Test 4 - Network Graphs | Electronics and Communication (ECE)
A comprehensive quiz covering network analysis, circuit theory, two-port networks, and graph theory concepts in Electronics and Communication Engineering
Questions
In the following graph, the number of trees (P) and the number of cut-set (Q) are
- P = 2 Q = 2
- P = 2 Q = 6
- P = 4 Q = 6
- P = 4 Q = 10
The equivalent inductance measured between the terminals 1 and 2 for the circuit shown in the figure is
- L1 + L2 + M
- L1 + L2 – M
- L1 + L2 + 2M
- L1 + L2 – 2M
How much current will flow in a 100 Hz series RLC circuit, if VS = 20 V, RT = 66 ohms and XT = 47 ohms?
- 1.05 A
- 303 mA
- 247 mA
- 107 mA
In the circuit shown below, the network N is described by the following Y matrix:
Y = $\left[
\begin{array}
\ 0.1S & -0.01S \\
0.01S & 0.1S
\end{array}
\right]$. The voltage gain $\dfrac{V_2}{V_1}$is
- 1/90
- -1/90
- -1/99
- -1/11
The Thevenin equivalent impedance Zth between the nodes P and Q in the following circuit is
- 1
- 1 + s + $\dfrac{1}{s}$
- 2 + s + $\dfrac{1}{s}$
- $\dfrac{s^2 + s + 1}{s^2 + 2s +1}$
A square pulse of 3 volts amplitude is applied to C - R circuit shown in figure. The capacitor is initially uncharged. The output voltage v0 at time t = 2 sec is
- 3 V
- -3V
- 4 V
- - 4V
Two series resonant filters are as shown in the figure. Let the 3-dB bandwidth of Filter 1 be B1 and that of Filter 2 be B2. The value of $\dfrac{B_1}{B_2}$ is
- 4
- 1
- $\dfrac{1}{2}$
- $\dfrac{1}{4}$
For the circuit shown in the figure, the time constant RC = 1 ms. The input voltage is v1 (t) = $\sqrt 2$sin 103t. The output voltage v0 (t) is equal to
- sin (103t – 450)
- sin (103t + 450)
- sin (103t – 530)
- sin (103t + 530)
For the lattice shown in the figure, Za = j2$\Omega$ and Zb = 2$\Omega$. Calculate the values of the open circuit impedance parameter [z] = $\left[
\begin{array}
\ Z_{11} & Z_{12} \\
Z_{21} & Z_{22}
\end{array}
\right]$.
- $\left[ \begin{array} \ 1-j & 1+j \\\\ 1+j & 1+j \end{array} \right]$
- $\left[ \begin{array} \ 1-j & 1+j \\\\ -1+j & 1-j \end{array} \right]$
- $\left[ \begin{array} \ 1+j & 1+j \\\\ 1-j & 1-j \end{array} \right]$
- $\left[ \begin{array} \ 1+j & -1+j \\\\ -1+j & 1+j \end{array} \right]$
The impedance parameters Z11 and Z12 of the two-port network in figure are
- Z11 = 2.75 $\Omega$and Z12 = 0.25 $\Omega$
- Z11 = 3 $\Omega$and Z12 = 0.5 $\Omega$
- Z11 = 3 $\Omega$and Z12 = 0.25 $\Omega$
- Z11 = 2.25 $\Omega$and Z12 = 0.5 $\Omega$
In the circuit given below, what value of RL maximizes the power delivered to RL?
- 2.4 $\Omega$
- $\dfrac{8}{3}$$\Omega$
- 4$\Omega$
- 6$\Omega$
In the figure shown below, assume that all the capacitors are initially uncharged. If vi (t) = 10u (t ) Volts, v0 (t) is given by
- 8e-0.004t Volts
- 8 (1- e-0.004t) Volts
- 8u (t) Volts
- 8 Volts
The transfer function H(s) = $\dfrac{V_0 (s)}{V_i (s)}$ of an RLC circuit is given by
H(s) = $\dfrac{10^6}{s^2 + 20s + 10^6}$
The Quality factor (Q-factor) of this circuit is
- 25
- 50
- 100
- 5000
A two port network is represented by ABCD parameters given by
$\left[
\begin{array}
\ V_1 \\
I_1
\end{array}
\right]
$$\left[
\begin{array}
\ A & B\\
C & D
\end{array}
\right]
$$\left[
\begin{array}
\ V_2 \\
-I_2
\end{array}
\right]
$
If port-2 is terminated by RL, the input impedance seen at port-1 is given by
- $\dfrac{A + BR_L}{C + DR_L}$
- $\dfrac{AR_L + C}{BR_L + D}$
- $\dfrac{DR_L + A}{BR_L + C}$
- $\dfrac{B + AR_L}{D + CR_L}$
The maximum power that can be transferred to the load resistor RL of 100$\Omega$ from the voltage source of 5 V is __________
- 1 W
- 10 W
- 0.25 W
- 0.5 W
If R1 = R2 = R4 and R3 = 1. 1R in the bridge circuit shown in figure, then the reading in the ideal voltmeter connected between a and b is
- 0.238 V
- 0.138 V
- -0.238 V
- 1 V
The first and the last critical frequencies (singularities) of a driving point impedance function of a passive network having two kinds of elements, are a pole and a zero respectively. The above property will be satisfied by
- RL network only
- RC network only
- LC network only
- RC as well as RL networks
The impedance looking into nodes 1 and 2 in the given circuit is
- 50 $\Omega$
- 100 $\Omega$
- 5 k$\Omega$
- 10. 1k$\Omega$
For the circuit shown in figure, Thevenin's voltage and Thevenin's equivalent resistance at terminals a - b is
- $5V \ and \ 2\Omega$
- $7.5V \ and \ 2.5\Omega$
- $4V \ and \ 2\Omega$
- $3V \ and \ 2.5\Omega$
Twelve 1$\Omega$ resistances are used as edges to form a cube. The resistance between two diagonally opposite corners of the cube is
- $\dfrac{5}{6}$ $\Omega$
- $\dfrac{1}{6}$ $\Omega$
- $\dfrac{6}{5}$ $\Omega$
- $\dfrac{3}{2}$ $\Omega$
The circuit shown in the figure is used to charge the capacitor C alternately from two current sources as indicated. The switches S1 and S2 are mechanically coupled and connected as follows:
For 2nT $\le$ t $\le$ (2n + 1) T, (n = 0, 1, 2, ...) S1 to P1 and S2 to P2
For (2n + 1) T $\le$ t $\le$ (2n + 2) T, (n = 0, 1, 2, ...) S1 to Q1 and S2 to Q2
Assume that the capacitor has zero initial charge. Given that u (t) is a unit step function, the voltage vc (t) across the capacitor is given by
- $\displaystyle \sum_{n=1}^\infty (-1)^n \ tu \ (t - nT)$
- u (t) + 2$\displaystyle \sum_{n=1}^\infty (-1)^n \ u \ (t - nT)$
- t u (t) + 2$\displaystyle \sum_{n=1}^\infty (-1)^n \ u \ (t - nT)$(t - nT)
- $\displaystyle \sum_{n=1}^\infty [0.5 - e^{-(t-2nT)} + 0.5 e^{-(t-2nT)} ]$
An input voltage v(t) = 10$\sqrt 2$ cos(t+100) + 10$\sqrt 3$ cos (2t+10o)V is applied to a series combination of resistance R = 1 $\Omega$ and an inductance L = 1 H. The resulting steady state current i(t) in ampere is
- 10 cos (t + 550) + 10 cos (2t + 100 + tan-12)
- 10 cos (t + 550) + 10 $\sqrt{ \dfrac{3}{2} }$ cos (2t + 550)
- 10 cos (t - 350) + 10 cos ( 2t + 100 - tan-12)
- 10 cos (t - 350) + 10 $\sqrt{ \dfrac{3}{2} }$cos (2t - 350)
