Chapter 31: Mesh and Nodal Analysis

Date: 21-7-19

31.1 Mesh Current Analysis

In mesh circuit analysis, the currents assigned to each loop are drawn as circulating around the loop in the same direction. Where a branch separates two loops, the current in that branch is the difference between the two circulating currents. For instance, if two loop currents $I_1$ and $I_2$ share a component $Z_2$, the current flowing through $Z_2$ is $(I_1 – I_2)$. This method of assigning loop currents is called Maxwell’s theorem.

The simultaneous equations are formed using Kirchhoff’s Voltage Law (KVL) around each mesh. For a three-loop network, the equations would look like:

I1(Z1 + Z2) – I2Z2 = E1
I2(Z2 + Z3 + Z4) – I1Z2 – I3Z4 = 0
I3(Z4 + Z5) – I2Z4 = -E2

These simultaneous equations can then be solved using determinants, given the known values of the impedances and voltage sources.

31.2 Node Analysis

A node is a point in a network where two or more branches join. A principal node is a point in a network where three or more branches join. A node voltage is the voltage at a specific node with respect to a reference node. For example, if node 3 is chosen as the reference node, then $V_1$ is the voltage at node 1 with respect to node 3 ($V_{13}$), and $V_2$ is the voltage at node 2 with respect to node 3 ($V_{23}$).

The purpose of node voltage analysis is to determine the values of voltages at all principal nodes with respect to the reference node. Once these voltages are found, the current flowing through any branch can be calculated.

By applying Kirchhoff’s Current Law (KCL) at each principal node, we assume currents are leaving the node. For node 1, the equation is formed by summing the currents:

(V1 – Vx)/ZA + (V1 – V2)/ZB + V1/ZD = 0

Similarly, for node 2:

(V2 – V1)/ZB + V2/ZE + (V2 + Vy)/ZC = 0

We make the current leaving the node a positive equation value; therefore, any voltage source that opposes this direction will be subtracted. We are simply summing the currents at the node to zero. This can be rewritten in terms of admittances (Y = 1/Z):

(YA + YB + YD)V1 – YBV2 – YAVx = 0
-YBV1 + (YB + YC + YE)V2 + YCVy = 0

Given the source voltages $V_x$ and $V_y$, these equations can be solved for the unknown node voltages $V_1$ and $V_2$ using determinants.