DMRC JE Electrical 9th- April- 2018 Paper With Solution and Explanation

Common-Base Configuration:

In this circuit configuration of a transistor, the base terminal is used as a common terminal between input and output as depicted in Fig. The input signal is applied between the emitter and base terminals. The output will be taken from the collection and base terminals. To explain the operation of NPN and PNP transistors, the common base configuration is used.

Common-Base dc Current Gain

The common-base dc current gain (α) is defined as the ratio of the collector current, I_c, and the emitter current I_E and it is represented by α. The dc current gain α can be expressed as

$\alpha = \dfrac{{{I_C}}}{{{I_E}}}$

Since the collector current is always less than the emitter current, the dc current gain α is always less than unity. Usually, the value of α is about 0.98. The dc current can be made closer to unity by controlling the width and doping level of base region as less as possible. Practically, α varies between 0.95 to 0.998. From the above equation, we get the collector current I_C = αI_E.

The emitter current is

I_E = I_B + I_C

I_E = I_B + αI_C

as I_C = αI_E

I_B = (1 – α)I_E

A common-base configuration offers the following characteristics:

• The transistor circuit will have a very low-input impedance
• Very high output impedance
• Unity (or less) current gain
• High-voltage gait and very good high-frequency response and hence find dominant applications in RF amplifiers and high-frequency circuits.

Note:- For common emitter configuration current amplification factor is β

For common collector configuration, the current amplification factor is Γ

Since the two capacitor are connected in the parallel combination therefore there equivalent capacitance is

C_eqv = C₁ + C₂

= 25 + 25

C_eqv = 50 F

• Damper windings are windings that are wound to the rotor poles of the machine (winding it similar to that of an induction machine) which help in two ways.
• We all know that a synchronous machine is not self-starting. Thus providing damper windings helps synchronous machines act as an induction motor ( only at starting). Which helps the machine to self-start.
• We all know that hunting is a persistent phenomenon when it comes to synchronous machines.

HUNTING

If the torque output of a prime mover pulsates, it may cause the rotor of the alternator to be pulled ahead of, and then behind, its normal running position. A diesel engine is one example of a prime mover that has a pulsating torque output. If an alternator is used with a diesel engine, the alternator rotor periodically will move slightly faster and then slower.

This pulsating or oscillating effect is called hunting. It causes the current to surge back and forth between alternators operating in parallel. This condition is unsatisfactory and may become cumulative, resulting in such a large increase in the current between the alternators that the overload relays open the circuit.

Hunting can be corrected by the use of a heavy flywheel. A damper winding is often used in the rotating field structure to minimize the pulsating torque. Figure. showed a damper winding, which is often called an amortisseur winding. This winding is embedded in slots in each of the pole faces of the rotating field.

The armortisseur winding consists of heavy conductors that are brazed or welded to two end rings. At the instant that hunting develops, the path of the armature flux changes so that it cuts the short-circuited conductors of the amortissuer winding. This change in the flux path produces induced currents in the damper winding. The currents oppose the force-producing them (by Lenz’s law). The proper design of the damper winding ensures that the effects of hunting are canceled by the induced currents in the short-circuited conductors.

Note:- 

The AVR(Automatic voltage regulator) consists basically of a circuit fed from the alternator output voltage which detects small changes in voltage and feeds a signal to an amplifier which changes the excitation to correct the voltage. Stabilizing features are also incorporated in the circuits to avoid ‘hunting’ (constant voltage fluctuations) or overcorrecting.

Consider a pure capacitive circuit connected to an AC supply. When a pure capacitor is connected to AC source, a changing value of the applied voltage causes the capacitor to charge and discharge alternatively. The charge that flows through the capacitor is proportional to the capacitance (size of the capacitor) and the applied voltage across the capacitor.

Power in the Purely capacitive circuit

In the case of the pure capacitive circuit, the current leads the applied voltage by 90°

Let the voltage V = V_m sinωt is applied to a pure capacitive circuit. Then the resulting current will be

I = I_m sin(ωt + π/2)

The power curve can be derived by multiplying the instantaneous values of the applied voltage by instantaneous currents.

P = VI = V_m sinωt × I_m sin(ωt + π/2)

i.e P = V_m I_m sinωt sin(ωt + π/2)

However sin(ωt + π/2) = cosωt

∴ P =  V_m I_m sinωt cosωt

or

P = ½ V_m I_m (2sinωt cosωt)

∴ P = ½ V_m I_msin2ωt

or

P = Vm sinωt × Im sin(ωt + φ)

P = V_mI_m sinωt . sin(ωt + φ)

Multiplying and dividing the above equation by 2 we get and as we know that 
2sinA.sinB = cos(A − B) − cos(A + B)

$\begin{array}{l}P = \dfrac{{{V_m}{{\mathop{\rm I}\nolimits} _m}}}{2}2\sin \omega t.\sin (\omega t + \Phi )\\\\ = \dfrac{{{V_m}{{\mathop{\rm I}\nolimits} _m}}}{2}\left[ {\cos (\omega t – \omega t – \Phi ) – \cos (\omega t + \omega t + \Phi )} \right]\\\\ = \dfrac{{{V_m}{{\mathop{\rm I}\nolimits} _m}}}{2}\left[ {\cos \Phi – \cos (2\omega t + \Phi )} \right]\end{array}$

Clearly, the instantaneous power P is composed of two terms.

1. A component that is independent of time, and depends on the RMS values of the voltage, the current and the phase angle separating them.
2. The second term is a time-varying sinusoid whose frequency is equal to twice the angular frequency of the supply due to the 2ω part of the term.

During the first quarter cycle, the power curve is above the horizontal axis and is positive (0 to 90). The circuit draws energy from the source and the capacitor is charged. The energy is stored in the electric field of the capacitor.

During the second quarter cycle, the power curve is below the horizontal and is negative (90 to 180). The capacitor is discharged and the energy from the dielectric field is returned to the source.

The energy stored in the electric field during the first quarter cycle is equal to the energy returned to the source during the second quarter cycle in a purely capacitive circuit. Therefore the total active energy during each cycle of the current is zero. The active power over the complete cycle of current in a purely capacitive circuit is zero.

Note:- Whether is a pure resistive circuit or a pure capacitive circuit the frequency of instantaneous power is always double the frequency of the applied voltage

One Meg ohm Rule for IR Value for Equipment

Based upon equipment rating:

< 1K V = 1 MΩ minimum
>1KV = 1 MΩ /1KV

As per IE Rules-1956

At a pressure of 1000 V applied between each live conductor and earth for a period of one minute the insulation resistance of HV installations shall be at least 1 Mega-ohm or as specified by the Bureau of Indian Standards.

Medium and Low Voltage Installations- At a pressure of 500 V applied between each live conductor and earth for a period of one minute, the insulation resistance of medium and low voltage installations shall be at least 1 Mega ohm or as specified by the Bureau of Indian Standards] from time to time.

At a pressure of 2.5 kV, DC applied between each live conductor and earth for a period of one minute, the insulation resistance of high voltage equipment shall be at least 5 MEGA-OHM or as specified in the relevant Indian Standard.

The effective or r.m.s. value of alternating current is that steady current (d.c.) which when flowing through a given resistance for a given time produces the same amount of heat as produced by the alternating current when flowing through the same resistance for the same time. It is also called the virtual value of a.c. and is represented by I_rms or I_eff or I_v.

An instantaneous voltage is the value of an AC voltage at a particular instant. For example, the maximum voltage shown in Figure is an instantaneous voltage; the positive maximum voltage occurs at 90°, and the negative maximum voltage occurs at 270°. Electricians generally consider that the most important instantaneous AC voltages are the maximum (crest or peak) voltage and the effective voltage.

Let us try to explain what is meant by the effective voltage and then observe why it is so important. The effective voltage is 70.7 percent of the peak voltage and occurs at 45°. Practically all AC voltmeters read the effective value of an AC voltage.

The effective value is also called the root-mean-square -(RMS) value because it is the square root of the average of the squared values between zero and maximum. The effective value of an AC current is stated in terms of an equivalent DC current. The phenomenon used for standard comparisons is the heating effect of the current.

Since nearly all AC voltmeters indicate effective values, this particular value is always understood when an electrician speaks of 120 volts, 240 volts, etc. The word effective is understood, although it is not commonly stated in electrical diagrams and instruction sheets. If an electrician wishes to speak of a crest or peak voltage, he will state “169 crest volts” or “169 peak volts,” for example. An effective voltage of 120 volts has a crest voltage of approximately 169 volts.

In a DC machine, the condition of maximum efficiency occurs when the variable losses (copper Losses Pc) is equal to the constant loss (Iron Losses I2LRa) (since iron loss occurs in the armature of a DC machine because the armature core is made of iron and it rotates in a magnetic field.

Hence from the above discussion, it is clear that

I²_LR_a = P_c

$\begin{array}{l}{I^2}_L = \dfrac{{{P_c}}}{{{R_a}}}\\\\{I_L} = \sqrt {\frac{{{P_c}}}{{{R_a}}}} \\i.e\\{I_L} = \sqrt {\dfrac{{{\rm{Variable losses}}}}{{{\text{Armature Resistance}}}}} \end{array}$

In simplex lap winding, the number of parallel paths is equal to the number of poles and all the conductors of anyone parallel path lie under one pair of poles.

If the flux produced by all the poles is exactly the same, the emf induced in each parallel path and hence the current carried by each one of them will be equal.

But in spite of best efforts, this condition cannot be obtained due to different properties of the steel used for the construction of poles and variation of air gap between the poles and armature. Hence, there is always a slight difference in the induced emf in various parallel paths which results in large circulating currents in the armature winding.

The various reason of circulating current in a Lap wound DC machine are

1. Armature Position: The armature may not be exactly centered with respect to the pole faces (because of imperfections in assembling the machine parts, or because of wear of the bearings). In either case, the air gap will not be uniform and some of the poles will, therefore, carry more flux than the others, thereby introducing inequalities in the EMFs of the several parallel armature circuits.
2. Field Winding Placement:- A simplex lap winding has as many parallel paths as poles and each parallel path is made up of coil sides, which lie under two adjacent poles. These poles may have different strengths because of errors in placing field windings or may be due to the difference in the reluctance of iron used for poles. Thus, the flux under different poles may not be exactly identical, and this results in inequalities in the emf of several parallel circuits.
3. Unequal Resistance:- The armature circuits may not all have identical resistance, or the winding as a whole may not be exactly symmetrical.

This circulating current causes a continuous I²R loss in the winding and tends to heat the armature. It may also cause an undue amount of sparking a the brushes (since the current circulates from one path to another through brushes) and may lead to overloading of the brushes and conductors in the armature. Circulating current also introduces commutation difficulty in a dc machine.

Due to heavy sparking at the brushes which not only heats up the machine but also causes undue burning and wear of commutator and brushes. If it is carried too far, it may result in a flashover from positive to negative brushes.

To overcome this detrimental effect, usually, equalizing connections are provided in all lap-wound armatures. These connections are made with the help of very low resistance (almost zero) wires called equalizers.

An equalizer is a copper wire having very low resistance and is used to connect the points of different parallel paths of armature winding which should have the same potential under ideal conditions. These equalizer rings carry the circulating current and relieve the brushes to carry this circulating current. Under this condition, the brushes are to carry only the load current.

The desired arrangement is to connect every coil of the armature to an equalizer. But it is not possible in practical, as the number of such connections would be very large. It is sufficient to connect every third or fourth coil, and usually, 10 to 20 equalizer rings are used. The functions of the equalizer rings are

1. To save the brushes from handling the circulating currents
2. To reduce the magnetic flux unbalance that causes the potential difference in various parallel paths

For making equalizer connections, the number of coils is should be multiple of the number of poles and at the same time, the number of coils per pole should be divisible by a small number 2, 3 or 4.

Number of equalizer rings = Number of coils- sides Per Pole ⁄ 2

The number of connections to one equalizer rings = Number of pairs of poles

Control rods are used in nuclear reactors to control the fission rate of uranium and plutonium. During the fission of U-235 nuclei, for every one slow neutron absorbed, three fast-moving neutrons are produced. Assuming that these three neutrons are slowed down, they will produce an uncontrolled nuclear fission reaction as their number will multiply exponentially. This will cause a nuclear explosion.

In order to sustain the nuclear reaction in a controlled manner only one neutron per fission reaction is needed. Thus, the excess neutrons must be absorbed by suitable materials.

Boron and cadmium are suitable materials for capturing excess neutrons and are called absorbers The absorbers are used in the form of rods o rectangular strips and are commonly called controlling rods.

When control rods are fully inserted into fuel within the reactor, they absorb all the neutrons. This in turn completely stops chain reaction and hence nuclear fire dies. This particular position of controlling rods is called the shut down position of the nuclear reactor.

During actual operation, the controlling rods are slowly taken out from the nuclear fuel within the reactor, till they absorb the right amount of neutrons and leave only as many neutrons which are just necessary to sustain controlled nuclear fission. When this stage is reached, the reactor is said to have gone critical.

FIELD AND ARMATURE CONFIGURATIONS

There are two arrangements of fields and armatures:

1. Revolving armature and the stationary field
2. Revolving field and stationary armature.

 ADVANTAGES OF ROTATING FIELD IN AN ALTERNATOR

In large alternators, the rotating field arrangement is usually forward due to the following advantages.

1. Ease of Construction: Armature winding of large alternators is complex, the connection and bracing of the armature windings can be easily made for the stationary stator.
2. The number of Slip Rings: If the armature is made rotating, the number of slip rings required for power transfer from the armature to the external circuit is atleast three. Also, heavy current flows through brush and slip rings causing problems and requiring more maintenance in large alternators. Insulation required for slip rings from the rotating shaft is difficult with the rotating armature system.
3. High voltage generation:- Voltages can be generated as high as 11,000 and 13,800 V. These values can be reached because the stationary armature windings do not undergo vibration and centrifugal stresses.
4. High current Rating:-  Alternators can have relatively high current ratings. Such ratings are possible because the output of the alternator is taken directly from the stator windings through heavy, well-insulated cables to the external circuit. Neither slip rings nor a commutator is used.
5. Better Insulation to Armature: Insulation arrangement of armature windings can easily be made from the core on the stator.
6. Reduced Rotor Weight and Rotor Inertia:  Since the field system is placed on the rotor, hence the insulation requirement is less (for low dc voltage). Also, rotational inertia is less. It takes lesser time to gain full speed.
7. Improved Ventilation Arrangement: The cooling can be provided by enlarging the stator core with radial ducts. Water cooling is much easier if the armature is housed in the stator.

Hence in almost all of the alternators, the armature is housed in the stator while the dc field system is placed in the rotor.

Extrinsic Semiconductor

In order to change the properties of intrinsic semiconductors a small amount of some other material is added to it. The process of adding other material to the crystal of intrinsic semiconductors to improve its conductivity is called doping.

The impurity added is called dopant. The doped semiconductor material is called extrinsic semiconductors. The doping increases the conductivity of the basic intrinsic semiconductors hence the extrinsic semiconductors are used in practice for the manufacturing of various electronic devices such as diodes, transistors etc.

Depending upon the type of impurities, the two types of extrinsic semiconductors are,

1. N-TYPE
2. P-TYPE

N-TYPE Impurity

The impurity material having five valence electrons is called the pentavalent atom. When this is added to an intrinsic semiconductor, it is called donor doping as each impurity atom donates one free electron to an intrinsic material. Such an impurity is called donor impurity. Examples of such impurities are arsenic, bismuth, phosphorous, Antimony, etc. This crests an extrinsic semiconductor with a large number of free electrons called an n-type semiconductor.

Germanium and Silicon are tetravalent. The impurity atoms may be either pentavalent or trivalent, i.e., for groups V and III of the periodic table. If a small quantity of a pentavalent impurity (having 5-electrons in tt outermost orbit) like Arsenic (As), Antimony (Sb), or Phosphorus (P) is introduced in Germanium, it replaces the equal number of Germanium atoms without changing the physical state of the crystal.

Each of the four out of five valency electrons of impurity says Arsenic enters into covalent bonds with Germanium, while the fifth valence electron is set free to moo from one atom to the other as shown in Fig. The impurity is called donor impurity as it donates an electron and the crystal is called an N-type semiconductor.

A small amount of Arsenic (impurity) injects billions of free electrons into Germanium thus it creasing its conductivity enormously. In an N-type semiconductor, the majority carriers of charge are the electrons and holes as minority carriers. This is because when donor atoms are added to a semiconductor, the extra free electrons give the semiconductor a greater number of free electrons than it would normally have.

The generated power in an electrical system is the sum of the load power and power losses(Mainly losses of transmission line and transformer)

P_G = P_D + P_L

Now the given generated power P_G = 558 and the given load Power P_D = 443 therefore

P_L = P_G − P_D = 558 − 443

P_L = 115000 W

Power Loss = I²R

R = P_L ⁄ I²

R = 115000 ⁄ 50²

R = 46Ω

The synchronous speed of an alternator is given as

Ns = 120f/P
= 120 × 50/2

Ns = 3000 RPM

The above circuit is a simple series parallel connection

The series resistance 5Ω and 5Ω are in parallel with the series resistance 20Ω and 5Ω

Now resistance 10Ω and 25Ω are in parallel therefore equivalent resistance become

R_eqv = (25 × 10) ⁄ (25 + 10)

R_eqv = 50 ⁄ 7

Hence current supplied by the battery through AB terminal is

I = V ⁄ R_eqv  = 10 × 7 ⁄ 50 = 1.4 A

I = 1.4 A

An RC time constant tells us about the relation between time, resistance, and capacitance. The time taken by the capacitor is directly proportional to the amount of resistance and capacitance in the circuit.

The time constant reflects the time required for the capacitor to charge up to 63.2% of the applied voltage.

So 63.2% of 100 V = 63.2 Volt