Introduction to Electric Circuits

152 Nodal and mesh analysis

Finally

V1 = A1/A

--[(Vs1Gll(G 3 -Jr- a 4 "Jr- (~5) -lt- (Vs2G3G5)I/(GI -[- G2 Jr- (~3)((~3 -]- a 4 --[- a5)

-- (G321]

Example 7.10

Is~

Vs,(

200 V

Figure 7.4

2A

Node2 \ Is~ (~t--~ Is I. Node^3

, , ~\,L la [ ] ~ ,L/";s___l .l.___ 4

R1 = 4~ I2 R3 = 2f~ R5 = 5~.

(20 ~"~) 4R4 (25 ~~)

I

5

Is2

"~ Vs2

,; 22ov

For the circuit of Fig. 7.4 determine (1) the potential difference across the
resistor R2, (2) the current supplied by the voltage source Vs2.

Solution

The nodes are identified as 1, 2, 3, 4 and 5, their respective voltages being V~, V2,

V3, V4 and Vs. We choose node 5 to be the reference node and let V5 = 0. Also

(V 1 - V5) = (V 1 - 0) = V 1 -- Vs1 -- 200V

and

(V 4 - V5)- (V 4 -O)= V 4 -- Vs2--220 V

We therefore have two unknown voltages (V2 and V3) and we need two
independent equations to solve for them.
Applying KCL to node 2 and assuming the currents to be flowing as shown,
we have

11 -F/2 +I3--/s--0

(V 2 - V1)/R , + (V 2 - Vs)/R 2 + (V 2 - V3)/R 3 - I s = 0

Using conductances we have

(V2 - V1)G1 + G2(V2 - Vs) + G3(V2 - V3) -/s = 0

G,V~_- GIVs, + G=V2- G2V~ + G~V2- G~V~ = &

(c, + o= + Q)v2- o~v~ = (& + O, Vs,)

Note that we could have written this equation down immediately without