UPPCL JE 2018 Electrical question paper with Explanation 27-Aug-2018

Ques.11. A capacitor consists of two similar plates each measuring 10 cm x 10 cm mounted parallel and opposite to each other. The Value of capacitance when the distance between them is 1 cm and the dielectric used in the air will be

  1. 8.854 F
  2. 8.854 pF
  3. 8.854 μF
  4. 8.854 mF

The capacitance of the parallel plate capacitor is given as

C =εoA/D


ε= Permittivity of free space 8.854 × 10−12

A = Area of the plate = 10 × 10  cm = 100−4m

D = Distance between the plate = 1 cm = 1 × 10−2m

C = 100−4 × 8.854 × 10−12 ⁄ 1 × 10−2

C = 8.854 × 10−12 = 8.854 pF


Ques.12. A balanced three-phase star-connected load draws power from a 440 V supply. The two connected wattmeters, W1 and W2, indicate 5 kW and 1200 W. Calculate the total power.

  1. 5 kW
  2. 6,200 kW
  3. 62 kW
  4. 6,200 W

The sum of two wattmeters gives the total power consumption in the three-phase load i.e

Total Power = P1 + P2

Reading of wattmeter 1

W1 = 5 kW = 5000 watts

Reading of wattmeter 2

W2 = 1200 watts

Total power = 5000 + 1200 = 6200 watts


Ques.13. Three identical impedances, each of (9.8 + j10)Ω. are connected to a 400 V, 50 Hz, AC pow supply. The power supplied to the load is measured by the two-wattmeter method. If the impedances are connected in a delta, find the readings of the two wattmeters.

  1. W1 = −4.4 kW and W2 = 1 kW
  2. W1 = 1 kW and W2 = −4.4 kW
  3. W1 = 18.9 kW and W2 = 5.06 kW
  4. W1 = 4.4 kW and W2 = 1 kW

Given (9.8 + j10)Ω

Impedance Z = (9.8 + j10)Ω

Resistance R = 9.8 Ω

Line voltage VL = 400 V

Impedance per phase ZP

ZP = √(R2 + X2L)

ZP = √(9.82 + 102)

ZP = 14Ω

In delta connected system line voltage is equal to the phase voltage

VP = VL = 440 V

IPh = Vp/Zp

IP = 400/14

IP = 28.57A

The power consumed by the three-phase delta connected circuit is

P = 3I2p Rp

P = 3 × 28.572 × 9.8

23997.6 W

W1 + W2 = 23997.6——-1

Power factor = Rp/Zp = 9.8/14

Cosφ = 0.7 = φ = 45°

Power factor of the two wattmeter is

tanØ =  √3[(W1 – W2) / (W1 + W2)]

tan45° = √3[(W1 – W2) ⁄ 23997.6

13871.09 = W1 – W2

 W= 13871.09 + W2

Putting the value of W1  in equation 1 we get

13871.09 + W2 + W2 = 23997.6

W=  5063.2 W = 5.06 kW

W1 = 18934.2 W = 18.9 kW


Ques.14. A 3-phase 10 kVA load has a power factor of 0.342. The power is measured by the two-wattmeter method. Find the reading of each wattmeter When the power factor is leading.

  1. W1 = −1 kW and W2 = 4.4 kW
  2. W1 = 1 kW and W2 = −4.4 kW
  3. W1 = 19 kW and W2 = 4.9 kW
  4. W1 = 4.4 kW and W2 = 1 kW


Apparent power S = 10 kVA

Power factor cosφ = 0.342

The volt-ampere input of a three-phase circuit is given by

S = √3.VL.IL

10 × 103 = √3.VL.IL

VL.I= 5.77 kVA

cosφ = 0.342 = φ = 70°

Reading of each wattmeter when the power factor is leading

W1 = VL.IL.Cos(30 + φ)

W1 = 5.77 × 103 .Cos(30 + 70°) = −1kW

W2 = VL.IL.Cos(30 − φ)

W2 = VL.IL.Cos(30 − 70) = 4.42 kW


Ques.15. Which of the following statements is FALSE?

  1. A single-phase system has many advantages over a three-phase system
  2. The supply frequency of a single-phase AC system in India is 50 Hz.
  3. To develop a polyphase system, the armature winding in a generator is divided into the required number of plates
  4. A three-phase system is found to be more economical

The single-phase system is used in almost all domestic and commercial applications such as in bulbs, lamps, fans, electric irons, refrigerators, washing machines, TV sets, computers, exhaust fans, etc But due to having limitations of its own in the field of generation, transmission and distribution and industrial application, it has been replaced by polyphase system.

Advantages of Three-phase system over single-phase system

Following are the main advantages of a three-phase system over a single-phase system:

(i) The output of a three-phase machine is nearly 1.5 times the output of a single-phase machine of the same size. Three-phase alternator occupies less space for a given size and voltage and also, costs less as compared to a single-phase machine of the same rating.

(ii) In the three-phase system, the power delivered is almost constant when the loads are balanced, whereas single-phase power pulsates. Even when the voltage and current are in phase, the power becomes zero twice in each cycle.

(iii) For transmission and distribution, the three-phase system requires only 75% of the weight of conducting material required by a single-phase system for given volt-amperes and voltage rating, and thus, the transmission becomes very much economical.

(iv) Three-phase induction motors are self-starting because it is possible to produce a rotating magnetic field with stationary coils by using the three-phase system. Whereas the single-phase motor has no starting torque without using auxiliary means.

(v) Three-phase induction motors have higher power factor and efficiency than that of single-phase induction motors.

(vi) Same rating machines like motors, generators, and transformers are simple in construction and have a smaller size as well as better operating characteristics than those of the single-phase machines.

(vii) The instantaneous power is a function of time in n single-phase system, hence fluctuates with respect to time and results in vibration in the single-phase motors. Therefore, the performance of the single-phase motor is poor whereas instantaneous power in the symmetrical three-phase system is constant.

(viii) The voltage regulation of a three-phase transmission line is better than that of a single-phase line. Moreover, a single-phase can be obtained from a three-phase but three-phase cannot be obtained from single-phase.

(ix) The rectifier converts AC to DC, the DC output voltage becomes smoother if the number of phases is increased.

Hence, it is found that the three phases are optimum to get all the above-said advantages. Any increase in the number of phases causes complications. Therefore, the three-phase system is stated to be standard globally.


Ques.16. Which of the following compound is widely used in the manufacture of ferrites?

  1. Fe2O3
  2. CuO
  3. FeO
  4. MgO

A ferrite is a ceramic material made by mixing and firing large proportions of iron(III) oxide (Fe2O3, rust) blended with small proportions of one or more additional metallic elements, such as barium, manganese, nickel, and zinc.

They are both electrically non-conductive, meaning that they are insulators, and ferrimagnetic, meaning they can easily be magnetized or attracted to a magnet.



Ques.17. The term “XL” is called the inductive reactance and is given by

  1. 1/(2πfL)
  2. (2πfC)
  3. (2πfL)
  4. 1/(2πfC)

The effect of inductance in an ac circuit is called Inductive reactance (XL) and is measured in ohms because it “resists” the flow of current in the circuit.  The inductive reactance of any given circuit is a function of the ac frequency and the inductance of that circuit.

The inductive reactance in a circuit is proportional to the Inducbmce of the circuit and the frequency of the alternating current. As the inductance is increased, the induced voltage (which opposes the applied voltage) is increased: hence the current flow is reduced.

Likewise. when the frequency of the circuit is increased. the rate of current change in the inductance coil is also increased; hence the induced (opposing) voltage is higher and the inductive reactance is increased. As inductive reactance increases, the current in the circuit flow is reduced.

The formula for the inductive reactance is

XL = 2πfL


f = frequency

L = inductance


Ques.18. The two-wattmeter method is used to measure the total power in a balanced circuit powered by a 415 V, 50 Hz, three-phase balanced power supply. If one wattmeter reads 4.5 kW and the other reads zero, the total power calculated will be

  1. 4.5 kW
  2. 0 W
  3. −4.5 kW
  4. 9 kW

The reading of two wattmeters can be expressed as

W1 =  VLILcos(30 − φ)
W2 =  VLILcos(30 + φ)

When PF is 0.5 (φ = 60°)

W1 =  VLILcos30° = Positive
W2 =  VLILcos90° = 0°

W1 + W= W1 = Total Power = 4.5 kW

∴ total power is measured by wattmeter W1 alone

 One wattmeter shows zero reading for cosφ = 0.5. For all power factors between 0 to 0.5 W2 shows negative and W1 shows positive, for lagging p.f.

Hence the reading of the wattmeter is 4.5 kW, 0


Ques.20. What happens when the ferromagnetic material is heated above the Curie temperature?

  1. It becomes anti-ferromagnetic
  2. It becomes diamagnetic
  3. It becomes ferromagnetic
  4. It becomes paramagnetic

In physics and materials science, the Curie temperature or Curie point is the temperature at which a ferromagnetic or a ferrimagnetic material becomes paramagnetic on heating the effect is reversible. A magnet will lose its magnetism if heated above the Curie temperature. The alignment of the domains becomes disordered under the action of both physical shock and being heated above the Curie temperature for that material.

Ferromagnetism decreases with rising temperature. If we heat a ferromagnetic substance, then at a definite temperature the ferromagnetic property of the substance suddenly disappears and the substance becomes paramagnetic. The temperature above which a ferromagnetic substance becomes paramagnetic is called the Curie temperature of the substance.

Scroll to Top