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.
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