In the h-parameter model of the 2-port network given in the figure shown, the value of $h_{22}$ (in S) is

1.25

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Solution

The correct option is A 1.25
When two, two port networks are connected in parallel then individual Y-parameters are added. Therefore from the given network,

For network (1) Y-parameter is,

$Y_{1} = ⎡ ⎢ ⎢ ⎣ \begin{matrix} \frac{1}{3} + \frac{1}{3} & - \frac{1}{3} - \frac{1}{3} & \frac{1}{3} + \frac{1}{3} \end{matrix} ⎤ ⎥ ⎥ ⎦$

Similarly for network (2),

$Y_{2} = ⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & - \frac{1}{2} - \frac{1}{2} & 1 \end{matrix} ⎤ ⎥ ⎥ ⎦$

Thus,
$Y = Y_{1} + Y_{2} + Y_{2} = ⎡ ⎢ ⎢ ⎣ \begin{matrix} \frac{5}{3} & - \frac{5}{6} - \frac{5}{6} & \frac{5}{3} \end{matrix} ⎤ ⎥ ⎥ ⎦$

$∴ I_{1} = Y_{11} V_{1} + Y_{12} V_{2}$

$= \frac{5}{3} V_{1} - \frac{5}{6} V_{2} . . . . . (i)$

$I_{2} = Y_{21} V_{1} + Y_{22} V_{2}$

$= - \frac{5}{6} V_{1} + \frac{5}{3} V_{2} . . . . . (i i)$

$A l s o h_{22} = {\begin{matrix} \frac{I_{2}}{V_{2}} \end{matrix} ∣ ∣ ∣}_{I_{1} = 0}$

$∴$ From equation (i), we get,

$V_{1} = \frac{1}{2} V_{2} . . . . . . (i i i)$

and from equation (ii) and (iii), we get,

$h_{22} = \frac{I_{2}}{V_{2}} = \frac{15}{12} = 1.25$

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