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).
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
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 _______.
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 ______.
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
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
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
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
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}
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
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