Chapter 31: Mesh and Nodal Analysis
31.1 Mesh Current Analysis
In mesh circuit analysis the currents considered to flow around the loops are drawn as circulating around the loops in the same direction, where current from one loop is subtracted from the other.[1516-1524] Simultaneous equations are formed using KVL.[1528, 1529]
The current flowing through Z2 is given by the phasor sum of (I1 – I2). This method of loop current analysis is called Maxwell’s theorem.[1530-1535]
A typical network produces these equations:[1536]
I2(Z2 + Z3 + Z4) – I1Z2 – I3Z4 = 0[1537]
I3(Z4 + Z5) – I2Z4 = -E2[1539]
Solve using determinants given the values for Z1, Z2, Z3, Z4, Z5 and E.[1540-1543]
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 3 or more branches join.[1556-1562]
A node voltage is the voltage at the node with respect to a reference node. For instance, if node 3 is chosen as the reference, then V1 is short for V13, which is the voltage at 1 with respect to 3.[1563-1572]
The purpose of node voltage analysis is to determine the values of voltages at all principal nodes with respect to the reference node.[1572] Then the current flowing in any branch can be found.[1574]
At node 1, summing the currents leaving the node yields:[1586-1588]
Alternatively, using Admittances (Y = 1/Z):
-YBV1 + (YB + YC + YE)V2 + YCVY = 0[1617]
Given VX and VY, the equations can be solved for V1 and V2 using determinants.[1618, 1619]
Worked Problems
Problem 4
Find VAB.[1335, 1340] Given a current source I = 20∠0° A.[1341] Node B is the reference node.[1345]
V1(1/16 + 1/(4+j3)) = 20∠0°[1347-1353]
V1 = 20 / (0.1225 – j0.12) = 69.63 + j37.5 V[1357-1359]
IY = V1 / (4+j3) = 15.64 – j2.37 A[1360]
VAB = IY × 4 = (62.59 – j9.39) = 63.29∠-8.53° V[1362]
Problem 5
In the above problem, we have to find VXY. Taking Y as the reference node, arrange for VX which is equal to VXY since Y is the reference.[1372, 1383, 1384] Node voltage analysis assumes current leaving the node opposes the reference.[1388-1392]
Problem 6
Find I1 and I2.[1411] Principal nodes are node 1 and node 2, with node 2 taken as the reference.[1413, 1414]
V1(1/20 + 1/25 + 1/10) = 100/25 + j50/10[1417]
V1(95/500) = 4 + j5[1418]
V1 = (21.05 + j26.32) V[1420]
Currents can now be found:
I2 = (21.05 + j26.32) / j20 = 1.69∠51.35° A[1422]
Problem 8
Determine VXY using nodal analysis.[1490] Node 3 is taken as the reference node, giving two principal node voltages to be solved for, V1 and V2.[1491-1499]
V1(1/(4+j3) + 1/5) – V2/5 = -25[1501-1506]