Question Hint. We know that: \[ \begin{array}{ll} V_{1}=z_{11} I_{1}+z_{12} I_{2} \\ V_{2}=z_{21} I_{1}+z_{22} I_{2} \end{array} \quad \text { and } \quad \begin{array}{l} I_{1}=y_{11} V_{1}+y_{12} V_{2} \\ I_{2}=y_{21} V_{1}+y_{22} V_{2} \end{array} \] Hint. We know that: V1​=z11​I1​+z12​I2​V2​=z21​I1​+z22​I2​​ and I1​=y11​V1​+y12​V2​I2​=y21​V1​+y22​V2​​

a) z_{11} :whe know, v_{1}=z_{11} I_{1}+Z_{12} I_{2}\Rightarrow \quad Z_{11}=v_{1} /\left.I_{1}\right|_{I_{2}=0}I_{2}=0 implies, V_{2} port is open circuited.\Rightarrow v_{1}=I_{1} \text { (Req) }where, Req =51145=\frac{15 \times 45}{15+45}=\frac{15 \times 45}{60}=11.25\therefore y_{11}=\frac{v_{1}}{I_{1}}=11.25 \mathrm{\Omega}y_{11}=11.25 \Omegaz_{22}:-since, v_{2}=z_{22} I_{2}+z_{21} I_{1}\Rightarrow z_{22}=V_{2} /\left.I_{2}\right|_{I_{1}=0 \text {. }}Now, open circuit V_{1} port and find the equivalent Mesistance hogen v_{2} port,ie. z_{22}=\operatorname{Req}=\frac{V_{2}}{I_{2}}\begin{array}{l}\text { Now, Req }=[301160+20] 1140 u=\left[\frac{30 \times 60}{90}+20\right] 1140 \\=401140 \\=\frac{40 \times 40}{40+40}=20 \\\therefore \frac{V_{2}}{I_{2}}=20 n=Z_{22} \\\Rightarrow y_{22}=20 r \\\end{array}c) 4_{11} i-we kerow, I_{1}=y_{11} v_{1}+y_{12} v_{2}\Rightarrow y_{11}=I_{1}\left|v_{1}\right|_{v_{2}=0}ie., short circuit ortput port and find input conductance. circuit is:-where, y_{11}=1 / Reor.\begin{aligned}R_{\text {eq }}=1511(15+601 / 20) & =15 / 1\left(15+\frac{60 \times 20}{60+20}\right) \\=15(1(15+15) & =151 / 30 \\& =\frac{15 \times 30}{15+30}=10 \\\therefore R_{\text {q }} & =10 \Omega\end{aligned}\begin{aligned}\Rightarrow \quad y_{11} & =\frac{1}{R_{0 V}}=\frac{I_{1}}{v_{1}}=\frac{1}{10} \\& \therefore y_{11}=0.1 \mathrm{v}\end{aligned}d) y_{22} i^{-}Since,\begin{array}{l} I_{2}=y_{21} v_{1}+y_{22} v_{2} \\\Rightarrow \quad y_{22}=\left.\frac{I_{2}}{V_{2}}\right|_{v_{1}=0}\end{array}ie., short ciacuit the irput pout and find the conductance ferom the output port:Circuit is:-Now, y_{22}=\frac{I_{2}}{v_{2}}=\frac{1}{R e q}.\begin{array}{l}=40 \|(20+12) \\=4011(32)=\frac{40 \times 32}{40+32}=17.778 \Omega \\\therefore R_{e q}=17.778 \Omega \\\Rightarrow \quad y_{22}=1 / 17.778=0.05825 \\\therefore y_{22}=0.056257 \\\end{array} ...