Introduction to Electric Circuits

vs, l(

200V

Figure 7.5

1 Y1 11 2 I3 Y3

) - (

4

7.3 Nodal voltage analysis 155

I Vs2 Y1 =(-j/3)S

210/-30V Y2 = (1/10)S

Y3 = (-j/5)s

V1- V4-- V1 = Us1--200/_0o V

and

V3- V4-- V3-- Us 2 --210/_-30o V

We therefore need one equation in order to determine the one unknown

voltage, V2. Applying KCL to node 2 we have

I1+I2+I3=0

YI(V2- Vl) "+- Y2(V2- V4) "[- Y3(V2- V3)= 0

FlU2- FlUs1 + Y2 V2 + Y~ V2 - Y~ U~2 = 0

(Y1 -[- Y2 -[- Y3)U2- Y1Usi -[- Y3Us2

V2- (YiVsi + Y3Vs2)/(Y1 + Y2 + Y3)

Now

Yi - l/j3 - 1/(3/_90 ~ - 0.33/_-90~ S

Therefore

YIVsl = 0.33/--90 ~ • 200/-0 ~ = 66.7/--90 ~ A = (0 - j66.7) A

Also

V 3 "-- l/j5-- 1/5/__90 ~ 0.2/_--90~ S

Therefore

Y3Vs2 = 0.2/-90 ~ x 210/-30 ~ = 42/-120 ~ = (-21 -j36.3) A

Thus

Y~Vsl + Y3Vs2 = (-21 -j103) A = 105.1/-101.5 ~ A

Also we have

Y,+Y2+~

Thus

= (l/j3) + (1/10) + (l/j5) = -j0.33 + 0.1 -j0.2

-- (0.1 -j0.53) S = 0.539/__-79.3 S