Introduction to Electric Circuits

9.2 The admittance or y-parameters 209

Example 9.2

Determine the admittance parameters for the T-circuit shown in Fig. 9.3.

Solution

From Equation (9.11), y~ = I1/V~ with V2 - 0. Short circuiting the output port
terminals makes V2 = 0 and then Z2 and Z3 are in parallel. Now

V1- I,[Z, + {Z2Z3/(Z2 q- Z3)I]

so that

11 = V1/[Z 1 q-{Z2Z3/(Z2 q- Z3)}]

Putting in the impedance values we have

11- V1/(10 + 2.5)= V~/12.5

Therefore

Yll = I~/V1 = (1,/12.5)S

From Equation (9.12), Y12 = I1/V2 with V1 = 0. Short circuiting the input
terminals to make V~ = 0 places Z~ and Z3 in parallel. Now

V 2 = &[Z2 q {Z1Z3/(Z1 -Jr- Z3)}]

so that

6 = V2/[Z2 -[- {Z1Z3/(Z1 -]- Z3)I]

Putting in the impedance values we have

/2 = V2/[5 + (50/15)] = V2//[(75 + 50)/15] = 15V2//125

By current division

11 = -I2[Z3/(Z1 + Z3)] = -(15V2/125)(5/15)= -1/2/25

SO

Y12-- I,/V2 = (-1/25)S

From Equation (9.13), Y21- I2/V1 with V2 = 0. We have seen above that,

with the output port short circuited to make V2 - 0, 11 = V1//12.5. By current

division,

12 = -I1[Z3/(22 + 23) ] = -(V1/12.5)(5/10)- (-V1/25)

Therefore

Y:I = I2/V1 = (-1/25)S.

Note that Y12 = Y21-