JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD

JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech II NETWORK THEORY Previous Question Paper (July/August - 2021 - R13)

B.Tech | II | II | R13 | July/August - 2021

Answer PaperPaper Code: 114ACMaximum Marks: 75Time: 3 hours

Answer any Five Questions. All Questions Carry Equal Marks.

Main

115 marks

Derive the relation between phase and line values of a three phase star connected balanced system.

8 marks

Answer

Consider a balanced three-phase star-connected system with phase voltages V_AN, V_BN, V_CN and line voltages V_AB, V_BC, V_CA. The line currents are I_A, I_B, I_C and phase currents are I_AN, I_BN, I_CN. In a star connection, the line current is equal to the phase current. Therefore, I_A = I_AN, I_B = I_BN, and I_C = I_CN.

The line voltage is the voltage between any two lines. For example, the line voltage V_AB is given by:

V_AB = V_AN - V_BN

In a balanced three-phase system, the phase voltages are equal in magnitude and displaced by 120 degrees from each other. Let the phase voltage magnitude be V_ph. The line voltage magnitude V_L can be derived as follows:

V_L² = |V_AN - V_BN|²

Using phasor representation, let V_AN = V_ph∠0° and V_BN = V_ph∠-120°.

V_AB = V_ph∠0° - V_ph∠-120°

V_AB = V_ph(cos(0°) + j sin(0°)) - V_ph(cos(-120°) + j sin(-120°))

V_AB = V_ph(1 + j0) - V_ph(-0.5 - j0.866)

V_AB = V_ph(1 + 0.5 + j0.866)

V_AB = V_ph(1.5 + j0.866)

The magnitude of V_AB is:

|V_AB| = V_ph * sqrt(1.5² + 0.866²)

|V_AB| = V_ph * sqrt(2.25 + 0.75)

|V_AB| = V_ph * sqrt(3)

Thus, the relation between line voltage (V_L) and phase voltage (V_ph) is V_L = √3 * V_ph.

The phase angle of V_AB is arctan(0.866/1.5) = 30°.

Therefore, in a balanced three-phase star-connected system:

Line current = Phase current (I_L = I_ph)
Line voltage = √3 * Phase voltage (V_L = √3 * V_ph)
The line voltage leads the corresponding phase voltage by 30°.

A balanced 3-phase, 200V, 50Hz supply is given to a load consisting of 3-impedances (4+j2), (1+j3) and (2+j4) ohms connected in star. Calculate the voltages across and currents in three phases of load.

7 marks

Answer

Given:

Supply Voltage (Line Voltage), V_L = 200 V
Frequency, f = 50 Hz
Load impedances connected in star:
Z_AN = (4 + j2) Ω
Z_BN = (1 + j3) Ω
Z_CN = (2 + j4) Ω

Since the system is balanced, the line voltage is distributed equally among the phases. The phase voltage (V_ph) is given by:

V_ph = V_L / √3

V_ph = 200 V / √3 ≈ 115.47 V

We can assume the phase voltages to be:

V_AN = 115.47∠0° V

V_BN = 115.47∠-120° V

V_CN = 115.47∠120° V

Voltages across the load impedances:

In a star connection, the voltage across each impedance is the phase voltage.

Voltage across Z_AN = V_AN = 115.47∠0° V
Voltage across Z_BN = V_BN = 115.47∠-120° V
Voltage across Z_CN = V_CN = 115.47∠120° V

Currents in the three phases of the load:

The current in each phase is calculated using Ohm's Law (I = V/Z).

Phase A current (I_A):

I_A = V_AN / Z_AN = (115.47∠0° V) / (4 + j2 Ω)

Convert Z_AN to polar form: |Z_AN| = √(4² + 2²) = √20 ≈ 4.47 Ω. Angle = arctan(2/4) ≈ 26.57°.

Z_AN ≈ 4.47∠26.57° Ω

I_A = (115.47 / 4.47) ∠(0° - 26.57°) A ≈ 25.83∠-26.57° A

Phase B current (I_B):

I_B = V_BN / Z_BN = (115.47∠-120° V) / (1 + j3 Ω)

Convert Z_BN to polar form: |Z_BN| = √(1² + 3²) = √10 ≈ 3.16 Ω. Angle = arctan(3/1) ≈ 71.57°.

Z_BN ≈ 3.16∠71.57° Ω

I_B = (115.47 / 3.16) ∠(-120° - 71.57°) A ≈ 36.54∠-191.57° A

Note: -191.57° is equivalent to 168.43° (-191.57° + 360°).

I_B ≈ 36.54∠168.43° A

Phase C current (I_C):

I_C = V_CN / Z_CN = (115.47∠120° V) / (2 + j4 Ω)

Convert Z_CN to polar form: |Z_CN| = √(2² + 4²) = √20 ≈ 4.47 Ω. Angle = arctan(4/2) ≈ 63.43°.

Z_CN ≈ 4.47∠63.43° Ω

I_C = (115.47 / 4.47) ∠(120° - 63.43°) A ≈ 25.83∠56.57° A

Summary of Voltages and Currents:

Parameter	Phase A	Phase B	Phase C
Voltage across Load	115.47∠0° V	115.47∠-120° V	115.47∠120° V
Current in Load	25.83∠-26.57° A	36.54∠168.43° A	25.83∠56.57° A

Show that in the two-wattmeter method of 3-Φ power measurement, the sum of the readings of the two wattmeters gives the total power consumed in 3-Φ circuit. Hence prove:

Φ = tan^-1(√3(W₁-W₂) / (W₁+W₂))

Where Φ is the phase angle of the load and W₁ & W₂ are the readings of the wattmeters.

15 marks

Answer

Derive the expression for current when a dc voltage V is applied suddenly (i.e. at time = 0) by closing a switch in a series R-L circuit by using Laplace transform.

15 marks

Answer

Find the expression for current of a series R-L-C circuit fed by constant DC voltage of 20 V with R = 4 Ω, L = 1 H and C = 0.25 F. Assume initial conditions to be zero.

15 marks

Answer

For the circuit shown in Figure below, obtain the Driving point Admittance.

Y₁₁ = I₁ / V₁

15 marks

Answer

Explain in detail Time Domain Response from Pole-Zero plot with an example.

15 marks

Answer

715 marks

Derive the relation between ‘Y’ and ABCD parameters.

7 marks

Answer

A two port network has the following parameters: Z₁₁ = 4Ω, Z₁₂ = 1Ω, Z₂₁ = 3Ω and Z₂₂ = 3Ω. Calculate transmission parameters.

8 marks

Answer

Design a high pass filter with a cut-off frequency of 1 KHz with a terminated design impedance of 800 Ω.

15 marks

JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech II NETWORK THEORY Previous Question Paper (July/August - 2021 - R13)

Main

Answer

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Voltages across the load impedances:

Currents in the three phases of the load:

Summary of Voltages and Currents:

Answer

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