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

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

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)

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:

Circuit for Mesh-Current

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

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

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

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

Conclusion Mesh-Current Method is simpler than Substitution

Example 4.4 via Mesh-Current

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

Example 4.4 via Mesh-Current (1) 2*ia + 8*(ia - ib) - 40 = 0 (2) 6*ib + 6*(ib - ic) + 8*(ib - ia) = 0 (3) 4*ic + 20 + 6*(ic - ib) = 0

Example 4.4 via Mesh-Current (1) 2*ia + 8*(ia - ib) - 40 = 0 (2) 6*ib + 6*(ib - ic) + 8*(ib - ia) = 0 (3) 4*ic + 20 + 6*(ic - ib) = 0 . . . (1’) ia = 4 + 8*ib / 10 (3’) ic = -2 + 6*ib / 10 Substitute both (1’) and (3’) in (2)

Example 4.4 via Mesh-Current (1) 2*ia + 8*(ia - ib) - 40 = 0 (2) 6*ib + 6*(ib - ic) + 8*(ib - ia) = 0 (3) 4*ic + 20 + 6*(ic - ib) = 0 . . . (1’) ia = 4 + 8*ib / 10 (3’) ic = -2 + 6*ib / 10 Substitute both (1’) and (3’) in (2) ia = 5.6 A ib = 2.0 A ic = -0.8 A

Next: Example 4.5 via Mesh-Current

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

Example 4.5 via Mesh-Current 20*(i1 - i3) - 50 + 5*(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

Example 4.5 via Mesh-Current 20*(i1 - i3) - 50 + 5*(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*i2 - 20*i3 (2’) 0 = -5*i1 + 10*i2 - 4*i3 (3’) 0 = -5*i1 - 4*i2 + 9*i3

Example 4.5 via Mesh-Current 20*(i1 - i3) - 50 + 5*(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*i2 - 20*i3 (2’) 0 = -5*i1 + 10*i2 - 4*i3 (3’) 0 = -5*i1 - 4*i2 + 9*i3 i2 = 26 A i3 = 28 A P4 = 2*2*4 = 16 W