Two passive two-port network $P$ and $Q$ are connected as shown in the figure. The impedance matrix of network $P$ is

Z_{P}=\left[\begin{array}{cc}40 \Omega & 60 \Omega \\ 80 \Omega & 100 \Omega\end{array}\right]

. The admittance matrix of network $Q$ is

Y_{Q}=\left[\begin{array}{cc}5 S & -2.5 S \\ -2.5 S & 1 S\end{array}\right]

. Let the $A B C D$ matrix of the two-port network $R$ in the figure be

\left[\begin{array}{ll}\alpha & \beta \\ \gamma & \delta\end{array}\right]

. The value of $\beta$ in $\Omega$ is _______ (rounded off to 2 decimal places)

🚀 Solve in Practice Mode

📖 Explanation

Z_{P}=\left[\begin{array}{cc} 40 & 60 \\ 80 & 100 \end{array}\right] ; \quad Y_{Q}=\left[\begin{array}{cc} 5 S & -2.5 S \\ -2.5 S & 1 S \end{array}\right]

From $Z_{p}$ matrix

\begin{aligned} & V_{1}=40 I_{1}+60 I_{2} \\ & V_{2}=80 I_{1}+100 I_{2} \end{aligned}

Convert Z-parameters in to ABCD parameters

\begin{aligned} V_{1} & =A V_{2}-B I_{2} \\ I_{1} & =C V_{2}-D I_{2} \end{aligned}

From equation (ii),

\begin{aligned} & I_{1}=\frac{V_{2}}{80}-\frac{100}{80} I_{2} \\ & I_{1}=\frac{V_{2}}{80}-\frac{5}{4} I_{2} \end{aligned}

Sub equation (iii) in equation (i),

\begin{aligned} & V_{1}=40\left(\frac{V_{2}}{80}-\frac{5}{4} I_{2}\right)+60 I_{2} \\ & V_{1}=\frac{V_{2}}{2}+10 I_{2} \end{aligned}

From equation (iii) and (iv),

\begin{aligned} {\left[\begin{array}{ll} A_{1} & B_{1} \\ C_{1} & D_{1} \end{array}\right] } & =\left[\begin{array}{cc} 0.5 & -10 \\ \frac{1}{80} & \frac{5}{4} \end{array}\right] \\ {[Y]_{Q} } & =\left[\begin{array}{cc} 5 & -2.5 \\ -2.5 & 1 \end{array}\right] \end{aligned}

\begin{aligned} & I_{1}=5 V_{1}-2.5 V_{2} \\ & I_{2}=-2.5 V_{1}+V_{2} \end{aligned}

Convert $Y$ -parameters to $\mathrm{ABCD}$ parameters From equation (vi),

\begin{aligned} & V_{1}=\frac{V_{2}}{2.5}-\frac{I_{2}}{2.5} \\ & V_{1}=0.4 V_{2}-0.4 I_{2} \end{aligned}

Sub equation (vii) in equation ( $\mathrm{v}$ ),

\begin{aligned} I_{1} & =5\left(0.4 V_{2}-0.4 I_{2}\right)-2.5 V_{2} \\ & =2 V_{2}-2 I_{2}-2.5 V_{2} \\ I_{1} & =-0.5 V_{2}-2 I_{2} \end{aligned}

From equation (vii) and equation (viii),

\begin{aligned} {\left[\begin{array}{ll} A_{2} & B_{2} \\ C_{2} & D_{2} \end{array}\right] } & =\left[\begin{array}{cc} 0.4 & 0.4 \\ -0.5 & 2 \end{array}\right] \\ T & =\left[\begin{array}{ll} A_{1} & B_{1} \\ C_{1} & D_{1} \end{array}\right] *\left[\begin{array}{ll} A_{2} & B_{2} \\ C_{2} & D_{2} \end{array}\right] \\ & =\left[\begin{array}{cc} 0.5 & -10 \\ \frac{1}{80} & \frac{5}{4} \end{array}\right] *\left[\begin{array}{cc} 0.4 & 0.4 \\ -0.5 & 2 \end{array}\right] \\ \beta & =0.5 \times 0.4+(-10) \times 2 \\ & =-19.8 \Omega \end{aligned}

Q2GATE 2023NAT

The admittance parameters of the passive resistive two-port network shown in the figure are $\mathrm{y}_{11}=5 \mathrm{~S}, \mathrm{y}_{22}=1 \mathrm{~S}, \mathrm{y}_{12}=\mathrm{y}_{21}=-2.5 \mathrm{~S}$ The power delivered to the load resistor $R_{L}$ in Watt is ____ (Round off to 2 decimal places).

🚀 Solve in Practice Mode

📖 Explanation

$Y =[Y]_{A}+[Y]_{B}$

\begin{aligned} Y_{A} & =\left[\begin{array}{cc} \frac{1}{3} & -\frac{1}{3} \\ -\frac{1}{3} & \frac{1}{3} \end{array}\right] \\ Y_{B} & =\left[\begin{array}{cc} 5 & -2.5 \\ -2.5 & 1 \end{array}\right] s \\ Y & =\left[\begin{array}{cc} \frac{1}{3} & -\frac{1}{3} \\ -\frac{1}{3} & \frac{1}{3} \end{array}\right]+\left[\begin{array}{cc} 5 \\ -2.5 & 1 \end{array}\right] \\ & =\left[\begin{array}{cc} \frac{16}{3} & -\frac{8.5}{3} \\ -\frac{8.5}{3} & \frac{4}{3} \end{array}\right] \\ I_{1} & =\frac{16}{3} V_{1}-\frac{8.5}{3} V_{2} \\ I_{2} & =-\frac{8.5}{3} V_{1}+\frac{4}{3} V_{2} \end{aligned}

Put $I_{2}=0$ and $V_{1}=20 \mathrm{~V}$ $0=-\frac{8.5}{3} \times 20+\frac{4}{3} V_{2}$

\begin{aligned} V_{\text {th }} & =V_{2}=\frac{8.5 \times 20}{4}=42.5 \mathrm{~V} \\ R_{\mathrm{th}} & =\frac{V_{2}}{I_{2}} \\ I_{2} & =\frac{4}{3} V_{2} \\ \frac{V_{2}}{I_{2}} & =\frac{3}{4} \Omega \end{aligned}

Equivalent circuit $I=\frac{42.5}{\frac{3}{4}+6}=6.296 \mathrm{~A}$

\begin{aligned} P & =I^{2} R_{L}=(6.296)^{2} \times 6 \\ & =238 \mathrm{~W} \end{aligned}

Q4GATE 2017MCQ

Two passive two-port networks are connected in cascade as shown in figure. A voltage source is connected at port 1. Given $V_{1}=A_{1}V_{2}+B_{1}I_{2}$ $I_{1}=C_{1}V_{2}+D_{1}I_{2}$ $V_{2}=A_{2}V_{3}+B_{2}I_{3}$ $I_{2}=C_{2}V_{3}+D_{2}I_{3}$ $A_{1},B_{1},C_{1},D_{1},A_{2},B_{2},C_{2} \; and \; D_{2}$ are the generalized circuit constants. If the Thevenin equivalent circuit at port 3 consists of a voltage source $V_T$ and impedance $Z_T$ connected in series, then

🚀 Solve in Practice Mode

📖 Explanation

For two port network, we can write,

\begin{bmatrix} A &B \\ C & D \end{bmatrix}=\begin{bmatrix} A_1 & B_1\\ C_1 & D_1 \end{bmatrix}\begin{bmatrix} A_2 & B_2\\ C_2 & D_2 \end{bmatrix}

$A=A_1A_2+B_1C_2\;\;...(i)$ $B=A_1B_2+B_1D_2\;\;...(ii)$ $V_1=AV_2-BI_2\;\;...(iii)$ To get, $V_T(I_2=0)$ ; from equation (iii) $V_2=V_T=\frac{V_1}{A}=\frac{V_1}{A_1A_2+B_1C_2}$ To get $Z_T(V_1=0)$ ; from equation (iii) $V_1=AV_2-BI_2$ $0=AV_2-BI_2$ $Z_T=\frac{V_2}{I_2}=\frac{B}{A}=\frac{A_1B_2+B_1D_2}{A_1A_2+B_1C_2}$

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