# Two Port Network and Network Functions

 Question 1
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 1 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 2
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 2 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 3
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 3 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 4
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 4 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, $I_s+8I_s=\frac{V_s-2I_s}{3}+\frac{V_s-2I_s}{6}$
$\Rightarrow \; 54I_s=2V_s-4I_s+V_s-2I_s$
$3V_s=60I_s$
$\frac{V_s}{I_s}=20\Omega$
 Question 5
The driving point impedance Z(s) for the circuit shown below is A $\frac{s^{4}+3s^{2}+1}{s^{3}+2s}$ B $\frac{s^{4}+3s^{2}+4}{s^{2}+2}$ C $\frac{s^{2}+1}{s^{4}+s^{2}+1}$ D $\frac{s^{3}+1}{s^{4}+s^{2}+1}$
GATE EE 2014-SET-3   Electric Circuits
Question 5 Explanation: Driving point impedance, Z(s) is ,
$Z(s)=s+\left ( \frac{\left ( s+\frac{1}{s} \right )\times \frac{1}{s}}{s+\frac{1}{s}+\frac{1}{s}} \right )$
$\;\;=s+\left ( \frac{s^2+1}{s^2} \right )\times \frac{s}{s^2+2}$
$\;\;=s+\frac{s^2+1}{s(s^2+2)}$
$\;\;=\frac{s^2(s^2+2)+^2+1}{s^3+2s}$
$Z(s)=\frac{s^4+3s^2+1}{s^3+2s}$
 Question 6
With 10 V dc connected at port A in the linear nonreciprocal two-port network shown below, the following were observed:

(i) 1 $\Omega$ connected at port B draws a current of 3 A
(ii) 2.5 $\Omega$ connected at port B draws a current of 2 A For the same network, with 6 V dc connected at port A, 1 $\Omega$ connected at port B draws 7/3 A. If 8 V dc is connected to port A, the open circuit voltage at port B is
 A 6V B 7V C 8V D 9V
GATE EE 2012   Electric Circuits
Question 6 Explanation:
$(i)$

$V_1=10V,V_2=3V, I_2=-3A$
$V_1=AV_2-BI_2$
$10=3A+3B$

$(ii) V_2=5V, I_2=-2A$
$10=5A+2B$
$A=\frac{10}{9}$
$B=\frac{20}{9}$
Given, $V_1=8V$
$(V_2)_{OC}=?$
$I_2=0$
$V_1=AV_2-BI_2$
$8=A(V_2)_{OC}-0$
$(V_2)_{OC}=\frac{8}{A}=\frac{8}{10/9}=7.2V$
 Question 7
With 10 V dc connected at port A in the linear nonreciprocal two-port network shown below, the following were observed:

(i) 1 $\Omega$ connected at port B draws a current of 3 A
(ii) 2.5 $\Omega$ connected at port B draws a current of 2 A With 10 V dc connected at port A, the current drawn by 7 $\Omega$ connected at port B is
 A 3/7A B 5/7A C 1A D 9/7A
GATE EE 2012   Electric Circuits
Question 7 Explanation:
Given, $V_1=10V, V_2=(-7I_2)$
$V_1=AV_2-BI_2$
$10=-7I_2A-BI_2$
$=-\frac{70}{9}I_2-\frac{20}{9}I_2$
$I_2=-1A$
-ve sign is signifies that current is drawn.
 Question 8
The two-port network P shown in the figure has ports 1 and 2, denoted by terminals (a,b) and (c,d) respectively. It has an impedance matrix Z with parameters denoted by $Z_{ij}$. A 1 $\Omega$ resistor is connected in series with the network at port 1 as shown in the figure. The impedance matrix of the modified two-port network (shown as a dashed box ) is A $\begin{pmatrix} Z_{11}+1 & Z_{12}+1\\ Z_{21}& Z_{22}+1 \end{pmatrix}$ B $\begin{pmatrix} Z_{11}+1 & Z_{12}\\ Z_{21}& Z_{22}+1 \end{pmatrix}$ C $\begin{pmatrix} Z_{11}+1 & Z_{12}\\ Z_{21}& Z_{22} \end{pmatrix}$ D $\begin{pmatrix} Z_{11}+1 & Z_{12}\\ Z_{21}+1& Z_{22} \end{pmatrix}$
GATE EE 2010   Electric Circuits
Question 8 Explanation: $V_1^{ab}=Z_{11}I_1+Z_{12}I_2$
$V_2=Z_{21}I_1+Z_{22}I_2$
As, $1\Omega$ resistor is connected in series with the network as port-1.
$V_2$ does not get affected,
$V_1^{ef}=V_1^{ab}+I_1 \times 1$
$\;\;=Z_{11}I_1+Z_{12}I_2+I_1$
$\;\;=(Z_{11}+1)I_1+Z_{12}I_2$
Modified Z-parameter $=\begin{bmatrix} Z_{11}+1 & Z_{12}\\ Z_{21} &Z_{22} \end{bmatrix}$
 Question 9
The parameter type and the matrix representation of the relevant two port parameters that describe the circuit shown are A z parameters, $\begin{bmatrix} 0 & 0\\ 0&0 \end{bmatrix}$ B h parameters, $\begin{bmatrix} 1 & 0\\ 0&1 \end{bmatrix}$ C g parameters, $\begin{bmatrix} 0 & 0\\ 0&0 \end{bmatrix}$ D z parameters, $\begin{bmatrix} 1 & 0\\ 0&1 \end{bmatrix}$
GATE EE 2006   Electric Circuits
Question 9 Explanation:
$I_1=g_{11}V_1+g_{12}I_2$
$V_2=g_{21}V_1+g_{22}I_2$ Since port-1 is open circuit, $I_1=0$
Port-2 is short-circuit, $V_2=0$
$g_{11}=\left.\begin{matrix} \frac{I_1}{V_1} \end{matrix}\right|_{I_2=0}=\frac{0}{V_1}=0$
$g_{12}=\left.\begin{matrix} \frac{I_1}{I_2} \end{matrix}\right|_{V_1=0}=\frac{0}{I_2}=0$
$g_{21}=\left.\begin{matrix} \frac{V_2}{V_1} \end{matrix}\right|_{I_2=0}=\frac{0}{V_1}=0$
$g_{22}=\left.\begin{matrix} \frac{V_2}{I_1} \end{matrix}\right|_{V_1=0}=\frac{0}{I_2}=0$
So, g-parameters,
$=\begin{bmatrix} g_{11} & g_{12}\\ g_{21}& g_{22} \end{bmatrix}=\begin{bmatrix} 0 &0 \\ 0 & 0 \end{bmatrix}$
 Question 10
The parameters of the circuit shown in the figure are
$R_{i}=1M\Omega, R_{o}=10\Omega , A=10^{6}V/V$.
If $v_{i}=1 \mu V$, then output voltage, input impedance and output impedance respectively are A 1V, $\infty$,10 $\Omega$ B 1 V, 0, 10 $\Omega$ C 1 V, 0, $\infty$ D 10 V, $\infty$, 10 $\Omega$
GATE EE 2006   Electric Circuits
Question 10 Explanation:
Output voltage $V_0=AV_i$
$V_0=10^6 \times 1 \times 10^{-6}=1V$
To calculate input impedance, $V_{dc}$ source is connected at input port Input impedance
$Z_i=\frac{V_{dc}}{I_i}$
as loop is not closed, $I_i=0$
So, $Z_i=\frac{V_{dc}}{0}=\infty$
To calculate output impedance, $V_{dc}$ source is connected at output port. Output impedance,
$Z_o=\frac{V_{dc}}{I_o}=\frac{I_oR_o+AV_i}{I_o}$
As $V_i=0 Z_o=R_o=10\Omega$
There are 10 questions to complete.