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

The capacitor in the series connection is given as

1/C = 1/C₁ + 1/C₂

or

C = (C₁ × C₂) ⁄ (C₁ + C₂)

Given

C = 5μF

C₁ = 15 μF

C₂ =?μF

5 = (15 × C₂) ⁄ (15 + C₂)

75 + 5C₂ = 15C₂

C₂ = 75/10 = 7.5μF

Energy consumption or power consumption refers to the electrical energy per unit time is given as

E = Power × Time = P × T

Shift 1 energy consumption

E₁ = 30 x 8 = 240 kW

Shift 2 energy consumption

E₂= 15 x 8 = 120 kW

Shift 3 energy consumption

E₃ = 25 x 8 = 200 kW

Total energy consumption

E = E₁ + E₂ + E₃

E = 240 + 120 + 200 = 560 kW

The energy stored in the capacitor is given as

U =CV²/2

Given

C₁ = 2μF =  2 × 10⁻⁶

C₂ = 4μF =  2 × 10⁻⁶

Voltage V = 100 V

The energy stored in capacitor 1

U₁ = C₁V²/2

U₁ = 2 × 10⁻⁶ × 100² ⁄ 2

U₁ = 0.01 J

The energy stored in capacitor 2

U₂ = C₂V²/2

U₂ = 4 × 10⁻⁶ × 100² ⁄ 2

U₁ = 0.02 J

Given (9.8 + j10)Ω

Impedance Z = (9.8 + j10)Ω

Resistance R = 9.8 Ω

Line voltage V_L = 400 V

Impedance per phase Z_P

Z_P = √(R² + X²_L)

Z_P = √(9.8² + 10²)

Z_P = 14Ω

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

V_P = V_L/√3 = 400/√3 = 230.94V

I_Ph = V_p/Z_p

I_P = 230.94/14

I_P = 16.5A

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

P = 3I²_p R_p

P = 3 × 16.5² × 9.8

8004.15 W

W₁ + W₂ =  8004.15——-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) ⁄ 8004.15

4626.87 = W₁ – W₂

 W₁ = 4626.87 + W₂

Putting the value of W₁  in equation 1 we get

4626.87 + W₂ + W₂ =8004.15

W₂ =  1688 W = 1.68 kW

W₁ = 6315 W = 6.31 kW

When the capacitance is connected in the series the total capacitance is equal to

C = 1/C₁ + 1/C₂ + 1/C₃ + ————-C_N

C = 1/1 + 1/1 +1/ 1———- + 1/1₁₀

C = 10 C

Let us Suppose that the three Resistance L₁, L₂, L₃ of 1 Henry, 2 Henry, and 3 Henry are connected in series respectively.

Now add all the inductance of the circuit we get

L_eq = L₁ + L₂ + L₃
L_eq = 1 + 2 + 3 = 5H

Now from the above result, we can conclude that the equivalent Inductance in the series combination is greater than the largest Inductance in the combination (since the largest Inductance was 3H).

Therefore, Option.4. is correct.

Nickel Iron and Edison batteries

The nickel-iron battery is one of the few systems which may be developed into a high-energy-density battery for electric vehicles. For long the two main designs for this battery have been the tubular positive type and the flat pocket plate type although cells with sintered type negative are also being manufactured.

Nickel-iron cells are arranged and suitably connected to form a battery for specific service; the main parts of nickel-iron consist of:

The active material consists of the positive and negative plates

Positive plate active material: Nickel hydroxide Ni(OH)₄ or apple green nickel peroxide NiO₂ for the positive plate. About 17 percent of graphite is added to increase conductivity. It also contains an activating additive barium hydroxide which is about 2 percent of the active material. The conductivity increases by the addition of flakes of pure nickel or graphite.

Negative plate active material:- Negative plate is made up of iron or its oxide. In order to increase the conductivity of the small quantity of mercuric oxide is added. The mixture is closely perforated in steel pockets. The plates of opposite polarity are closely interleaved. The plates of the same polarity are welded to a common strap to form a group.

Separators: Plates of opposite polarity are separated by using hard rubber strips or an insulating rod of ebonite to prevent short circuits.

Electrolyte: The electrolyte is a mixture of 21% solution of potassium hydroxide (KOH) in distilled water with a specific gravity of about 1.2.

Value of current

I = 10 sin314t

Comparing the above equation with the standard equation

I = Im sinωt

Here peak value Im = 10 A

Therefore the RMS value is

I_RMS = 0.707 × 10

I_RMS = 7.07 A

Ferroelectricity, the property of certain nonconducting crystals, or dielectrics, that exhibit spontaneous electric polarization (separation of the center of positive and negative electric charge, making one side of the crystal positive and the opposite side negative) that can be reversed in direction by the application of an appropriate electric field.

In certain dielectric materials, polarization is not a linear function of the applied electric field (same as magnetization curve). Such materials exhibit hysteresis curves similar to that of ferromagnetic materials and are known as ferroelectric materials. The hysteresis curve exhibited by a ferroelectric material is shown in Fig.

Hysteresis Properties:-  When an increasing electric field is applied to a ferroelectric material, it results in an increase in polarisation and it reaches a maximum value for particular field strength.

On the other hand, if we decrease the electric field, the polarisation decreases. When the field strength is zero, i.e., E = 0, a small amount of polarization exists in the material.

This polarisation is known as remanent polarisation. Now the ferroelectric material is said to be spontaneously polarised. In order to reduce the value of polarization to zero, an electric field strength (-E_c) should be applied. This field is known as the coercive field.

The capacitance of the parallel plate capacitor is given as

C =ε_oA/D

where

ε_o = Permittivity of free space 8.854 × 10⁻¹²

A = Area of the plate = 0.01m²

D = Distance between the plate = 2.5 cm = 2.5 × 10⁻²m

C = 0.01 × 8.854 × 10⁻¹² ⁄ 2.5 × 10⁻²

C = 3.54  × 10⁻¹² = 3.54  pF