The reluctance of any part of a magnetic circuit may be defined as the ratio of the drop in magnetomotive force to the flux produced in that part of the circuit. It is measured in ampere-turns/Weber and is denoted by S.

Reluctance = m.m.f ⁄ flux

Ques.22. Which of the following losses is together called iron losses?

Hysteresis loss and frictional loss

Hysteresis loss and copper loss

Eddy current loss and hysteresis loss

Eddy current loss and frictional loss

Answer.3. Eddy current loss and hysteresis loss

Explanation:-

Core Losses

Core losses are also known as iron losses or Magnetic losses. These Losses are constant and independent of the load. They are induced in the machine due to hysteresis and eddy currents produced in the core produced when the magnetization is changing. They mainly occur in the armature teeth and core as also in the pole shoe.

Hysteresis losses occur due to the reversal of the magnetization of the armature core. So, the hysteresis loss can be obtained as:

Hysteresis Loss = K_{h} × B_{M}^{1.67} × f × v watts

where
Kh = Hysteresis constant depends upon the material
Bm = Maximum flux density
f = frequency
v = Volume of the core

The eddy current loss exists due to eddy currents. When the armature core rotates, it cuts the magnetic flux and e.m.f. gets induced in the core. This induced e.m.f. sets up eddy currents which cause power loss. Ibis loss is given by,

Where K_{e} = Eddy current constant
t = thickness of the core

Ques.23. In the speed-time curve of a train, speed in and time are plotted.

km/h; sec

km/s; sec

km/h; hour

km/s; hour

Answer.1. km/h; sec

Explanation:-

The curve that shows the instantaneous speed of the train in kmph along the ordinate and time in seconds along the abscissa is known as the speed-time curve.

The area under the speed-time curve gives the distance traveled during, given time internally and slope at any point on the curve toward abscissa gives the acceleration and retardation at the instance, out of the two speed—time curve is more important.

Ques.24. Identify the important feature of a DC series motor.

High starting torque

Medium starting torque

Zero starting torque

Low starting torque

Answer.1. High starting torque

Explanation:-

In DC series motor Torque (Ta) increase as the Square of armature current (Ia) T_{a} ∝ I_{a}^{2}. So DC motor provides high starting torque.

Ques.25. 4F2D is a/an number

Binary

Hexadecimal

Octal

Decimal

Answer.2. Hexadecimal

Explanation:-

The hexadecimal number system is also called base-16, a numeration system in which all numbers are represented using the symbols 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, and F only. The system is base of 16. The hexadecimal numbers are used to represent binary numbers because of ease of conversion and compactness.

Ques.26. Three lamps are connected in series across a 120V supply and take a current of 1.5A. If the resistance of two of the lamps is 30Ω, what is the resistance of the third lamp?

20Ω

30Ω

40Ω

10Ω

Answer.1. 20 Ω

Explanation:-

Method 1

Supply Voltage = 120 V

Current I = 1.5 A

Total Resistance R = V/I = 120/1.5 = 80 Ω

Now resistance of 3 lamp be R_{1}, R_{2}, & R_{3}

R_{eq} = R_{1} + R_{2} + R_{3}

80 = 30 + 30 + R_{3}

R_{3 }= 20 Ω

Method 2

Applying KVL on the above circuit

120 = 1.5(R_{1} + R_{2}) + 1.5R_{3}

120 = 1.5(30 + 30) + 1.5R_{3}

120 = 90 + 1.5R_{3}

R_{3 }= 30/1.5

R_{3 }= 20 Ω

Ques.27. In the circuit shown, the pressure coils of two wattmeters are connected to

B

R

Y

Neutral Wire

Answer.1. B

In Two wattmeter method the current coils of the wattmeter are connected with any two lines, say R and Y and the potential coil of each wattmeter is joined across the same line, the third line i.e. B as shown below in the figure . In the circuit B is the common point for both the wattmeter.

Ques.28. Two bulbs of rating 230V, 60W, and 230V, 100W are connected in parallel across supply mains, Identify the correct statement.

The 60W bulb will glow brighter

Neither bulb will glow

Both will glow equally bright

The 100W bulb will glow brighter

Answer.4. The 100W bulb will glow brighter

Explanation:-

For Bulb 1

Power = 60 W

Voltage = 230 V

R = V^{2}/P

R = (230 × 230)/60

R = 881.66 Ω

For Bulb 2

Power = 230 W

Voltage = 120 V

R = V^{2}/P

R = (230 × 230)/100

R = 529 Ω

In a parallel connection, the voltage across each element is the same. So when a 60W bulb and 1o0 W bulb are connected in parallel, the voltage across them will be the same i.e 230 V in the given case. To find which bulb will glow brighter we need to find the power dissipation across each of them. From the relation

P = V^{2}/R

Since the voltage is the same we can say that power dissipation will be higher for the bulb with lower resistance i.e. 100W bulb, or the bulb with the higher rating will glow brighter.

In a series connection, the current flowing across each element is the same. So when the 60W bulb and 100W bulb are connected in series, the same current will flow through them. To find which bulb will glow brighter we need to find the power dissipation across each of them. From the relation

P = I^{2}R

Ques.29. Two coupled coils with L_{1} = L_{2} = 0.5H have a coupling coefficient of K = 0.75. The turn ratio N_{1} ⁄ N_{2 }= ?

0.5

4

1

2

Answer.3. 1

Explanation:-

The self-inductance is given as

L = μN^{2}A/I

L ∝ N^{2}

where

N is the number of turns of the solenoid

A is the area of each turn of the coil

l is the length of the solenoid

and μ is the permeability constant

L_{1}/L_{2} = N^{2}_{1}/N^{2}_{2}

0.5/0.5 = N^{2}_{1}/N^{2}_{2}

N_{1}/N_{2} = 1

Ques.30. Is it possible to have current in a transmission line, under no-load conditions?

Yes, because of the capacitance effect

No, because of the proximity effect

Yes, because of the corona effect

Yes, because of skin effect

Answer.1. Yes, because of the capacitance effect

Explanation:-

During the no-load condition, the current flowing is only charging current due to line capacitance. It increases the capacitive var in the system. Since the line is under no load the line inductance will be less. Therefore, the capacitive var becomes greater than inductive var during no load or light load condition. Due to this phenomenon, the receiving end voltage becomes greater than the sending end voltage. This effect is also called the Ferranti effect.

In the case of short lines, the effect is negligible, but it increases rapidly with the increase in the length of the line. Therefore, this phenomenon is observable only in medium and long lines. For long high voltage and EHV transmission lines, shunt reactors are provided to absorb a part of the charging current or shunt capacitive VAr of the transmission line under no load or light load conditions, in order to prevent the overvoltage on the line.