DMRC JE Electrical 10th- April- 2018 Paper With Solution and Explanation

Ground Resistance Measurements

To maintain sufficiently low resistance values of grounding systems, periodic testing is required. The testing involves measurement to ensure that they do not exceed design limits. The methods of measuring and testing the ground resistance and soil resistivity are as follows:

• Two-point method
• Three-point method
• Fall-of-potential method
• Ratio method
• Four-point method
• Touch Potential Method
• Clamp-On method

Two-Point Method

This method may be used to measure the resistance of a single driven ground rod. It uses an auxiliary ground rod whose resistance is either known or can be measured. The resistance value of the auxiliary ground rod also must be very small compared to the resistance of the driven ground rod so that the measured value can be assumed to be wholly contributed by the drivel ground rod. For example, this test might be applicable in the measurement o resistance of the single driven ground rod for a residence or in congested areas where finding room to drive two auxiliary rods may be a problem.

In this case, the municipal metallic water supply line can be assumed as the auxiliary ground rod whose resistance value is approximately 1Ω or less. This value is quite small compared to the value of a single driven ground rod, whose value is in the order of 25Ω. The reading obtained is that of the two grounds in the series. The lead resistances will also be measured and should be deducted from the final measurements. This method is usually adequate where a go, no-go type of test is required.

 Three-Point Method

This method is similar to the two-point method except it uses two auxiliary rods. To obtain accurate values of resistance measurements, the resistance of the auxiliary electrodes should be approximately equal to or less than that of the electrode under test. The connections for the three-point method are shown in Figure

Either AC of 50Hz or DC may be used for making this test. The advantage of using AC is that it minimizes the effects of stray currents on measurement readings. However, if stray currents happen to be of the same frequency, the error will be introduced in the readings.

The use of DC for making this test will totally eliminate the AC stray currents. The use of direct current for making this test will totally eliminate the ac stray currents. However, stray direct current and formation of gas around the electrode will introduce errors in the readings when using direct current for this test. The
average of two readings will give an accurate test value.

Fall of Potential Method

This method measures grounding electrode resistance based upon the principle of the potential drop across the resistance. It also uses two auxiliary electrodes (one current rod and the other potential rod) that are placed at a sufficient distance from the test electrodes.

A current of known magnitude is passed through the electrode under the test and one of the auxiliary electrodes (current rod). The drop in potential between the electrode under the test and the second auxiliary electrode (potential rod) is measured.

• The Potier triangle method is also called as Zero power factor (ZPF) method.
• According to this method in any alternator armature resistance drop IR_a and the armature leakage reactance drop IX_L are basically the e.m.f quantities while armature reaction is considered as m.m.f quantities.
• This method is based on the separation of the armature leakage reactance and armature reaction effect.
• The armature leakage reactance X_L is called as Potier reactance hence this method is also called as Potier reactance method.

For Maximum Power to be transferred the Load Resistance R_L and the internal Resistance Rth should be Equal.

R_th = R_L

Thevenin’s resistance can be found by replacing 10 V source with a short-circuit.

As seen from Fig. , R_th = 2 + 2 + (6 || 6)  = 7Ω.

We will remove the load resistance R_L and find the equivalent Thevenin’s source for the circuit to the left of terminals A and B.

As seen from Fig. V_th equals the drop across the vertical resistor of 6Ω because no current flows through 2Ω and 2Ω resistors

Now from voltage division Rule

Vth = 10 × 6 ⁄ (6 + 6) = 5 V

Now the Load Current I_L from the  above circuit is

I_L = V_th ⁄ (R + R_L) = 5 ⁄ (7 + 7)

= 5 ⁄ 14

I_L = 0.357 A

Power = I²R_L

P = (0.357)² × 7
= 0.1274 × 7

= 0.892 Watt or 892 mW

Kanthal is the trademark for a family of iron-chromium-aluminum (FeCrAl) alloys used in a wide range of resistance and high-temperature applications. Kanthal FeCrAl alloys consist of mainly iron, chromium (20–30%) and aluminum, and cobalt (4–7.5 %). The first Kanthal FeCrAl alloy was developed by Hans von Kantzow in Hallstahammar, Sweden.

The alloys are known for their ability to withstand high temperatures and have intermediate electric resistance. As such, it is frequently used in heating elements.

A general conceptual definition of a time constant can be given as follows:

A time constant is a measure of the time it takes to observe significant changes in a given process.

When the capacitor is discharging

When the capacitor discharge the time constant can be defined as the time required for the charging current of the capacitor to fall to 0.368 ≅ 0.37 of its initial maximum value, starting from its maximum value.

Hence At 1 time constant ( 1T ) Vc = 0.37Vc.

Therefore, Vc = 0.37 x 50V = 18.5 V

COULOMB’S LAW AND CONCEPT OF PERMITTIVITY

Coulomb’sfirst law states that like charges repel each other while opposite charges attract each other. Coulomb’s second law states that the Force of attraction between two opposite charges or force of repulsion between two like charges is

• Directly proportional to the product of the charges, the distance between them being the same;
• Inversely proportional to the square of the straight distance between them, the magnitude of the charges being constant.

If we assume the charges to be of magnitude (Q) and (q) separated by a straight distance d, then from Coulomb’s second law we can write

F ∝ Q.q ⁄ d²,

(F) being the force of repulsion if both charges are alike or force of attraction if the charges have opposite polarity. The force (F) is measured in Newtons when the magnitude of the charges is expressed in Coulombs and the distance in meters.

$F = K\dfrac{{Q.q}}{{{d^2}}}$

where k is Coulomb’s constant (k = 9 ×10⁹ N m² C⁻²)

If Q = q = 1 coulomb and d = 1 meter

$\begin{array}{l}F = 9 \times {10^9}\dfrac{{1 \times 1}}{{{1^2}}}\\\\F = 9 \times {10^9}\end{array}$

If Q = q = 1 coulomb and d = 1 meter. This gives rise to the definition of unit charge (i.e. 1 Coulomb which means that it is such a charge that when placed at one meter apart from another similar charge experiences a force of 9 x 10⁹ Newton in the vacuum.

Force Law For Current-Carrying Conductors

The force (F) per unit length (L) between two wires carrying currents I₁ and I₂ separated by a distance d in a vacuum is given by:

$\dfrac{F}{l} = \dfrac{{{\mu _o}{I_1}{I_2}}}{{2\pi d}}$

Where μo is Vacuum permeability or permeability of free space =  4π × 10^-7 H/m

Now the given data

Current in the wire = I₁ = _I2 = 1A

The distance between the wire d = 1m

Putting the value in the above equation

$\begin{array}{l}\dfrac{F}{l} = \dfrac{{{\mu _o}{I_1}{I_2}}}{{2\pi d}}\\\\\dfrac{F}{l} = \dfrac{{4\pi \times {{10}^{ – 7}} \times 1 \times 1}}{{2\pi \times 1}}\\\\\dfrac{F}{l} = 2\pi \times {10^{ – 7}}\end{array}$

Magnetic field Intensity (H) is also called Magnetic field Strength, Magnetic Intensity, Magnetic field, Magnetic Force, and Magnetization Force.

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

Suppose that a current of I amperes flows through a coil of N turns wound on a toroid of length I meters. The MMF is the total current linked to the magnetic circuit i.e  IN ampere-turns. If the magnetic circuit is homogeneous and of a uniform cross-sectional area, the MMF per meter length of the magnetic circuit is termed the magneticfield strength, magnetic field intensity, or magnetizing force. It is represented by the symbol H and is measured in ampere-turns per meter (At/m).

$H = \dfrac{F}{L} = \dfrac{{IN}}{l}$

Hence the magnetic field Intensity can be defined as the ratio of applied MMF to the length of the path that it acts over.

If  N conductors are grouped together to form a coil or a cable then field strength due to current I passing through each conductor of the group can also be calculated by using the same expression. The only change will be the field strength calculated above will get multiplied by N. The value of L of the circular path is 2πr

So magnetic field strength at a point ‘r’ meters away from the center of such group of “N” conductors each carrying the current I ampere is

$H = \dfrac{{IN}}{{2\pi r}}$

The SI unit of Magnetic field intensity is Ampere/meter whereas in the CGS system (the centimeter–gram–second ) the unit of Magnetic field intensity is  Oersted

Parallel operation of Transformer

The need for the operation of two or more transformers in parallel often arises due to:

• Load growth, which exceeds the capacity of an existing transformer
• Lack of space (height) for one large transformer
• A measure of security (the probability of two transformers failing at the same time is very small)
•  The adoption of a standard size transformer throughout an installation
•  To maximize the electrical power efficiency, availability, reliability, and flexibility.

When two or more transformers run in parallel, they must satisfy the following conditions.

1. Same voltage ratio of transformers
2. Same percentage impedance
3. Same Polarity
4. Same phase sequence
5. Same voltage ratio

Same Voltage Ratio

An equal voltage ratio is necessary to avoid a no-load circulating current. Circulating current may be defined as that current that flows in transformers operating in parallel when they do not supply the load. Consider two transformers connected in parallel on their primary sides only.

If the voltage readings on the secondaries are not the same, there will be circulating currents between the secondaries when they are connected in parallel. This also creates circulating currents in primaries even when the load is not connected.  As the internal impedance of the transformer is small, a small voltage difference may cause a high circulating current which increases copper losses. These circulating currents produce undesirable effects as:

1. They reduce the permissible output of the parallel combined group because circulating current produces unequal load sharing.
2. They increase the power loss.
3. They may overload one of the transformers.

Same percentage impedance

The current shared by two transformers running in parallel should be proportional to their MVA rating. And current carried by these transformers is inversely proportional to their internal impedance. Hence the impedance of the transformer running in parallel is inversely proportional to their MVA Rating Each transformer has a particular value of impedance which may be different from the other transformer i.e their resistance to reactance is different.

Consider two transformers whose ratings are in the ratio of 4: 1. It is obvious that the first transformer must have one-fourth of the impedance of the second transformer hence the current drawn by the first transformer will be 4 times to the second transformer.

Due to the difference in the quality of percentage impedance, there will be a divergence of the phase angle of the two currents so that one transformer will be working with a lower and the other with a higher power factor than the combined load.

Same Polarity.

The polarity of all transformers running in parallel must be the same. The polarity of the transformer refers to the instantaneous direction of Induced EMF in the secondary. If the instantaneous direction of induced secondary EMF in two transformers are opposite to each other when the same input power is fed to both of the transformers, then the transformer is said to be in opposite polarity.

If the instantaneous direction of induced secondary EMF in two transformers is sane when the same input power is fed to both of the transformers, then the transformer is said to be in the same polarity.

The difference in the polarity causes the huge circulating current that flows in the transformer and even may lead to a dead short circuit.

In a synchronous machine, there are two types of rotor construction

1. Salient Pole Type
2. Cylindrical Type (non-salient pole)

A salient pole synchronous machine is distinguished from the round rotor machine by constructional features of field poles that project with a large interpolar air gap.

This type of construction is commonly employed in machines coupled to hydroelectric turbines which are inherently slow-speed ones so the synchronous machine has multiple pole pairs as different from machines coupled to high-speed steam turbines (3,000/1,500 rpm) which have a two- or four-pole structure. Salient pole machine analysis is made through the two-reaction theory.

A synchronous machine with salient or projecting poles has a non-uniform air gap due to which its reactance varies with the rotor position. Consequently, a cylindrical rotor machine possesses one axis of symmetry (pole axis or direct axis) whereas a salient-pole machine possesses two axes of geometric symmetry field poles axis, called the direct axis or d-axis, and axis passing through the center of the interpolar space, called the quadrature axis or q axis, as shown in Fig

The permeance offered to an MMF wave is highest when it is aligned with the field pole d-axis. Obviously, two mmf act on the d-axis of a salient-pole synchronous machine ie. field m.m.f. and armature m.m.f.

The permeance offered is lowest when it is oriented at 90° to the field pole q-axis only one m.m.f., ie. armature mmf acts on the q-axis because the field has no component in the q-axis.

The direct axis flux path includes two small air gaps under pole faces only whereas the quadrature axis flux path has two larger air gaps in the inter-polar regions. Hence with direct axis flux path has minimum reluctance and the quadrature axis flux path has maximum reluctance.

According to the two-reaction theory, the armature mmf F_a produced by stator l_a is resolved into two components, one along with the direct or d-axis and another along the quadrature or q-axis.

The direct axis component of F_a is F_ad and the quadrature axis component is F_aq. The effect of F_ad is either magnetizing at demagnetizing (depending on whether the p.f. is leading or lagging), whereas the effect of F_aq is entirely cross-magnetizing.

Similarly, flux Φ_a produced by stator current l_a is resolved into two components Φ_ad and Φ_aq

Φ_ad induces direct axis armature reaction voltage E_ad and Φ_aq induces quadrature axis armature reaction voltage E_aq.

• An ideal transformer is a perfectly coupled loose-less transformer with infinite high core magnetic permeability.
• It is purely an imaginary transformer that has many similar properties related to the practical transformer and it is only used for study purposes.
• The energy in an ideal transformer is completely utilized i.e its efficiency is exactly 100%.

Ideal transformer Systematic Circuit Diagram

• Let us assume an ideal transformer with primary winding is connected to an AC supply and secondary winding is open-circuited i.e no load is applied to it.
• Now when an alternating voltage V₁ is applied to the primary winding due to electromagnetic induction an emf E₁ is induced on the primary side. Since there is no voltage drop in the ideal transformer, therefore, counter emf is equal and opposite to applied voltage V₁
• Since the primary coil is pure inductive and there is no load connected to the secondary side, therefore, primary winding draws a small magnetizing current Im.
• In the case of pure inductance the current lags behind the applied voltage by 90^o
• This magnetizing current produces flux Φ which is directly proportional to the current and therefore in phase with it.
• The flux passes through both the primary and secondary winding of the transformer so according to Faraday’s law of electromagnetic induction, the voltage V₂ gets induced in the secondary winding. This V₂ produced a counter emf E₂ (mutually induced emf)
• At any instant (if the turns ratio is the same) the value of V₁ is always equal to V₂in the case of an ideal transformer.
• Both the emf E₁ and E₂ always lag behind the flux by 90^o and their magnitudes depend upon the rate of change of flux and the number of turns in the primary and secondary winding.