SSC JE electrical question paper 2019 with solution SET-2

Answer.3. 4

Explanation:-

Lap Winding:- In a lap winding the number of parallel paths (a) is always equal to the number of poles (p) and also to the number of brushes. In the given question number of poles is 4 hence the number of brushes will be 4.

Wave Winding:- In wave windings, the number of parallel paths (a) is always two and there may be two or more brush positions.

Answer.4. DC Series Motor

Explanation:-

• A DC motor can be operated from a single-phase AC supply.
• The direction of the torque produced by a DC machine is determined by the direction of current flow in the armature conductors and by the polarity of the field.
• Torque is developed in a DC machine because the commutator arrangement permits the field and armature currents to remain in phase, thus producing torque in a constant direction.
• A similar result can be obtained by using an AC supply, and by connecting the armature and field windings in series.
• A series DC motor connected in this configuration can therefore operate on a single-phase AC supply and is referred to as a universal motor. An additional consideration is that, because of the AC excitation, it is necessary to reduce AC core losses by laminating the stator; thus, the universal motor differs from the series DC motor in its construction features.

• Since the load current is sinusoidal and therefore reverses direction each half-cycle; however, the torque generated by the motor is always in the same direction, resulting in a pulsating torque, with nonzero averaging value.
• In a dc series motor when the direction of current changes, the direction of the field flux also changes. Thus by Fleming’s left-hand rule, we can see that the force is always unidirectional thus generating unidirectional torque.

Answer.3. RL series circuit

Explanation:-

A choke coil is simply a coil having a large inductance, but a small resistance.  A device that can regulate the current in an A.C. circuit without much loss of energy is called as ‘choke coil’.

• If the choke coil were not used the voltage across the resistor would be the same as the applied voltage. Thus by using the choke coil the voltage across the resistor is reduced by a factor $\frac{R}{{\sqrt {{R^2} + {\omega ^2}{L^2}} }}$ which is the power factor (cos φ) for practical inductance coil.
• The ohmic resistance of a practical choke coil is not exactly zero. This is the reason, that some electrical energy is always lost in the form of heat energy.
• The choke coil can be used only in A.C. circuits but not in D.C. circuits because for direct current (ω = 0) the inductive part of reactance (= ω_L) of the coil is zero and only ohmic resistance is left which is already very small.

Answer.1. Nφ

Explanation:-

The coefficient of self-inductance is given as

L = Nφ/I

Magnetic flux (φ) = MMF/ Reluctance = NI/S

$\begin{array}{l}L = \dfrac{N}{I} \times \dfrac{{NI}}{S}\\\\L = \dfrac{{{N^2}}}{S}\end{array}$

Self-inductance of long solenoid

Let L be the length of a long solenoid and the total number of turns are N. The number of tuns per unit length is N/I. When a current I flows through it, the magnetic field inside the solenoid,

$B = {\mu _o}{\mu _R}\dfrac{{NI}}{l}$

Where

μ_o = permeability of free space

μ_r = Relative permeability

If A is the area of each turn (cross-section area), then the magnetic flux through each turn,

${\Phi _B} = BA = {\mu _o}{\mu _R}A\dfrac{{NI}}{l}$

∴  total magnetic flux linked through the solenoid is

$N{\Phi _B} = BA = {\mu _o}{\mu _R}A\dfrac{{{N^2}I}}{l}$

Since self-inductance

$L = \frac{{N{\Phi _B}}}{I}$

So self-inductance of the solenoid is

$\begin{array}{l}L = {\mu _o}{\mu _R}A\dfrac{{{N^2}I}}{{I.l}}\\\\L = {\mu _o}{\mu _R}A\frac{{{N^2}}}{l}\end{array}$

Answer.3. 14.4 V

Explanation:-

Given

N = 360 turns

φ = 200 μWb

Induced EMF

$e = N \times \dfrac{{d\Phi }}{{dt}}$

dφ = φ₂ − φ₁

dφ = 200 − (−200)……(since the flux is reversing)

dφ = 400 × 10⁻⁶ Wb

L = 360

time (t) = 0.01 sec = 10⁻² sec

$e = 360 \times \dfrac{{400 \times {{10}^{ – 6}}}}{{{{10}^{ – 2}}}}$

e = 36 × 0.4 = 14.4 V

Answer.4. Apparent power

Explanation:-

Apparent Power (S): It is defined as the product of r.m.s value of voltage (V) and current (1). It is denoted by S.

S = V/I Volt Ampere

Note:- Active power (VI cosφ) is actually used up in the circuit to do a useful work e.g. producing torque in motors and supplying heat, light etc. On the other hand, reactive power (= VI sinφ) is neither consumed nor it does any useful work in the circuit.

Active power is unidirectional i.e. it flows from source to load only. However, reactive power flows back and forth between the source and reactance (L or C).

Active power. It is the power that is actually consumed or utilized in the circuit. It is also defined as the product of apparent power and power factor. It is denoted by P and measured in unit watt or kilowatt (kW) P= VI cos (φ).

Answer.2. $\dfrac{1}{2r}At/m$

Explanation:-

Magnetic Field Strength (H)  gives the quantitative measure of strongness or weakness of the magnetic field.

H = B/μ_o

Where

B = Magnetic Flux Density

μ_o = Vacuum Permeability

The magnetic Field strength at the center of circular loop carrying current I is given by

B = μ_oI/2r

B/μ_o = I/2r

H = I/2r At/m

Where r = Radius

Answer.3. 186 meters

Explanation:-

“Conduit” means rigid or flexible metallic tubing or mechanically strong and fire-resisting non-metallic tubing into which a cable or cables may be drawn for the purpose of affording it or them mechanical protection.

Since the size of the conduit is 62 meters. In conduit, there is a 2-phase wire and 1 neutral wire. Therefore the total number of conducting wire is 3 in a conduit

Total length of cable = 62 × 3 = 186 meters.

Answer.1. √3/2

Explanation:-

The factor by which, the induced E.M.F gets reduced due to short pitching is called pitch factor or coil span factor denoted by K_c. It is given as

K_c = cosα/2

given

α = 60°

K_c = cos60/2

K_c = cos30°

K_c = √3/2

Answer.4. Shaded and unshaded portion of the pole

Explanation:-

SHADED-POLE MOTORS

The shaded-pole motor has the lowest starting torque of all the single-phase motor. It is relatively inexpensive and is used to turn very small fan blades connected directly to the shaft of the motor. The basic construction of this motor is very simple since no start winding is present. The imbalance in the magnetic field needed to produce rotation is obtained by shading a portion of the run winding with heavy copper wire or band.

When the motor is energized, the strength of the magnetic field is different in the area of the main poles than in the area of the shaded poles, allowing the rotor to turn.

The direction of rotation of a shaded-pole motor is determined by the orientation of the main poles and the shaded poles. The rotor will turn in the direction indicated by the arrows in Figure. Shaded-pole motors are used for very small applications and are rated in watts instead of horse-power. They normally range in size from 6 watts to roughly 35 watts. This is the equivalent of 1/125 horse-power to 1/20 horsepower. The direction of rotation is from the unshaded portion of the pole to the shaded portion of the pole.