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).
\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).
238 | |
452.25 | |
632.12 | |
145.25 |
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}
\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).
0.25 | |
0.5 | |
0.75 | |
0.85 |
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
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
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}} | |
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}} | |
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}} | |
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}} |
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}
\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 _______.
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 _______.
18.25 | |
24.35 | |
28.45 | |
35.55 |
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 ______.
10 | |
20 | |
30 | |
40 |
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
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.