Introduction to Electric Circuits

7.3 Nodal voltage analysis 149

Setting the reference voltage to zero, V3 = 0, and then we have

I- V~/R~ + V1/R2 + (V, - Vz)/R3

Using conductances

I = G1 V1 .Jr_ GzV2 + G3(V 1 _ V2 )

Finally

(G1 + G2 + G3)V1- G3V2 = I

Applying KCL to node 2:

(7.2)

&=&+h

(V 1 -- V2)/R 3 = (V 2 - V3)/R 4 + (V 2 - V3)/R 5

It is vitally important to get the signs correct here. For a current leaving a node,

the node voltage is positive with respect to the other node. Thus we write

(V2- V3)/R4 for/4 which leaves node 2 and (V2- V3)/R5 for 15 which leaves

node 2. However, for 13 which enters node 2 we must write (V1 - Vz)/R3.

Using conductances and putting ~ = 0 we get

G3( V 1 -- V2) : G4 V 2 + Gs V 2

Rearranging, we have

(G 3 --]- G 4 -~- G5)V 2 - G3V , --" 0 (7.3)

We see from Equations (7.2) and (7.3) that for node 1:

9 the coefficient of I/1 is the sum of all the conductances connected to node 1;

9 the coefficient of V2 is (-1) times the conductance connected between node

2 and node 1;

9 the right-hand side of the equation is the current flowing into the node from

the source I;

and for node 2:

9 the coefficient of V2 is the sum of all the conductances connected to node 2;

9 the coefficient of V1 is (-1) times the conductance connected between node

1 and node 2;

9 the right-hand side of the equation is zero because there is no source

feeding current directly into node 2.

In matrix form Equations (7.2) and (7.3) may be written

-G3 (G3 + G4 + Gs) Vz