Ques.21. What capacitance must be placed in a series with a 15-μF capacitor to obtain a total capacitance of 5μF?

25 μF

7.5 μF✓

10 μF

4 μF

The capacitor in the series connection is given as

1/C = 1/C_{1} + 1/C_{2}

or

C = (C_{1} × C_{2}) ⁄ (C_{1} + C_{2})

Given

C = 5μF

C_{1} = 15 μF

C_{2} =?μF

5 = (15 × C_{2}) ⁄ (15 + C_{2})

75 + 5C_{2 }= 15C_{2}

C_{2 }= 75/10 = 7.5μF

Ques.22. A factory runs in 3 shifts of 8 hours each, in which it consumes 30 kW, 15 kW, and 25 kW respectively. Calculate the energy consumed by the factory per day.

186.67 kW

373 kW

560 kW✓

746.67 kW

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

E_{1} = 30 x 8 = 240 kW

Shift 2 energy consumption

E_{2}= 15 x 8 = 120 kW

Shift 3 energy consumption

E_{3} = 25 x 8 = 200 kW

Total energy consumption

E = E_{1} + E_{2} + E_{3}

E = 240 + 120 + 200 = 560 kW

Ques.23. Two capacitors of 2 μF and 4 μF are connected in a parallel across a 100-V DC supply. The energy stored in the capacitors will respectively be ____

0.01 J and 0.02 J✓

0.1 J and 0.2 J

0.001 J and 0.002 J

1 J and 2 J

The energy stored in the capacitor is given as

U =CV^{2}/2

Given

C_{1} = 2μF = 2 × 10^{−6}

C_{2} = 4μF = 2 × 10^{−6}

Voltage V = 100 V

The energy stored in capacitor 1

U_{1} = C_{1}V^{2}/2

U_{1} = 2 × 10^{−6 }× 100^{2} ⁄ 2

U_{1} = 0.01 J

The energy stored in capacitor 2

U_{2} = C_{2}V^{2}/2

U_{2} = 4 × 10^{−6 }× 100^{2} ⁄ 2

U_{1} = 0.02 J

Ques.24. Three identical impedances, each of (9. 8+j 10)Ω. are connected across a 400 V, 50 Hz AC supply. The power supplier to the load is measured by the two-wattmeter method. Find the readings of the two wattmeters if the impedances are connected in star.

W_{1} = 6.31kW and W_{2} = 1.68kW✓

W_{1} = 1 kW and W_{2} = −4.4 kW

W_{1} = −6.36 and W_{2} = 1.64 kW

W_{1} = 4.4 kW and W_{2} = 1 kW

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} = √(R^{2} + X^{2}_{L})

Z_{P} = √(9.8^{2} + 10^{2})

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 = 3I^{2}_{p} R_{p}

P = 3 × 16.5^{2} × 9.8

8004.15 W

W_{1} + W_{2} = 8004.15——-1

Power factor = Rp/Zp = 9.8/14

Cosφ = 0.7 = φ = 45°

Power factor of the two wattmeter is

tanØ = √3[(W_{1} – W_{2}) / (W_{1} + W_{2})]

tan45° = √3[(W_{1} – W_{2}) ⁄ 8004.15

4626.87 = W_{1} – W_{2}

W_{1 }= 4626.87 + W_{2}

Putting the value of W_{1 }in equation 1 we get

4626.87 + W_{2} + W_{2} =8004.15

W_{2 }= 1688 W = 1.68 kW

W_{1} = 6315 W = 6.31 kW

Ques.25. What will be the total capacitance of 10 capacitor of equal capacitance C connected in parallel?

90C/10

C

C/90

10C✓

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

C = 1/C_{1} + 1/C_{2} + 1/C_{3} + ————-C_{N}

C = 1/1 + 1/1 +1/ 1———- + 1/1_{10}

C = 10 C

Ques.26. In a series combination of several inductors, the equivalent inductance is ______

Equal to the largest inductance of the combination

Lower than the largest inductance of the combination

Lower than the smallest inductance of the combination

Greater than the largest inductance of the combination✓

Let us Suppose that the three Resistance L_{1}, L_{2}, L_{3} of 1 Henry, 2 Henry, and 3 Henry are connected in series respectively.

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.

Ques.27. The active material of a nickel-iron battery is___

21% solution of H_{2}SO_{4}

Powdered NaCl and its oxide

98% solution of KCl

Nickel Hydroxide✓

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 NiO_{2} 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.

Ques.28. An alternating current is given by I =10 sin314t A. Its RMS value will be

0.707A

7.07 A✓

5A

10A

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

Ques.29. In ferroelectric materials, the hysteresis loop is the _____ function of the applied electric field.

Parabolic

Non-linear✓

Exponential

Linear

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.

Ques.30. The plate area of a parallel-plate capacitor is 0.01 m^{2}. The distance between the plates is 2.5 cm. If the insulating medium is air, its capacitance will be

35.4 × 10^{−11}

3.54 × 10^{−11}

3.54 × 10^{−12}✓

35.4 × 10^{−12}

The capacitance of the parallel plate capacitor is given as

C =ε_{o}A/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^{−2}m