Introduction to Electric Circuits

7.3 Nodal voltage analysis 157

Figure 7.7

Vs, )

(5V)

Is = 5A ,,.._ v-

'~

1 12 ~2 4 Q 3

R2 = 5fl I3

R1 = 2s R3 = 4~

1. I=
   4

R4 = 1~

Solution

The voltages of the four nodes are V1, V2, V3 and V4. The reference node is node
4 and its voltage is V4 = 0. Also

V 1 - V4 = Vl-- Vs1-- -5V

Nodes 2 and 3, together with the voltage source Vs2, constitute a supernode.
Applying KCL to the supernode we have

I2+I3+I4=Is

(V2- V~)/R2 + (V3- V4)/R4 + (V2- V4)/R3 : Is

Using conductances and with V4 = 0 and V~ - Vs~ we have

GzV 2 -- G2Vsl + G4V 3 + G3V 2 = I s

(G2 + G3)V2 + G4V3 : Is + GzVs~

But V 3 -- Vs2 + V2, so

(G2 + G3)V2 + G4(Vs2 + V2) = Is + GzVs~

(G 2 + G 3 -~- G4)V2-- Is + G2Vs1- G4Vs2

V2 = (Is + GzVsl- G4Vsz)/(G2 + G3 + G4)

Putting in the values,

Is - 5 A, G2 = (1/5) S, G3 = (1/4) S, G4 = (1/1) S, Vsl = -5 V and Vs2 = 2 V

V2 = (5 - 1 - 2)/(0.2 + 0.25 + 1) : 2/1.45 : 1.38 V

The potential difference across the resistor R 2 is

V2 - 1/1 : 1.38 - (-5) = 6.38 V.

The potential difference across the resistor R4 is

V3- V4- (Vs2 + V2) - V4 = 2 + 1.38- 0 = 3.38 V