia + 8*(ia - ib) = 0
(2) 6*ib + 6*(ib - ic) + 8*(ib - ia) = 0
(3) 4*ic *(ic - ib) = 0

14. Example 4.4 via Mesh-Current
(1) 2*ia + 8*(ia - ib) = 0
(2) 6*ib + 6*(ib - ic) + 8*(ib - ia) = 0
(3) 4*ic *(ic - ib) = 0
. . .
(1’) ia = *ib / 10
(3’) ic = *ib / 10
Substitute both (1’) and (3’) in (2)

15. Example 4.4 via Mesh-Current
(1) 2*ia + 8*(ia - ib) = 0
(2) 6*ib + 6*(ib - ic) + 8*(ib - ia) = 0
(3) 4*ic *(ic - ib) = 0
. . .
(1’) ia = *ib / 10
(3’) ic = *ib / 10
Substitute both (1’) and (3’) in (2)
ia = 5.6 A
ib = 2.0 A
ic = -0.8 A

16. Next: Example 4.5 via Mesh-Current

17. Example 4.5 via Mesh-Current
Use the Mesh-Current Method to compute the power dissipated in the 4 Ω resistor
To this end, compute the currents i1, i2, and i3 in the 3 meshes
With 3 unknowns we need 3 equations
And need to express the branch current that controls the dependent voltage source as a function of the other 3 currents
Then we can express the power consumed in the 4 Ω resistor

18. Example 4.5 via Mesh-Current
20*(i1 - i3) *(i1 - i2) = 0
5*(i2 - i1) + i2 + 4*(i2 - i3) = 0
4*(i3 - i2) + 15*iα + 20*(i3 - i1) = 0
express iα as a function of the other currents
iα = i1 – i3

19. Example 4.5 via Mesh-Current
20*(i1 - i3) *(i1 - i2) = 0
5*(i2 - i1) + i2 + 4*(i2 - i3) = 0
4*(i3 - i2) + 15*iα + 20*(i3 - i1) = 0
express iα as a function of the other currents
iα = i1 – i3
(1’) 50 = 25*i1 - 5*i *i3
(2’) 0 = -5*i1 + 10*i2 - 4*i3
(3’) 0 = -5*i1 - 4*i *i3

20. Example 4.5 via Mesh-Current
20*(i1 - i3) *(i1 - i2) = 0
5*(i2 - i1) + i2 + 4*(i2 - i3) = 0
4*(i3 - i2) + 15*iα + 20*(i3 - i1) = 0
express iα as a function of the other currents
iα = i1 – i3
(1’) 50 = 25*i1 - 5*i *i3
(2’) 0 = -5*i1 + 10*i2 - 4*i3
(3’) 0 = -5*i1 - 4*i *i3
i2 = 26 A i3 = 28 A
P4 = 2*2*4 = 16 W