MESH CURRENT METHOD

Mesh analysis (or the mesh current method) is a method that is used to solve planar circuits for the currents (and indirectly the voltages) at any place in the circuit. Planar circuits are circuits that can be drawn on a plane surface with no wires crossing each other. A more general technique, called loop analysis (with the corresponding network variables called loop currents) can be applied to any circuit, planar or not. Mesh analysis and loop analysis both make use of Kirchhoff’s voltage law to arrive at a set of equations guaranteed to be solvable if the circuit has a solution. Mesh analysis is usually easier to use when the circuit is planar, compared to loop analysis.

STEPS INVOLVED TO SOLVE 

Define a loop current around each loop in clockwise direction (although it could be arbitrary). Assume there are  independent loops in the circuit, then we have  loop currents as the unknown variables.

Apply KVL around each of the loops in the same clockwise direction to obtain  equations. While calculating the voltage drop across each resistor shared by two loops, both loop currents (in opposite positions) should be considered.

Solve the equation system with  equations for the  unknown loop currents.                                                                                                                                                                                                                    

What are some difficulties I might run into?

A current source in a mesh-current analysis can be a headache. If there is a current source between two meshes, then you must add an extra variable (call it Vx, the voltage across the source). Then you will need an extra equation as well. The extra equation is simply that the difference between the two mesh currents is the value of the source.
A dependent source can be a headache. Simply write the voltages as you normally would for an independent source (but writing down the dependency for the value). Then write an extra equation to describe the dependency in terms of the mesh currents.                                                                                                                                                                                                                                                                         LET US SOLVE SOME PROBLEMS                                                                                                                                                                                                                                                                                                                       

QUES1  Solve for the current through the 5 ohm resistor and the current through the 18V source using mesh current Analysis. First, label each mesh (window pane) with a mesh current. For consistency, make each mesh in a clock-wise direction. Now write KVL equations for each loop. KVL for i1: -18V + 5(i1-i2) + 4(i1-i3) + 1(i1) = 0 then gather terms: 10i1 - 5i2 - 4 i3 -18V = 0 Note that the i1 term is positive, and all other current terms are negative (because they are all clockwise, all other panes will contribute a negative term). Let's do the other two panes with terms gathered up directly (write the total resistance of the loop multiplied by the mesh current that goes through that total resistance): KVL for i2: -5i1 + 10i2 - 3i3 - 12 = 0 KVL for i3: -4i1 -3i2 +9i3 = 0 Now solve the three equations in three unknowns: i1 = 7.02A i2 = 6.28A i3 = 5.21A The current through the 5 ohm resistor is just i1 - i2, or 0.74A. The current through the 18V is i1, or 7.02A. All the branch currents are shown below from a Pspice simulation:

 

QUES2 Solve for the current through the 5 ohm resistor and the current through the 4V source using mesh current Analysis.

First, define a mesh current arround each mesh (window pane) of the circuit. Define each one in a clockwise direction.

Now write KVL equations for each loop. On the first loop, we run into a problem. We do not know the voltage across the 5A source. (This is one of the difficulties you can run into when using mesh current.) We add a new unknown to handle this problem. We run into the same problem with the 2A source, so two new unknowns (V5 and V2) are labeled:

Now we are ready to write the KVL equations.

KVL loop 1:

-V5 + 5(i1-i2) = 0
KVL loop 2:

5(i2-i1)+10(i2-i4)+2(i2-i3) = 0
or
-5i1 + 17i2 - 2i3 = 0
KVL loop 3:

2(i3-i2) + V2 = 0

KVL loop 4:

10i4 + 4V + 10(i4-i2) = 0
or
-10i2 + 20i4 = -4

We have two extra unknowns, so we need two more equations. These can be found by examining the loops with the sources:

i1= 5A
i3 = -2A

We can then solve for the remaining unknowns:

i2 = 1.58A
i4 = 0.59A

We can then find the current through the 5 ohm resistor as:
I = i1 - i2 = 3.42A

The current through the 4V source is simply i4, or 0.59A.