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Test 1 - Network Graphs | Electronics and Communication (ECE)

Description: A test for Network Graphs of Electronics and Communication (ECE)
Number of Questions: 20
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Tags: Network Graphs
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In the interconnection of ideal sources shown in the figure, it is known that the 60 V source is absorbing power.

Which of the following can be the value of the current source I?

  1. 10 A

  2. 13 A

  3. 15 A

  4. 18 A

Show answer
Correct Option: A
Explanation:

The RC circuit shown in the figure is

  1. a low-pass filter

  2. a high-pass filter

  3. a band-pass filter

  4. a band-reject filter

Show answer
Correct Option: C
Explanation:

A series RLC circuit has a resonance frequency of 1 kHz and a quality factor Q = 100. If each R, L and C is doubled from its original value, the new Q of the circuit is

  1. 25

  2. 50

  3. 100

  4. 200

Show answer
Correct Option: B
Explanation:

In the circuit shown below, the value of RL such that the power transferred to RL is maximum is

  1. 5$\Omega$

  2. 10$\Omega$

  3. 15$\Omega$

  4. 20$\Omega$

Show answer
Correct Option: C
Explanation:

The ABCD parameters of an ideal n : 1 transformer shown in figure are $ \left[ \begin{array} \ n & 0 \\ 0 & x \end{array} \right]

$. The value of X will be

  1. n

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

  3. n2

  4. $\dfrac{1}{n^2}$

Show answer
Correct Option: B
Explanation:

A circuit consists of a resistor, an inductor and a capacitor connected in series to a 150 V AC mains. For the circuit, R = 9 Ohms, XL = 28 Ohms and XC = 16 Ohms. What is the value of the current in the circuit?

  1. 10 A

  2. 15 A

  3. 20 A

  4. 25 A

Show answer
Correct Option: A
Explanation:

Z = $\sqrt{R^2 + (X_L - X_C)^2 }$ = $\sqrt{9^2 + (28 - 16)^2 }$ = 15 Ohms = I = 150/15 = 10 A

Impedance Z as shown in the figure is

  1. $j29\Omega$

  2. $j9\Omega$

  3. $j19\Omega$

  4. $j39\Omega$

Show answer
Correct Option: B
Explanation:

For the circuit shown in the figure, the initial conditions are zero. Its transfer function H(s) =$\dfrac{V_0(s)}{V_i(s)}$is

  1. $\dfrac{1}{s^2 + 10^6s + 10^6}$

  2. $\dfrac{10^6}{s^2 + 10^3s + 10^6}$

  3. $\dfrac{10^3}{s^2 + 10^3s + 10^6}$

  4. $\dfrac{10^6}{s^2 + 10^6s + 10^6}$

Show answer
Correct Option: D
Explanation:

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

  1. 3/7 A

  2. 5/7

  3. 1A

  4. 9/7 A

Show answer
Correct Option: C
Explanation:

 Using Y-admittance model to figure out current for port B: I4 = Y21.V1 + Y22.V2 (i) from the 1st condition, -3 = Y21.10 + Y22.3 (since V2 = - I4 *RL = -(-3)*1 = 3V) (ii) from the 2nd condition, -2 = Y21.10 + Y22.5 (since V2 = - I4 *RL = -(-2)*2.5 = 5V) Solving these 2 equations, we get : Y21 = -9/20 and Y22 = 1/2 Hence, I4 = (-9/20).V1 + (1/2).V2 Given, V1 = 10V and V2 = (-I4 * 7) Substituting, We get I4 = -1A Ans : Current drawn by the 7$\Omega$ load resistor = -I4 = 1A

For parallel RLC circuit, which one of the following statements is NOT correct?

  1. The bandwidth of the circuit deceases if R is increased.

  2. The bandwidth of the circuit remains same if L is increased.

  3. At resonance, input impedance is a real quantity.

  4. At resonance, the magnitude of input impedance attains its minimum value.

Show answer
Correct Option: D
Explanation:

With 10 V dc connected at port A in the linear non-reciprocal 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

  1. 6 V

  2. 7 V

  3. 8 V

  4. 9 V

Show answer
Correct Option: C
Explanation:

 on applying Thevenin's equation, for (i) and (ii), we get Vth = 3Rth +3.....(1) Vth = 2Rth +5...(2) subtracting (1) and (2), we get, Rth = 2ohms and Vth = 9V when input was 10V if the input is 6V then Vth = 7/3 * (2+1) =7V Vth = Vb=f(Vs) which implies Vth = mVs + c Vth(10V) = m(10) +C...(3)  Vth(6) = m(6)+ C...(4) Vth(7) = m(7) + C....(5) and Vth(8V) = m(8) + C.....(6) by substituting values in eq. (3 & 4) we get, 9 = 10m +C...(7) and 7 = 6m + C...(8) subtracting(7) and (8), we get, m = 0.5 and c = 4 now, Vth(8V) = m(8)+C = 8 volts  

The time domain behaviour of an RL circuit is represented by

L$\dfrac{d_i}{d_t} + R_i = V_0 (1 + Be^{-RT/L} sint ) \ u(t)$

For an initial current of i (0) = $\dfrac{V_0}{R}$, the steady state value of the current is given by

  1. i (t) $\rightarrow \dfrac{V_0}{R}$

  2. i (t) $\rightarrow \dfrac{2V_0}{R}$

  3. i (t) $\rightarrow \dfrac{V_0}{R}$(1 + B)

  4. i (t) $\rightarrow \dfrac{2V_0}{R}$(1 + B)

Show answer
Correct Option: A
Explanation:

The average power delivered to an impendence (4 -j3)$\Omega$by a current 5 cos(100$\Omega$t + 100) A is

  1. 44.2 W

  2. 50 W

  3. 62.5 W

  4. 125 W

Show answer
Correct Option: B

Assuming both the voltage sources are in phase, the value of R for which maximum power is transferred from circuit A to circuit B is

  1. 0.8 $\Omega$

  2. 1.4 $\Omega$

  3. 2$\Omega$

  4. 2.8 $\Omega$

Show answer
Correct Option: A

For the circuit shown in the figure, the Thevenin voltage and resistance looking into X - Y are

  1. $\dfrac{4}{3}$V, 2$\Omega$

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

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

  4. 4V, 2$\Omega$

Show answer
Correct Option: D
Explanation:

The driving point impedance Z(s) of a network has the pole-zero locations as shown in the figure below. If Z(0) = 3, then Z(s) is

  1. $\dfrac{3(s+3)}{s^2 + 2s + 3}$

  2. $\dfrac{2(s+3)}{s^2 + 2s + 2}$

  3. $\dfrac{3(s-3)}{s^2 - 2s - 2}$

  4. $\dfrac{2(s-3)}{s^2 - 2s - 2}$

Show answer
Correct Option: B
Explanation:

The minimum number of equations required to analyse the circuit shown in the figure below is

  1. 3

  2. 4

  3. 6

  4. 7

Show answer
Correct Option: B
Explanation:

Number of loops = b − n + 1 = minimum number of equations Number of branches = b = 8 Number of nodes = 5 Minimum number of equations = 8 − 5 + 1 = 4  Hence, (2) is the correct option.

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

Show answer
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

Show answer
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

Show answer
Correct Option: C
Explanation:

I = 20/√(66 x 66 + 47 x 47) I = 247 mA

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