Introduction to Electric Circuits

150 Nodal and mesh analysis

This is of the form Ax = B so that Cramer's rule may be used to solve for V~

and V2.

Circuits with voltage sources

The circuits considered in the previous examples have had current sources only.

If a voltage source exists it could be replaced by its equivalent current source

and then the analysis proceeds as described above. Otherwise the analysis is as

shown in the following examples.

Example 7.9

Obtain an expression for the voltage at node 1 in the circuit of Fig. 7.3.

Figure 7.3

Vs,(

5 R1 1 R3 2 R5 3

• [ ]- - I I I ] -

) R2 R4 ( Vs2

Solution

The nodes are identified as 1, 2, 3, 4 and 5, their voltages being V1, V2, V3, V4 and

Vs, respectively. Node 4 is taken to be the reference, so V4 = 0. Also we see that

V5 = Vs~ and that I/3 = Vs2. We therefore have two unknown node voltages,

and to solve for them we need two equations.

Applying KCL to node 1 and using conductances rather than resistances we

have, assuming all currents flow away from the node,

G3(V1 - g2) -+- G2(V1- g4) + GI(V1- V5)= 0

G3V1 - G3V2 + G2V1 - 0 + G1VI - G1Vsl = 0

(G~ + G2 + G3)V1- G3V2 = G~Vs~ (7.5)

Applying KCL to node 2, again using conductances,

G3(V2- V,) + G4(V2- V4) + Gs(V2- V3)= 0

G3V2 - G3VI + G4V2 - 0 + G5V 2 - GsVs2 = 0

(G3 + G4 + Gs)V2- G3V~ = GsVs2 (7.6)

From Equations (7.5) and (7.6) we see a similar pattern to that in Equations
(7.2) and (7.3) emerging: for node 1,