# Two Port Network and Network Functions

 Question 1
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).

 A 238 B 452.25 C 632.12 D 145.25
GATE EE 2023   Electric Circuits
Question 1 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}
 Question 2
In the two-port network shown, the $h_{11}$ parameter (where, $h_{11}=\frac{V_{1}}{I_{1}}$, when $V_{2}$ =0) in ohms is _____________ (up to 2 decimal places).
 A 0.25 B 0.5 C 0.75 D 0.85
GATE EE 2018   Electric Circuits
Question 2 Explanation:

By KCL,
$\frac{V_a-1}{1}+\frac{V_a}{1}+\frac{V_a+2I_1}{1}=0$
$3V_a+2I_1=1\;\;...(i)$
$I_1=\frac{1-V_a}{1}\;\;...(ii)$
Substitute equation (ii) in equation (i) [/latex]
$V_a=-1$
$I_1=\frac{1-V_a}{1}=\frac{1-(-1)}{1}=2$
$h_{11}=\frac{V_1}{I_1}=\frac{1}{2}=0.5\Omega$

 Question 3
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
 A $V_{T}=\frac{V_{1}}{A_{1}A_{2}}$, $Z_{T}=\frac{A_{1}B_{2}+B_{1}D_{2}}{A_{1}A_{2}+B_{1}C_{2}}$ B $V_{T}=\frac{V_{1}}{A_{1}A_{2}+B_{1}C_{2}}$, $Z_{T}=\frac{A_{1}B_{2}+B_{1}D_{2}}{A_{1}A_{2}}$ C $V_{T}=\frac{V_{1}}{A_{1}+A_{2}}$, $Z_{T}=\frac{A_{1}B_{2}+B_{1}D_{2}}{A_{1}A_{2}+B_{1}C_{2}}$ D $V_{T}=\frac{V_{1}}{A_{1}A_{2}+B_{1}C_{2}}$, $Z_{T}=\frac{A_{1}B_{2}+B_{1}D_{2}}{A_{1}A_{2}+B_{1}C_{2}}$
GATE EE 2017-SET-1   Electric Circuits
Question 3 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)
$V_1=AV_2-BI_2$
$0=AV_2-BI_2$
$Z_T=\frac{V_2}{I_2}=\frac{B}{A}=\frac{A_1B_2+B_1D_2}{A_1A_2+B_1C_2}$
 Question 4
The z-parameters of the two port network shown in the figure are
$z_{11}=40\Omega ,z_{12}=60\Omega ,z_{21}=80\Omega$ and $z_{22}=100 \Omega$.
The average power delivered to $R_{L}=20\Omega$, in watts, is _______.
 A 18.25 B 24.35 C 28.45 D 35.55
GATE EE 2016-SET-2   Electric Circuits
Question 4 Explanation:

Given, $Z_{11}=40\Omega ,Z_{12}=60\Omega$
$Z_{21}=80\Omega ,Z_{22}=100\Omega$
From the figure,
$V_2=-20I_2\;\;...(i)$
$V_1=40I_1+60I_2\;\;...(ii)$
$V_2=80I_1+100I_2\;\;...(iii)$
From equation (i) and (iii), we get
$-20I_2=80I_1+100I_2$
$\Rightarrow \; I_2=-\frac{2}{3}I_1\;\;...(iv)$
Using equation (ii) and (iv), we get
$V_1=40I_1+60I_2$
$\;\;=40I_1+60\left ( \frac{-2}{3} I_1\right )$
$V_1=0$
From the figure,
$20=10I_1+V_1$
Since, $V_1=0$
So, $I_1=2A$
$I_2=\frac{-4}{3}A$
Power dissipated in
$P_{R_L}=I_2^2R_L=\left ( \frac{4}{3} \right )^2 \times 20$
$\;\;=\frac{16}{9} \times 20=35.55W$
 Question 5
The driving point input impedance seen from the source $V_s$ of the circuit shown below, in $\Omega$, is ______.
 A 10 B 20 C 30 D 40
GATE EE 2016-SET-2   Electric Circuits
Question 5 Explanation:
To find impedance seen by $V_s$
$Z_s=\frac{V_s}{I_s}$
$V_1=2I_s$

Applying KCL at node A,
$I_s=4V_1=\frac{V_A}{3}+\frac{V_A}{6}$
$V_A=V_s-V_1$ and $V_1=2I_s$
So, $I_s+8I_s=\frac{V_s-2I_s}{3}+\frac{V_s-2I_s}{6}$
$\Rightarrow \; 54I_s=2V_s-4I_s+V_s-2I_s$
$3V_s=60I_s$
$\frac{V_s}{I_s}=20\Omega$

There are 5 questions to complete.