Test 4 - Network Graphs | Electronics and Communication (ECE)

Description: A test for Network Graphs of Electronics and Communication (ECE)
Number of Questions: 22
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Tags: Network Graphs GATE(ECE)
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In the following graph, the number of trees (P) and the number of cut-set (Q) are

  1. P = 2 Q = 2

  2. P = 2 Q = 6

  3. P = 4 Q = 6

  4. P = 4 Q = 10


Correct Option: C
Explanation:

The equivalent inductance measured between the terminals 1 and 2 for the circuit shown in the figure is

  1. L1 + L2 + M

  2. L1 + L2 – M

  3. L1 + L2 + 2M

  4. L1 + L2 – 2M


Correct Option: D
Explanation:

The sign of M is as per the sign of L if current enters or exits the dotted terminals of both the coils. The sign of M is the opposite of L if current enters in dotted terminal of a coil and exits from the dotted terminal of other coil. Thus, Leq = 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. 1.05 A

  2. 303 mA

  3. 247 mA

  4. 107 mA


Correct Option: C
Explanation:

I = 20/√(66 x 66 + 47 x 47) I = 247 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. 1/90

  2. -1/90

  3. -1/99

  4. -1/11


Correct Option: D
Explanation:

The Thevenin equivalent impedance Zth between the nodes P and Q in the following circuit is

  1. 1

  2. 1 + s + $\dfrac{1}{s}$

  3. 2 + s + $\dfrac{1}{s}$

  4. $\dfrac{s^2 + s + 1}{s^2 + 2s +1}$


Correct Option: A
Explanation:

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

  1. 3 V

  2. -3V

  3. 4 V

    • 4V

Correct Option: B
Explanation:

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

  1. 4

  2. 1

  3. $\dfrac{1}{2}$

  4. $\dfrac{1}{4}$


Correct Option: D
Explanation:

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

  1. sin (103t – 450)

  2. sin (103t + 450)

  3. sin (103t – 530)

  4. sin (103t + 530)


Correct Option: A
Explanation:

For the lattice shown in the figure, Za = j2$\Omega$ and Zb = 2$\Omega$. Calculate the values of the open circuit impedance parameter [z] = $\left[ \begin{array} \ Z_{11} & Z_{12} \\ Z_{21} & Z_{22} \end{array} \right]$.

  1. $\left[ \begin{array} \ 1-j & 1+j \\ 1+j & 1+j \end{array} \right]$

  2. $\left[ \begin{array} \ 1-j & 1+j \\ -1+j & 1-j \end{array} \right]$

  3. $\left[ \begin{array} \ 1+j & 1+j \\ 1-j & 1-j \end{array} \right]$

  4. $\left[ \begin{array} \ 1+j & -1+j \\ -1+j & 1+j \end{array} \right]$


Correct Option: D
Explanation:

The impedance parameters Z11 and Z12 of the two-port network in figure are

  1. Z11 = 2.75 $\Omega$and Z12 = 0.25 $\Omega$

  2. Z11 = 3 $\Omega$and Z12 = 0.5 $\Omega$

  3. Z11 = 3 $\Omega$and Z12 = 0.25 $\Omega$

  4. Z11 = 2.25 $\Omega$and Z12 = 0.5 $\Omega$


Correct Option: A
Explanation:

 

In the circuit given below, what value of RL maximizes the power delivered to RL?

  1. 2.4 $\Omega$

  2. $\dfrac{8}{3}$$\Omega$

  3. 4$\Omega$

  4. 6$\Omega$


Correct Option: C
Explanation:

In the figure shown below, assume that all the capacitors are initially uncharged. If vi (t) = 10u (t ) Volts, v0 (t) is given by

  1. 8e-0.004t Volts

  2. 8 (1- e-0.004t) Volts

  3. 8u (t) Volts

  4. 8 Volts


Correct Option: B
Explanation:

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

  1. 25

  2. 50

  3. 100

  4. 5000


Correct Option: B
Explanation:

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

  1. $\dfrac{A + BR_L}{C + DR_L}$

  2. $\dfrac{AR_L + C}{BR_L + D}$

  3. $\dfrac{DR_L + A}{BR_L + C}$

  4. $\dfrac{B + AR_L}{D + CR_L}$


Correct Option: D
Explanation:

 

The maximum power that can be transferred to the load resistor RL of 100$\Omega$ from the voltage source of 5 V is __________

  1. 1 W

  2. 10 W

  3. 0.25 W

  4. 0.5 W


Correct Option: C
Explanation:

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

  1. 0.238 V

  2. 0.138 V

  3. -0.238 V

  4. 1 V


Correct Option: C
Explanation:

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

  1. RL network only

  2. RC network only

  3. LC network only

  4. RC as well as RL networks


Correct Option: B
Explanation:

The impedance looking into nodes 1 and 2 in the given circuit is

  1. 50 $\Omega$

  2. 100 $\Omega$

  3. 5 k$\Omega$

    1. 1k$\Omega$

Correct Option: A

For the circuit shown in figure, Thevenin's voltage and Thevenin's equivalent resistance at terminals a - b is

  1. $5V \ and \ 2\Omega$

  2. $7.5V \ and \ 2.5\Omega$

  3. $4V \ and \ 2\Omega$

  4. $3V \ and \ 2.5\Omega$


Correct Option: B
Explanation:

Twelve 1$\Omega$ resistances are used as edges to form a cube. The resistance between two diagonally opposite corners of the cube is

  1. $\dfrac{5}{6}$ $\Omega$

  2. $\dfrac{1}{6}$ $\Omega$

  3. $\dfrac{6}{5}$ $\Omega$

  4. $\dfrac{3}{2}$ $\Omega$


Correct Option: A
Explanation:

For Current i, there are 3 similar paths. So current will be divide in 3 paths.

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

  1. $\displaystyle \sum_{n=1}^\infty (-1)^n \ tu \ (t - nT)$

  2. u (t) + 2$\displaystyle \sum_{n=1}^\infty (-1)^n \ u \ (t - nT)$

  3. t u (t) + 2$\displaystyle \sum_{n=1}^\infty (-1)^n \ u \ (t - nT)$(t - nT)

  4. $\displaystyle \sum_{n=1}^\infty [0.5 - e^{-(t-2nT)} + 0.5 e^{-(t-2nT)} ]$


Correct Option: C
Explanation:

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

  1. 10 cos (t + 550) + 10 cos (2t + 100 + tan-12)

  2. 10 cos (t + 550) + 10 $\sqrt{ \dfrac{3}{2} }$ cos (2t + 550)

  3. 10 cos (t - 350) + 10 cos ( 2t + 100 - tan-12)

  4. 10 cos (t - 350) + 10 $\sqrt{ \dfrac{3}{2} }$cos (2t - 350)


Correct Option: C
Explanation:

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