UPPCL AE Electrical Engineer Solved Paper 2019

UPPCL AE / March 15, 2022

Ques.61. An ammeter with a range of 0 to 100 µA has an internal resistance of 100 Ω.  For extending its range to 0 to 500 µA, the shunt resistance required is

  1. 20 Ω
  2. 22.22 Ω
  3. 50 Ω
  4. 25 Ω

Answer.4. 25Ω

Explanation:-

An ammeter with a range of 0 to 100 µA has an internal resistance of 100 Ω. 

Full scale deflection current (Im) = 100 µA = 1 × 10−4 A

Internal Resistance Rm = 100 Ω

After range extension, Ammeter current (I) = 500 µA = 5 × 10−4 A

Multiplication factor of ammeter is given by

m = I/Im = (5 × 10−4 A)/(1 × 10−4 A)

m = 5

The shunt resistance of the ammeter can be calculated as

Rsh = Rm/(m − 1) = 100/(5 − 1)

Rsh = 25Ω

 

Ques.62. A BJT current source is given in the figure. Assume the Si-PNP transistor to operate in the active region. The value of current I in mA is

A single-phase AC distribution line supplies two single-phase loads as shown in the figure below. The impedances of line segments A-B and B-C are j0.25 Ω and j0.35 Ω respectively. The voltage drop from A to C is

  1. 1.85
  2. 2.15
  3. 5.25
  4. 4.1

Answer.2. 2.15

Explanation:-

Applying KVL loop at the input side through Zener diode as it regulates the voltage across the emitter resistance.

VZ −IERE − VBE = 0

∵ For PNP transistor potential barrier is −0.7 V

5 − IE × 2 − 0.7 = 0

2 × 103 × IE = 4.3

IE  = 2.15 mA

∵ The value of β is not given we assume that β is very large

∴ IE = IC = I = 2.15 mA

 

Ques.63. A single-phase AC distribution line supplies two single-phase loads as shown in the figure below. The impedances of line segments A-B and B-C are j0.25 Ω and j0.35 Ω respectively. The voltage drop from A to C is

A single-phase AC distribution line supplies two single-phase loads as shown in the figure below.

 

  1. (30 + j 4.5) V
  2. (4.5 + j 30) V
  3. (30 − j 4.5) V
  4. (4.5 − j 30) V

Answer.2. (4.5 + j30) V

Explanation:-

VA = VC − (40 × j0.35) − (40 + 30∠−cos−1(0.8)) × j0.25

IB = 30∠−cos−10.8

= 30∠−36.87° A

IC = 40∠0° A

IAB = IB + IC

VC – A = − (40 × 0.35∠90°) − (40 + 30∠−36.86) × 0.25∠90°

VC – A = 16.6207∠74.29° + 14∠90°

VC – A = 30.34∠81.46°

or

VC – A = (4.5 + j30) V

 

Ques.64. A factory draws 100 kW at 0.8 p.f. lagging from a 3-phase, 11 kV supply. It is desired to raise the p.f. to unity using the capacitor bank. The total power rating of the capacitor bank is

  1. 75 kVAR
  2. 100 kVAR
  3. 62.5 kVAR
  4. 50 kVAR

Answer.1. 75 kVAR

Explanation:-

Given,

Active power (P) = 100 kW

Power factor (p.f) = 0.8 lag

Apparent power (S) = ? kVA

Reactive power (Q) = ? kVAR

⇒ The power rating of the capacitor is rated  as kVAR i.e reactive power Q

P = S × cosφ

100 = S × 0.8

S = 125 kVA

Reactive power is given by

Q = S × sinφ

∵ sin2φ + cos2φ =1

sin2φ + 0.82 =1

sinφ = 0.6

∴ Q = 125 × 0.6

Q = 75 kVAR

 

Ques.65. An open circuit test is conducted on a 1100/110 V, 50 Hz single-phase transformer with instruments connected on the low voltage side of the transformer. The voltmeter reads 110 V. The ammeter reads 2 A. The wattmeter reading is 65 W. The approximate core-loss resistance and magnetizing reactance, referred to as the low voltage side, are respectively.

  1. 1.8615 Ω, 0.5757 Ω
  2. 18.615 Ω, 5.757 Ω
  3. 1861552 Ω, 5757 Ω
  4. 186.15 Ω, 57. 575 Ω

Answer.4. 186.15 Ω, 57. 575 Ω

Explanation:-

Given

Vo = 110 V

Io = 2 A

Po = 65 W

Power can be expressed as

Po = VoIo Cosφ

Cosφ = Po/VoIo

Cosφ = 65/110 × 2

Cosφ = 0.295 

∴ sinφ = 0.955

The core loss component of the current can be expressed as

Iw = IoCosφ

Iw = 2 × 0.295

Iw = 0.6W

Magnetizing component of current can be expressed as

Im = Iosinφ

Im = 2 × 0.955

Im = 1.9 A

Cores loss Resistance

Ro = V/Iw = 110/0.6

Ro = 186.44 Ω

Magnetizing reactance

Xm = Vo/Im = 110/1.9

Xm =57.89 Ω

 

Ques.66. The Zener diode in the circuit has a Zener voltage Vz of 15 V and a power rating of 0.5 Watt. If the input voltage is 40 V. What is the minimum value of Rs that prevents the Zener diode from being destroyed?

The Zener diode in the circuit has a Zener voltage Vz of 15 V and a power rating of 0.5 Watt. If the input voltage is 40 V.

  1. 150 Ω
  2. 250 Ω
  3. 750 Ω
  4. 550 Ω

Answer.3. 750 Ω

Explanation:-

Given

VZ = 15 V

Vin = 40 V

PZ = 0.5 W = Vz . Iz

Iz = 0.5/15 = 1/30 A

The minimum value of RS that prevents Zener diode from being destroyed

Iz = (Vin − Vz)/Rs

1/30 = (40 − 15)/Rs

Rs = 750 Ω

 

Ques.67. In the circuit shown, assume that the voltage source and transformers are ideal. The AC voltage source is 10√2 sin(100πt) V. The RMS value of the current flowing through the 1 Ω resistor is approximately (rounded off till first decimal place).

In the circuit shown, assume that the voltage source and transformers are ideal. The AC voltage source is 10√2 sin(100πt) V.

  1. 0.1 A
  2. 9.1 A
  3. 10.0 A
  4. 0.9 A

Answer.1. 0.1 A

Explanation:-

AC voltage Source = 10√2 sin(100πt) V

Transferring the 10-ohm resistor to the primary side with the transformation ratio of 10:1.

R’ = 1000 Ω

Total equivalent resistance in the secondary terminal of the 1 :10 transformer is

R’ = 1 + 1000 = 10001 Ω

Now, referring the R’ to the primary side or at the source side.

R'(new) = 10.01 Ω

Then the RMS value of the current flowing in the 1 Ω resistor and source RMS voltage at the secondary side is 100 V.

I = 100/1001 = 0.0999 ≅ 0.1 A

 

Ques.68. For the circuit shown in the figure, find the node voltages V1 and V2.

For the circuit shown in the figure, find the node voltages V1 and V2.

  1. V1 = 8.33 V, V2 = 10.33 V
  2. V1 = 8 V, V2 = 10 V
  3. V1 = 6 V, V2 = 8 V
  4. V1 = −7.33 V, V2 = −5.33 V

Answer.4. V1 = −7.33 V, V2 = −5.33 V

Explanation:-

Applying KCL (super Node) in the above figure.

2 = (V1 − 0)/2 + (V2 − 0) + 7 = 0

8 = 2V1 + V2 + 28

V2 = −2V1 − 20—–(1)

Applying KVL at super node:

−V1 − 2 + V2 = 0

V2 = V1 + 2—–(2)

Equating equation (1) and (2)

V1  = −7.33 V

V2 = −5.33 V

 

Ques.69. In a boost converter shown in the figure, the duty cycle is 0.5. The inductor current is assumed to be continuous. Capacitor C is assumed to be very large. If the switching frequency is 20 kHz, the peak to peak inductor current ripple is

In a boost converter shown in the figure, the duty cycle is 0.5. The inductor current is assumed to be continuous.

  1. 0.45 A
  2. 0.35 A
  3. 0.25 A
  4. 0.15 A

Answer.3. 0.25 A

Explanation:-

Input Voltage (Vin) = 20 V

Frequency (f) = 20 kHz = 20 × 103 Hz

Duty cycle (α) = 0.5

Inductor (L) = 2 mH = 2 × 10−3

The peak to peak ripple inductor current is given as

ΔI = (α × Vin)/(f × L)

ΔI = (0.5 × 20)/(20 × 103 × 2 × 10−3)

ΔI =0.25 A

 

Ques.70. A sinusoidal AC voltage source feeds a pure inductor through a diode as shown in the figure. The duration of conduction (in degrees) of the diode in one input power cycle is

A sinusoidal AC voltage source feeds a pure inductor through a diode as shown in the figure. The duration of conduction

  1. 30°
  2. 90°
  3. 180°
  4. 360°

Answer.4. 360°

Explanation:-

The duration of conduction (in degrees) of the diode in one input power cycle is 360°.

 

Ques.71. A single-phase full-bridge voltage source inverter is operated in 180° mode with square wave output. If the input DC supply is 100 V, the RMS value of the fundamental output voltage is

  1. 80 V
  2. 50 V
  3. 90 V
  4. 70 V

Answer.3. 90 V

Explanation:-

The fundamental value of the output voltage is given as

Vout fundamental = (2√2 × Vdc)/π

= (2√2 × 100)/3.14

Vout fundamental = 90 V

 

Ques.72. A charge q1 = 7 μC is located at the origin, and a second charge q2 = -5 μC is located on the positive x-axis, 0.3 m from the origin. Find the electric field at point P, which has coordinates (0,0.4) m. Coulomb constant is given by Ke = 1/4πεo is the = 8.99 × 109 Nm2/C2, εo is the permittivity of free space; [katex](\widehat i,\widehat j)[/katex] are the unit vectors along x, y axes.

  1. [katex]- (1.1\widehat i + 2.5\widehat j) \times {10^5}N/C[/katex]
  2. [katex](1.1\widehat i + 2.5\widehat j) \times {10^5}N/C[/katex]
  3. [katex](2.5\widehat i + 1.1\widehat j) \times {10^5}N/C[/katex]
  4. [katex]- (2.5\widehat i + 1.1\widehat j) \times {10^5}N/C[/katex]

Answer.2. [katex](1.1\widehat i + 2.5\widehat j) \times {10^5}N/C[/katex]

Explanation:-

The Electric potential field is given by

[katex]E = \dfrac{Q}{{4\pi {\varepsilon _o}{r^2}}}{\widehat a_r}[/katex]

A charge q1 = 7 μC is located at the origin, and a second charge q2 = -5 μC is located on the positive x-axis,

 

Ques.73. The terminal voltage of an ideal DC voltage source is 12 V when connected to a 2 W resistive load. When the load is disconnected, the terminal voltage rises to 12.4V. What are the values of source voltage, Vs, and internal resistance Rs of the source?

  1. Vs =12 V, Rs = 4 Ω
  2. Vs =10 V, Rs = 8 Ω
  3. Vs =12.4 V, Rs = 2.4 Ω
  4. Vs =12 V, Rs = 2.4 Ω

Answer.3. Vs =12.4 V, Rs = 2.4 Ω

Explanation:-

When the load is disconnected i.e. open-circuited, the terminal voltage is 12.4 V. Therefore, the source voltage is 12.4

And, when the resistive load of 2 W, the terminal voltage seen is 12 V.

Therefore internal resistance Rs

Voc − Vs = I × Rs

P = V.I

2 = 12 × I

I = 1/6A

12.4 − 12 = Rs/6

Rs = 2.4 Ω

 

Ques.74. Incremental fuel costs in Rs/MWh for a plant consisting of two generating units are given by dF1/dP1 = 0.4P1 +400 and dF2/dP2 =  0.48P2 +320. The allocation of loads P1 and P2 between generating units 1 and 2, respectively, for the minimum cost of generation to serve a total load of 900 MW, neglecting losses, is

  1. 200 MW and 7000 MW
  2. 500 MW and 400 MW
  3. 300 MW and 600 MW
  4. 400 MW and 500 MW

Answer.4. 400 MW and 500 MW

Explanation:-

The incremental fuel cost is Rs/MWh are:

dF1/dP1 = 0.4P1 +400

dF2/dP2 =  0.48P2 +320

The total load shared by both generating units is:

P1 + P2 = 900 ———(1)

Equating both incremental cost equations, we get

0.4P1 +400 = 0.48P2 +320

0.48P2 =0.4P1 + 80

Substituting the value in equation 1

P1 = 400 MW

P2 = 500 MW

 

Ques.75. A power system has three synchronous generators. The turbine-governor characteristics corresponding to the generators are P1 =50(50 − f), P2 = 100(51 − f), P3 = 150(52 − f) Where, f denotes the system frequency in Hz, and P1, P2, P3 are the power outputs of the turbines in MW. Assuming generators and transmission network to be lossless, the system frequency for a load of 700 MW is

  1. 49 Hz
  2. 47.5 Hz
  3. 48 Hz
  4. 49.5 Hz

Answer.1. 49 Hz

Explanation:-

Turbine Governor Characteristics

P1 =50(50 − f)

P2 = 100(51 − f)

P3 = 150(52 − f)

P1 + P2 + P3 = 700

50(50 − f) + 100(51 − f) + 150(52 − f) = 700

2500 − 50f + 5100 − 100f + 7800 − 150f = 700

14700 = 300f

f = 49 Hz

For UPPCL JE 2016 Electrical paper with complete solution Click Here

For UPPCL JE 2016(Evening Shift) Electrical paper with complete solution Click Here

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