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

Metals contain a large number of free electrons and thus electric current can pass through them easily. For example, metals offer a little resistance to the flow of free electrons through them, i.e., they have low resistance.

Substances like manganin, constantan, nichrome, tungsten, etc., offer high resistance to the flow of electric current through them. Hence, they have high resistance.

Constantan (or Eureka) is an alloy of copper (55%) and nickel (45%). It is used in the construction of pyrometers and thermocouples. Manganin is an alloy of copper (83%), manganese (13%), and nickel (4%).

It is used in the construction of rheostats and resistors. Nichrome is an alloy of nickel (60%) and chromium (40%). It is used in the construction of the heating element.

The heating elements (or heating coils) of electrical heating appliances such as electric iron and toaster, etc., are made of an alloy rather than a pure metal because

(i) the resistivity of an alloy is much higher than that of pure metal

(ii) an alloy does not undergo oxidation (or burn) easily even at high temperatures when it is red hot.

For example, nichrome alloy is used for making the heating elements of electrical appliances such as electric iron, toaster, electric kettle, room heaters, water heaters (geysers, and hair dryers, etc., because:

(i) Nichrome has very high resistivity (due to which the heating element made of nichrome has a high resistance and produces a lot of heat on passing current).

(ii) Nichrome does not undergo oxidation (or burn) easily even at high temperatures. Due to this nichrome wire can be kept red-hot without burning or breaking in the air.

The resistivity of conductors (like metals) is very low. The resistivity of most of the metals increases with temperature. On the other hand, the resistivity of insulators like ebonite, glass, and diamond is very high and does not change with temperature. The resistivity of semiconductors like silicon and germanium is in-between those of conductors and insulators and decreases on increasing the temperature.

Given

Apparent power S = 10 kVA

Power factor cosφ = 0.342

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

S = √3.V_L.I_L

10 × 10³ = √3.V_L.I_L

V_L.I_L = 5.77 kVA

cosφ = 0.342 = φ = 70°

Reading of each wattmeter when the power factor is lagging

W₁ = V_L.I_L.Cos(30 + φ)

W₁ = 5.77 × 10³ .Cos(30 − 70°) = 4.42 kW

W₂ = V_L.I_L.Cos(30 − φ)

W₂ = V_L.I_L.Cos(30 + 70) = −1kW

As given in the above question

W₁ = 2000 W

W₂ = 500 W

The power factor of the two wattmeters is

tanØ =  √3[(W₁ – W₂) / (W₁ + W₂)]

tanØ = √3[(2000 – 500) ⁄ (2000 + 500)]

Φ = 45°

Power factor = cosφ = cos45° = 0.707

Materials that restrict or prohibit the flow of sound or heat or electricity through them are called insulators or insulating materials. They also avoid the accumulation of heat by proper distribution. 

Electrically the insulating material should have high resistivity to reduce the Leakage current and high dielectric strength to enable it to withstand higher voltage without being punctured or broken down. Further, the insulator should have a small dielectric loss.

Dielectric properties:-

The following important properties of insulation are taken into consideration in electrical insulating materials.

(a) High Resistivity:-  Electrical conductivity is the capacity of a material to conduct the electric current. Resistivity is the reciprocal of conductivity and is commonly measured as Insulation resistance. Thus a good insulator is that which possesses low conductivity or high resistivity.

(b) High Dielectric strength:- The Dielectric strength of an insulating material determines its ability to resist puncture or breakdown by electric potentials. Hence the limiting intensity by electric field above which the insulators get punctured or ruptured is called the dielectric strength of the dielectric and is commonly expressed in volts per mm of the thickness of the material or volts per centimeter.

The dielectric strength of a material can also be defined as the electrical field strength, that must be applied to cause a disruptive effect or discharge (arc or spark) or accumulate current through the body of the material. 

The dielectric strength is affected by a number of factors such as

(a) Dielectric strength decreases as the thickness of the material increases

(b) Dielectric strength decreases as the time of the application of electric current increases

(c) Dielectric strength decreases as the frequency of the applied electric field is increased

(d) Dielectric strength decreases in presence of humidity.

(e) Dielectric strength decreases with an increase in temperature in case of air as an insulator.

(c) Low Dielectric loss:- These are generally caused due to absorption of electrical energy and also by the leakage of current through the insulating material.

The absorption of electrical energy. under an alternating electric field gives rise to the dissipation of the electrical energy in the insulating material, while leakage takes place as a result of conduction. The conduction is appreciable, however, only at high temperatures, otherwise, it is negligible.

A good and ideal dielectric is that which does not absorb electrical energy and the charge of the capacitor is completely recovered when the electrical field is removed.

A dielectric material, however, always causes some electrical energy loss, which is measured by the phase difference between the phase angle and 90 angles of an ideal capacitor. A perfect insulator is that has a phase angle of 90 and a loss angle of zero.

As given in the above question

W₁ = 2000 W

W₂ = 500 W

V_L = 440 V

The power factor of the two wattmeters is

tanØ =  √3[(W₁ – W₂) / (W₁ + W₂)]

tanØ = √3[(2000 – 500) ⁄ (2000 + 500)]

Φ = 45°

Power factor = cosφ = cos45° = 0.707

Total power

W = W₁ + W₂ = 2000 + 500 = 2500W

Input power

P = √3.V_L.I_L.cosφ

2500 = √3 × 440 × I_L × 0.7

I_L = 4.64 A

Given

Voltage = 400 V

Inductance L = 0.0318 H

Inductive Reactance X_L = 2πfL

Frequency f = 50 Hz

X_L = 2 × π × 50 × 0.0318

X_L = 10Ω

Impedance per phase Z_P

Z_P = √(R² + X²_L)

Z_P = √(10² + 10²)

Z_P = 14.14Ω

|Zph| = 14.14Ω

also

V_P = V_L/√3 = 400/√3

V_P = 230V

I_P = V_p/Z_p

I_P = 230/14.14

I_P = 16.26 A

Insulator are the property of the material that opposes the flow of electrical current. i.e insulator are the bad conductor of the electricity. E.g of the insulator is plastic, wood, glass etc.

Battery cycle and depth of discharge One discharge and charge period are referred to as a battery cycle. A major factor affecting battery life is the depth of discharge i.e. how much the battery is discharged and how often.

The depth of discharge of a battery is a measure in the percentage of the amount of energy that can be removed from a battery during a cycle.

Limiting the depth of discharge will make the battery last longer. If discharged beyond its recommended depth of discharge, the life of a battery will be considerably reduced. 

As the rate of discharge is increased, the cell terminal voltage, that is, the voltage at the output terminals of the cell drops because the internal resistance results in voltage drops within the cell due to the flow of current from the cell. These losses result in increased internal heating within the cell and reduced available energy.

It is suggested that lead-acid batteries not be discharged beyond 80%, even if the battery is designed for deep discharge. Lead-acid batteries that are deeply discharged may suffer from a permanent loss of capacity due to sulphation, acid dilution, and a greater tendency for freezing. The fully charged lead-acid cell has 2.25 Volts. The 80% of this voltage is nearly 1.8 Volts.

In star-connected load the phase voltage between the neutral point and any one of the line connections is 1/√3 × V_L of the line voltage. 

Therefore the line voltage is √3 times the phase voltage.

i.e V_L = √3V_ph

In star connection, the line current is equal to the phase current.

I_L = I_ph