## The Mesh Current Method //

The mesh current method of circuit analysis discussed in this technical article **employs mesh currents as the independent variables**. The idea is to write the appropriate number of independent equations, using mesh currents as the independent variables. Subsequent application of Kirchhoff’s Voltage Law (KVL) around each mesh provides the desired system of equations.

**Figure 1:** The **current i**, defined as flowing from left to right, establishes the polarity of the voltage across **R**.

In the mesh current method, we observe that a current flowing through a resistor in a specified direction defines the polarity of the voltage across the resistor, as illustrated in Figure 1, and that the sum of the voltages around a closed circuit must equal zero, by KVL. Once a convention is established **regarding the direction of current flow around a mesh**, simple application of KVL provides the desired equation. Figure 2 illustrates this point.

**Figure 2:** Once the direction of current flow has been selected, KVL requires that:

**v _{1} – v_{2} – v_{3} = 0.**

The number of equations one obtains by this technique is equal to the number of meshes in the circuit. All branch currents and voltages may subsequently be obtained from the mesh currents, as will presently be shown. Since meshes are easily identified in a circuit, this method provides a very efficient and systematic procedure for the analysis of electric circuits.

The following box outlines the procedure used in

applying the mesh current method to a linear circuit.

In mesh analysis, it is important to be consistent in choosing the direction of current flow. To avoid confusion in writing the circuit equations, unknown mesh currents are defined exclusively clockwise when we are using this method. To illustrate the mesh current method, consider the simple two-mesh circuit shown in Figure 3.

This circuit is used to generate two equations in the two unknowns, **the mesh currents i _{1} and i_{2}**. It is instructive to first consider each mesh by itself. Beginning with

**mesh 1**, note that the voltages around the mesh have been assigned in Figure 4 according to the

**direction of the mesh current i**.

_{1}Recall that

as long as signs are assigned consistently, an arbitrary direction may be assumed for any current in a circuit. If the resulting numerical answer for the current is negative, then the chosen reference direction is opposite to the direction of actual current flow.

Thus, one need not be concerned about the actual direction of current flow in mesh analysis, once the directions of the mesh currents have been assigned. The correct solution will result, eventually.

**Figure 4:** Mesh 1 – KVL requires that: **v _{S} – v_{1} – v_{2} = 0**

where:

**v**_{1}= i_{1}R_{1},**v**._{2}= (i_{1}– i_{2})R_{2}

According to the sign convention, then, the **voltages v _{1} and v_{2}** are defined as shown in

**Figure 4**(see above).

Now, it is important to observe that while

mesh current iis equal to the current flowing through resistor_{1}R(and is therefore also the branch current through_{1}R), it is not equal to the current through_{1}R._{2}

The branch current through **R _{2}** is the difference between the

**two mesh currents i**. Thus, since the polarity of

_{1}− i_{2}**voltage v**has already been assigned, according to the convention discussed in the previous paragraph, it follows that the

_{2}**voltage v**is given by:

_{2}**v _{2} = (i_{1} − i_{2})R_{2 }**– equation 1

**Finally, the complete expression for mesh 1 is:**

**v _{S} − i_{1}R_{1} − (i_{1} − i_{2})R_{2} = 0** – equation 2

The same line of reasoning applies to the second mesh. Figure 5 (see below) depicts the voltage assignment around the second mesh, following the clockwise direction of **mesh current i _{2}**. The mesh current

**i**is also the branch current through

_{2}**resistors R**.

_{3}and R_{4}**Figure 5:** Mesh 2 – KVL requires that **v _{2} + v_{3} + v_{4} = 0** where:

**v**_{2}= (i_{2}– i_{1})R_{2}**v**_{3}= i_{2}R_{3}**v**_{4}= i_{2}R_{4}

However, the current through the resistor that is shared by the two meshes, denoted by **R _{2}**, is now equal to

**i**; the voltage across this resistor is:

_{2}− i_{1}**v _{2} = (i_{2} − i_{1})R_{2 }** – equation 3

**and the complete expression for mesh 2 is:**

**(i _{2} − i_{1})R_{2} + i_{2}R_{3} + i_{2}R_{4} = 0** – equation 4

**expression for v**obtained in equation 3 different from equation 1?

_{2}The reason for this apparent discrepancy is that the voltage assignment for each mesh was dictated by the (clockwise) mesh current. Thus, since the mesh currents flow through **R _{2}** in opposing directions, the voltage assignments for

**v**in the two meshes are also opposite. This is perhaps a potential source of confusion in applying the mesh current method.

_{2}

You should be very carefulto carry out the assignment of the voltages around each mesh separately.

Combining the equations for the two meshes, we obtain the following system of equations:

**(R _{1} + R_{2})i_{1} − R_{2}i_{2} = v_{S} − R_{2}i_{1} +(R_{2} + R_{3} + R_{4})i_{2} = 0** – equation 6

These equations may be solved simultaneously to obtain the desired solution, namely, **the mesh currents i _{1} and i_{2}**. You should verify that knowledge of the mesh currents permits determination of all the other voltages and currents in the circuit.

### Mesh-Current Method for Circuit Analysis (VIDEO)

The technique Michelle Munteanu is demonstrating here is a **simplified version of Mesh Current Method** that was relevant to the type of circuits we were analyzing in the class.

**Reference //** Fundamentals of electrical engineering by Giorgio Rizzoni (Get it from Amazon)

I remembered old times at the university

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