Question 1 |
A linear 2-port network is shown in Fig. (a). An ideal DC voltage source of 10 V is
connected across Port 1. A variable resistance R is connected across Port 2. As R is
varied, the measured voltage and current at Port 2 is shown in Fig. (b) as a V_2 versus -I_2 plot. Note that for V_2=5V,I_2=0mA, and for V_2=4V,I_2=-4mA.
When the variable resistance R at Port 2 is replaced by the load shown in Fig. (c), the current I_2 is _______ mA (rounded off to one decimal place).
When the variable resistance R at Port 2 is replaced by the load shown in Fig. (c), the current I_2 is _______ mA (rounded off to one decimal place).
3.2 | |
6.8 | |
4 | |
8 |
Question 1 Explanation:
CASE-1:
I_2=0
V_{oc}=5V
CASE-2: With help of linearity
\frac{4-5}{4-0}=\frac{I_{sc}-5}{I_{sc}-0}\Rightarrow I_{sc}=20mA
R_{th}=\frac{V_{oc}}{I_{sc}}=\frac{5}{20}k=250\Omega =0.25k\Omega
CASE-3
I_2=\frac{10-5}{1.25}=4mA
Question 2 |
Consider the two-port network shown in the figure.
The admittance parameters, in siemens, are
The admittance parameters, in siemens, are
y_{11}=2,\:y_{12}=-4,\:y_{21}=-4,\:y_{22}=2 | |
y_{11}=1,\:y_{12}=-2,\:y_{21}=-1,\:y_{22}=3 | |
y_{11}=2,\:y_{12}=-4,\:y_{21}=-1,\:y_{22}=2 | |
y_{11}=2,\:y_{12}=-4,\:y_{21}=-4,\:y_{22}=3 |
Question 2 Explanation:
Write KCL at V_{1}
\begin{aligned} I_{1}&=\frac{V_{1}}{1}+\frac{V_{1}-V_{2}}{1}-3 V_{2} \\ I_{1}&=2 V_{1}-4 V_{2} \\ \text{Write KCL at }V_{2}\qquad I_{2}&=\frac{V_{2}}{1}+\frac{V_{2}-V_{1}}{1}\\ I_{2}&=-V_{1}+2 V_{2} \\ [y]&=\left[\begin{array}{cc} 2 & -4 \\ -1 & 2 \end{array}\right] \mho \end{aligned}
Question 3 |
For a 2-port network consisting of an ideal lossless transformer, the parameter S_{21} (rounded
off to two decimal places) for a reference impedance of 10 \Omega, is _____.
1 | |
0.5 | |
1.5 | |
0.8 |
Question 3 Explanation:
For ideal transformer of n : 1, the scattering matrix is
\begin{bmatrix} S_{11} &S_{12} \\ S_{21} & S_{22} \end{bmatrix}=\begin{bmatrix} \frac{n^{2}-1}{n^{2}+1} &\frac{2n}{n6{}} \\ & \end{bmatrix}
S_{21}=\frac{2n}{n^{2}+1}=\frac{2(2)}{2^2+1}=\frac{4}{5}=0.8
\begin{bmatrix} S_{11} &S_{12} \\ S_{21} & S_{22} \end{bmatrix}=\begin{bmatrix} \frac{n^{2}-1}{n^{2}+1} &\frac{2n}{n6{}} \\ & \end{bmatrix}
S_{21}=\frac{2n}{n^{2}+1}=\frac{2(2)}{2^2+1}=\frac{4}{5}=0.8
Question 4 |
In the given circuit, the two-port network has the impedance matrix[Z] =\begin{bmatrix} 40 & 60\\ 60 & 120 \end{bmatrix}. The value of Z_L for which maximum power is transferred to the load is _________ \Omega.
48 | |
38 | |
40 | |
54 |
Question 4 Explanation:
From maximum power transfer theorem
Z_{L}=Z_{th}
Z_{th}=Z_{22}-\frac{Z_{12}\times Z_{21}}{R_{S}+Z_{11}}
For given data,
Z_{th}=120-\frac{60\times 60}{10+40}=48\Omega
Z_{L}=48\Omega
Z_{L}=Z_{th}
Z_{th}=Z_{22}-\frac{Z_{12}\times Z_{21}}{R_{S}+Z_{11}}
For given data,
Z_{th}=120-\frac{60\times 60}{10+40}=48\Omega
Z_{L}=48\Omega
Question 5 |
The ABCD matrix for a two-port network is defined by:
\begin{bmatrix} V_{1}\\ I_{1} \end{bmatrix}=\begin{bmatrix} A & B\\ C & D \end{bmatrix}=\begin{bmatrix} V_{2}\\ - l_{2} \end{bmatrix}
The parameter B for the given two-port network (in ohms, correct to two decimal places) is _______.
\begin{bmatrix} V_{1}\\ I_{1} \end{bmatrix}=\begin{bmatrix} A & B\\ C & D \end{bmatrix}=\begin{bmatrix} V_{2}\\ - l_{2} \end{bmatrix}
The parameter B for the given two-port network (in ohms, correct to two decimal places) is _______.
4.8 | |
5.5 | |
3.4 | |
6.2 |
Question 5 Explanation:
B=-\left.\frac{V_{1}}{I_{2}}\right|_{V_{2}=0}
When V_{2}=0 (i.e., when port-2 is short circuited)
\begin{aligned} I_{1}&=\frac{V_{1}}{2 \Omega+(5 \Omega \| 2 \Omega)}=\frac{7 V_{1}}{24 \Omega}\\ I_{2}&=-I_{1} \times \frac{5 \Omega}{5 \Omega+2 \Omega}=\frac{-5 V_{1}}{24 \Omega}\\ \text { So, } \quad B&=-\frac{V_{1}}{I_{2}}=\frac{24}{5} \Omega=4.80 \Omega \end{aligned}
When V_{2}=0 (i.e., when port-2 is short circuited)
\begin{aligned} I_{1}&=\frac{V_{1}}{2 \Omega+(5 \Omega \| 2 \Omega)}=\frac{7 V_{1}}{24 \Omega}\\ I_{2}&=-I_{1} \times \frac{5 \Omega}{5 \Omega+2 \Omega}=\frac{-5 V_{1}}{24 \Omega}\\ \text { So, } \quad B&=-\frac{V_{1}}{I_{2}}=\frac{24}{5} \Omega=4.80 \Omega \end{aligned}
Question 6 |
The z-parameter matrix \begin{bmatrix} z_{11} &z_{12} \\ z_{21}& z_{22} \end{bmatrix} for the two-port network shown is
\begin{bmatrix} 2 & -2\\ -2&2 \end{bmatrix} | |
\begin{bmatrix} 2 & 2\\ 2&2 \end{bmatrix} | |
\begin{bmatrix} 9 & -3\\ 6&9 \end{bmatrix} | |
\begin{bmatrix} 9 & 3\\ 6&9 \end{bmatrix} |
Question 6 Explanation:
Redrawing the circuit,
\begin{aligned} Z_{11} &=\frac{V_{1}}{I_{1}}=3 \| 6=2 \Omega \\ V_{2} &=-3 \times I_{3 \Omega}=-3 \times I_{1} \times \frac{6}{9}=-2 I_{1} \\ \therefore Z_{21} &=\frac{V_{2}}{I_{1}}=-2 \\ \text { Now, } Z_{22} &=\left.\quad \frac{V_{2}}{I_{2}}\right|_{I_{1}=0}=3 \| 6=2 \Omega \\ V_{1} &=-6 \times I_{6 \Omega} \\ &=-6 \times I_{2} \times \frac{3}{9} \\ &=-2 \\ \therefore[Z] &=\left[\begin{array}{cc} 2 & -2 \\ -2 & 2 \end{array}\right] \end{aligned}
\begin{aligned} Z_{11} &=\frac{V_{1}}{I_{1}}=3 \| 6=2 \Omega \\ V_{2} &=-3 \times I_{3 \Omega}=-3 \times I_{1} \times \frac{6}{9}=-2 I_{1} \\ \therefore Z_{21} &=\frac{V_{2}}{I_{1}}=-2 \\ \text { Now, } Z_{22} &=\left.\quad \frac{V_{2}}{I_{2}}\right|_{I_{1}=0}=3 \| 6=2 \Omega \\ V_{1} &=-6 \times I_{6 \Omega} \\ &=-6 \times I_{2} \times \frac{3}{9} \\ &=-2 \\ \therefore[Z] &=\left[\begin{array}{cc} 2 & -2 \\ -2 & 2 \end{array}\right] \end{aligned}
Question 7 |
The z-parameter matrix for the two-port network shown is
\begin{bmatrix} 2j\omega & j\omega \\ j\omega & 3+2j\omega \end{bmatrix}
where the entries are in \Omega. Suppose Z_{b}(j\omega )=R_{b}+j\omega. Then the value of R_{b} (in \Omega) equals ________
\begin{bmatrix} 2j\omega & j\omega \\ j\omega & 3+2j\omega \end{bmatrix}
where the entries are in \Omega. Suppose Z_{b}(j\omega )=R_{b}+j\omega. Then the value of R_{b} (in \Omega) equals ________
1 | |
2 | |
3 | |
4 |
Question 7 Explanation:
For T-network,
\begin{aligned} Z_{11} &=Z_{a}+Z_{c} \\ Z_{22} &=Z_{b}+Z_{c} \\ \text { and, } \quad Z_{12} &=Z_{21}=Z_{c} \\ \text { Given, } &[Z]=\left[\begin{array}{cc} 2 j \omega & j \omega \\ j \omega & 3+2 j \omega \end{array}\right] \\ \text { Therefore } Z_{12} &=j \omega \\ \text { and, } \quad Z_{22} &=3+2 j \omega \\ &=3+j \omega+j \omega \\ &=Z_{b}+Z_{c} \\ &=R_{b}+j \omega+Z_{c} \\ & R_{b}=3 \Omega \end{aligned}
\begin{aligned} Z_{11} &=Z_{a}+Z_{c} \\ Z_{22} &=Z_{b}+Z_{c} \\ \text { and, } \quad Z_{12} &=Z_{21}=Z_{c} \\ \text { Given, } &[Z]=\left[\begin{array}{cc} 2 j \omega & j \omega \\ j \omega & 3+2 j \omega \end{array}\right] \\ \text { Therefore } Z_{12} &=j \omega \\ \text { and, } \quad Z_{22} &=3+2 j \omega \\ &=3+j \omega+j \omega \\ &=Z_{b}+Z_{c} \\ &=R_{b}+j \omega+Z_{c} \\ & R_{b}=3 \Omega \end{aligned}
Question 8 |
Consider a two-port network with the transmission matrix: T=\begin{pmatrix} A & B\\ C& D \end{pmatrix}. If the network is reciprocal, then
T^{-1}=T | |
T^{2}=T | |
Determinant (T) = 0 | |
Determinant (T) = 1 |
Question 8 Explanation:
For reciprocal network AD-BC= 1
|T|=1
|T|=1
Question 9 |
The ABCD parameters of the following 2-port network are
\begin{bmatrix} 3.5+j2 &20.5 \\ 20.5 & 3.5-j2 \end{bmatrix} | |
\begin{bmatrix} 3.5+j2 &30.5 \\ 0.5 & 3.5-j2 \end{bmatrix} | |
\begin{bmatrix} 10 &2+j0 \\ 2+j0 & 10 \end{bmatrix} | |
\begin{bmatrix} 7+j4 &0.5 \\ 30.5 & 7-j4 \end{bmatrix} |
Question 9 Explanation:
\begin{aligned} \left[\begin{array}{l} V_{1} \\ I_{1} \end{array}\right] &=\left[\begin{array}{ll} A & B \\ C & D \end{array}\right]\left[\begin{array}{l} V_{2} \\ I_{2} \end{array}\right] \\ \text { When } I_{2} &=0 \\ V_{1} &=(5+j 4+2) I_{1} \\ V_{2} &=2 I_{1} \\ A &=\left.\frac{V_{1}}{V_{2}}\right|_{I_{2}=0}=\frac{7+j 4}{2}=3.5+j 2 \\ C &=\left.\frac{I_{1}}{V_{2}}\right|_{I_{2}=0}=\frac{1}{2}=0.5 \end{aligned}
Option (B) matches. From here only
Question 10 |
The 2-port admittance matrix of the circuit shown is given by
\begin{bmatrix} 0.3 & 0.2\\ 0.2& 0.3 \end{bmatrix} | |
\begin{bmatrix} 15 & 5\\ 5& 15 \end{bmatrix} | |
\begin{bmatrix} 3.33 & 5\\ 5& 3.33 \end{bmatrix} | |
\begin{bmatrix} 0.3 & 0.4\\ 0.4& 0.3 \end{bmatrix} |
Question 10 Explanation:
\begin{aligned} V_{1} &=6 I_{1}+4 I_{2} \\ V_{2} &=4 I_{1}+6 I_{2} \\ [Z] &=\left[\begin{array}{cc} 6 & 4 \\ 4 & 6 \end{array}\right] \\ Y&=\frac{1}{20}\left[\begin{array}{rr} 6 & -4 \\ -4 & 6 \end{array}\right]=\left[\begin{array}{rr} 0.3 & -0.2 \\ -0.2 & 0.3 \end{array}\right] \\ &\text { Ignoring negative sign: } \\ [Y]&=\left[\begin{array}{cc} 0.3 & 0.2 \\ 0.2 & 0.3 \end{array}\right] \end{aligned}
There are 10 questions to complete.