📘 12th Tamil - Quarterly Exam 2024 - Answer Key | Villupuram District

Electrical Engineering

B.Tech I Year I Semester (R13) | December 2015

Question Paper & Solutions

PART-A Answer all questions (10 x 2 = 20 Marks)

1. (a) State and explain Ohm's law.

Solution:

Statement: Ohm's law states that at a constant temperature and other physical conditions, the voltage (V) across a conductor is directly proportional to the current (I) flowing through it.

Explanation:

• Mathematically, this relationship is expressed as: \(V \propto I\)
• This can be written as an equation: \(V = R \cdot I\)
• Where:
  • V is the potential difference or voltage, measured in Volts (V).
  • I is the electric current, measured in Amperes (A).
  • R is the constant of proportionality, known as resistance, measured in Ohms (Ω). Resistance is the property of a material that opposes the flow of electric current.
• Limitation: Ohm's law is only applicable to linear or "ohmic" materials (like most metals) and under conditions of constant temperature, as resistance can change with temperature.

1. (b) Three resistors of 10Ω, 20Ω, and 30Ω are connected in star. Find the equivalent delta resistances.

Solution:

Given the star-connected resistors: \(R_1 = 10\Omega\), \(R_2 = 20\Omega\), and \(R_3 = 30\Omega\).

The formulas for converting a star network to an equivalent delta network are:

\[ R_{12} = R_1 + R_2 + \frac{R_1 R_2}{R_3} \] \[ R_{23} = R_2 + R_3 + \frac{R_2 R_3}{R_1} \] \[ R_{31} = R_3 + R_1 + \frac{R_3 R_1}{R_2} \]

Step-by-step calculation:

• Resistance between terminals 1 and 2 (\(R_{12}\)): \[ R_{12} = 10 + 20 + \frac{10 \times 20}{30} = 30 + \frac{200}{30} = 30 + 6.67 = 36.67 \Omega \]
• Resistance between terminals 2 and 3 (\(R_{23}\)): \[ R_{23} = 20 + 30 + \frac{20 \times 30}{10} = 50 + \frac{600}{10} = 50 + 60 = 110 \Omega \]
• Resistance between terminals 3 and 1 (\(R_{31}\)): \[ R_{31} = 30 + 10 + \frac{30 \times 10}{20} = 40 + \frac{300}{20} = 40 + 15 = 55 \Omega \]

The equivalent delta resistances are \(36.67 \Omega\), \(110 \Omega\), and \(55 \Omega\).

1. (c) Define the average value and RMS value of an alternating quantity.

Solution:

Average Value:

• The average value of an alternating quantity (like current or voltage) is defined as the arithmetic mean of all its instantaneous values over one half-cycle.
• It is calculated over a half-cycle because the average value over a complete symmetrical cycle is zero.
• For a sinusoidal voltage waveform, the average value is: \( V_{avg} = \frac{2V_m}{\pi} \approx 0.637 V_m \), where \(V_m\) is the peak value.

RMS (Root Mean Square) Value:

• The RMS value of an alternating current is the equivalent steady DC current that would produce the same amount of heat in a given resistor over a given time period. It is also known as the "effective" value.
• It is calculated by taking the square root of the mean of the squares of the instantaneous values over one complete cycle.
• For a sinusoidal voltage waveform, the RMS value is: \( V_{rms} = \frac{V_m}{\sqrt{2}} \approx 0.707 V_m \), where \(V_m\) is the peak value.

1. (d) A series RLC circuit has R=5Ω, L=0.2H, C=50µF. The applied voltage is 230V. Find the resonant frequency.

Solution:

The resonant frequency (\(f_r\)) of a series RLC circuit is the frequency at which the inductive reactance (\(X_L\)) equals the capacitive reactance (\(X_C\)). The formula is:

\[ f_r = \frac{1}{2\pi\sqrt{LC}} \]

Given values:

• Inductance (L) = 0.2 H
• Capacitance (C) = 50 µF = \(50 \times 10^{-6}\) F

Calculation:

\[ f_r = \frac{1}{2\pi\sqrt{0.2 \times (50 \times 10^{-6})}} \] \[ f_r = \frac{1}{2\pi\sqrt{10 \times 10^{-6}}} \] \[ f_r = \frac{1}{2\pi \times \sqrt{10} \times 10^{-3}} \] \[ f_r = \frac{1000}{2\pi \times 3.162} \] \[ f_r = \frac{1000}{19.87} \approx 50.33 \text{ Hz} \]

The resonant frequency of the circuit is approximately 50.33 Hz. (Note: The resistance R and applied voltage V are not needed to calculate the resonant frequency).

1. (e) Explain the principle of a DC generator.

Solution:

The principle of a DC generator is based on Faraday's Law of Electromagnetic Induction.

• The Law States: Whenever a conductor cuts through magnetic flux lines, a dynamically induced electromotive force (EMF) is generated in it.
• How it Works:
  1. A DC generator consists of armature conductors rotating within a stationary magnetic field.
  2. As the armature rotates (driven by a prime mover like a diesel engine or turbine), the conductors continuously cut the magnetic flux produced by the field poles.
  3. This relative motion between the conductors and the magnetic field induces an EMF in the conductors.
  4. The direction of the induced EMF is determined by Fleming's Right-Hand Rule.
  5. The induced EMF is initially alternating in nature. A mechanical rectifier called a commutator is used to convert this internal alternating EMF into a unidirectional (DC) EMF at the output terminals (brushes).

1. (f) What are the various losses in a DC machine?

Solution:

The losses in a DC machine can be categorized into three main types:

1. Copper Losses (or Electrical/Winding Losses): These are due to the resistance of the windings and are proportional to the square of the current (\(I^2R\)).
   • Armature Copper Loss: \(I_a^2 R_a\), where \(I_a\) is armature current and \(R_a\) is armature resistance.
   • Field Copper Loss: \(I_{sh}^2 R_{sh}\) (for shunt machines) or \(I_{se}^2 R_{se}\) (for series machines).
   • Brush Contact Loss: Voltage drop across the brushes.
2. Iron Losses (or Core/Magnetic Losses): These occur in the armature core due to the alternating magnetic flux. They are constant for a given speed.
   • Hysteresis Loss: Energy lost due to the repeated reversal of magnetization in the armature core material.
   • Eddy Current Loss: Power loss due to circulating currents induced in the armature core. It is minimized by using a laminated core.
3. Mechanical Losses: These are due to the physical rotation of the machine.
   • Friction Loss: Loss in bearings and at the brush-commutator interface.
   • Windage Loss: Air friction loss caused by the rotating armature.

Iron losses and mechanical losses are together known as Stray Losses or Rotational Losses.

1. (g) Define the regulation and efficiency of a single-phase transformer.

Solution:

Voltage Regulation:

Voltage regulation of a transformer is the percentage change in its secondary terminal voltage from a no-load condition to a full-load condition, with the primary voltage held constant. It is a measure of how well the transformer maintains a constant secondary voltage under varying loads.

\[ \text{Regulation (%)} = \frac{V_{\text{no-load}} - V_{\text{full-load}}}{V_{\text{full-load}}} \times 100\% \]

A lower regulation value is desirable, as it indicates better performance.

Efficiency (\(\eta\)):

The efficiency of a transformer is the ratio of the output power delivered to the load to the input power supplied by the source. It indicates how effectively the transformer transfers power.

\[ \eta = \frac{\text{Output Power}}{\text{Input Power}} \times 100\% \]

Since Input Power = Output Power + Losses, the formula can also be written as:

\[ \eta = \frac{\text{Output Power}}{\text{Output Power + Core Losses + Copper Losses}} \times 100\% \]

1. (h) Explain the principle of operation of a transformer.

Solution:

The operation of a transformer is based on the principle of mutual electromagnetic induction.

• A transformer consists of two coils, a primary winding and a secondary winding, wound on a common laminated magnetic core.
• When an alternating voltage (\(V_1\)) is applied to the primary winding, it drives an alternating current (\(I_1\)) through it.
• This alternating current produces a continuously changing magnetic flux (\(\phi\)) in the magnetic core.
• According to Faraday's Law of Induction, this changing flux induces a self-induced EMF (\(E_1\)) in the primary winding, which opposes the applied voltage.
• The same changing flux also links with the secondary winding. According to the principle of mutual induction, it induces an EMF (\(E_2\)) in the secondary winding.
• The magnitude of the induced EMF in each winding is proportional to the number of turns in that winding (\(E \propto N\)). Therefore, \(\frac{E_2}{E_1} = \frac{N_2}{N_1}\).
• If a load is connected to the secondary, the induced EMF \(E_2\) drives a current (\(I_2\)) through the load, thus transferring power from the primary to the secondary circuit without any direct electrical connection.

1. (i) Write the expression for the synchronous speed of an induction motor and explain the terms.

Solution:

The expression for the synchronous speed of an induction motor (and any synchronous machine) is:

\[ N_s = \frac{120 \times f}{P} \]

Explanation of terms:

• \(N_s\) is the synchronous speed of the rotating magnetic field, measured in revolutions per minute (RPM). This is the speed at which the motor's magnetic field rotates.
• \(f\) is the frequency of the AC power supply, measured in Hertz (Hz). In India, the standard is 50 Hz.
• \(P\) is the number of poles in the stator winding. This is always an even number (e.g., 2, 4, 6).

1. (j) What is the function of a moving iron ammeter?

Solution:

The primary function of a moving iron (MI) ammeter is to measure the magnitude of electric current in a circuit.

Key features related to its function:

• Measures both AC and DC: The deflecting torque in an MI instrument is proportional to the square of the current (\(T_d \propto I^2\)). This means the deflection is always in the same direction regardless of the current's polarity, allowing it to measure both AC (RMS value) and DC currents.
• Working Principle: It operates on the principle that a piece of soft iron, when placed in the magnetic field of a current-carrying coil, is either attracted into the field (attraction type) or repelled by it (repulsion type). The force of attraction or repulsion is calibrated to indicate the current on a scale.
• Applications: Due to their robust construction and ability to measure both AC and DC, they are widely used in industrial applications and laboratory panels where high accuracy is not the primary concern.

PART-B Answer one full question from each unit (5 x 10 = 50 Marks)

2. (a) Derive the expression for the equivalent resistance of a star-connected network into its equivalent delta-connected network.

Solution:

To derive the expressions for star-to-delta conversion, we equate the resistances between corresponding pairs of terminals for both networks, ensuring they are electrically equivalent.

Let the star network have resistors \(R_1, R_2, R_3\) connected to a common point N. Let the equivalent delta network have resistors \(R_{12}, R_{23}, R_{31}\) connected between terminals (1,2), (2,3), and (3,1) respectively.

Step 1: Calculate resistance between terminals 1 and 2.

• For the Star network, the resistance between terminals 1 and 2 is \(R_{1-2(Star)} = R_1 + R_2\). (Terminal 3 is open).
• For the Delta network, the resistance between terminals 1 and 2 is \(R_{12}\) in parallel with the series combination of \(R_{23}\) and \(R_{31}\). \[ R_{1-2(Delta)} = \frac{R_{12}(R_{23} + R_{31})}{R_{12} + R_{23} + R_{31}} \]

Equating these gives our first equation: \[ R_1 + R_2 = \frac{R_{12}(R_{23} + R_{31})}{R_{12} + R_{23} + R_{31}} \quad \cdots (1) \]

Step 2: Calculate resistance between terminals 2 and 3.

Similarly, we get:

\[ R_2 + R_3 = \frac{R_{23}(R_{12} + R_{31})}{R_{12} + R_{23} + R_{31}} \quad \cdots (2) \]

Step 3: Calculate resistance between terminals 3 and 1.

And again for terminals 3 and 1:

\[ R_3 + R_1 = \frac{R_{31}(R_{12} + R_{23})}{R_{12} + R_{23} + R_{31}} \quad \cdots (3) \]

Step 4: Solve the equations for \(R_{12}, R_{23}, R_{31}\).

Subtracting equation (2) from (1):

\[ (R_1 + R_2) - (R_2 + R_3) = \frac{R_{12}R_{23} + R_{12}R_{31} - R_{23}R_{12} - R_{23}R_{31}}{R_{12} + R_{23} + R_{31}} \] \[ R_1 - R_3 = \frac{R_{31}R_{12} - R_{23}R_{31}}{R_{12} + R_{23} + R_{31}} \quad \cdots (4) \]

Adding equations (3) and (4):

\[ (R_3 + R_1) + (R_1 - R_3) = \frac{R_{31}R_{12} + R_{31}R_{23}}{R_{12} + R_{23} + R_{31}} + \frac{R_{31}R_{12} - R_{23}R_{31}}{R_{12} + R_{23} + R_{31}} \] \[ 2R_1 = \frac{2 R_{12} R_{31}}{R_{12} + R_{23} + R_{31}} \] \[ R_1 = \frac{R_{12} R_{31}}{R_{12} + R_{23} + R_{31}} \quad \cdots (A) \]

By symmetry, we can write the expressions for \(R_2\) and \(R_3\):

\[ R_2 = \frac{R_{12} R_{23}}{R_{12} + R_{23} + R_{31}} \quad \cdots (B) \] \[ R_3 = \frac{R_{23} R_{31}}{R_{12} + R_{23} + R_{31}} \quad \cdots (C) \]

These are the formulas for Delta-to-Star conversion. Now we derive the Star-to-Delta formulas.

Step 5: Derive the Star-to-Delta formulas.

Consider the combinations from equations (A), (B), and (C):

\[ R_1 R_2 + R_2 R_3 + R_3 R_1 = \frac{R_{12}^2 R_{23} R_{31} + R_{12} R_{23}^2 R_{31} + R_{12} R_{23} R_{31}^2}{(R_{12} + R_{23} + R_{31})^2} \] \[ R_1 R_2 + R_2 R_3 + R_3 R_1 = \frac{R_{12} R_{23} R_{31} (R_{12} + R_{23} + R_{31})}{(R_{12} + R_{23} + R_{31})^2} \] \[ R_1 R_2 + R_2 R_3 + R_3 R_1 = \frac{R_{12} R_{23} R_{31}}{R_{12} + R_{23} + R_{31}} \quad \cdots (D) \]

Now, divide equation (D) by equation (C), (A), and (B) respectively:

Divide (D) by (C):

\[ \frac{R_1 R_2 + R_2 R_3 + R_3 R_1}{R_3} = \frac{R_{12} R_{23} R_{31}}{R_{12} + R_{23} + R_{31}} \times \frac{R_{12} + R_{23} + R_{31}}{R_{23} R_{31}} = R_{12} \] \[ \implies R_{12} = \frac{R_1 R_2 + R_2 R_3 + R_3 R_1}{R_3} = R_1 + R_2 + \frac{R_1 R_2}{R_3} \]

Similarly, dividing (D) by (A) gives \(R_{23}\), and dividing (D) by (B) gives \(R_{31}\):

\[ R_{23} = \frac{R_1 R_2 + R_2 R_3 + R_3 R_1}{R_1} = R_2 + R_3 + \frac{R_2 R_3}{R_1} \] \[ R_{31} = \frac{R_1 R_2 + R_2 R_3 + R_3 R_1}{R_2} = R_3 + R_1 + \frac{R_3 R_1}{R_2} \]

These are the required expressions for star-to-delta transformation.

OR

2. (b) Find the equivalent resistance between terminals A and B of the network shown.

Solution:

The given circuit is a bridge network which is not balanced. We can check the balance condition:

\[ \frac{R_{AC}}{R_{CB}} = \frac{10}{40} = 0.25 \] \[ \frac{R_{AD}}{R_{DB}} = \frac{30}{20} = 1.5 \]

Since \( \frac{R_{AC}}{R_{CB}} \neq \frac{R_{AD}}{R_{DB}} \), the bridge is unbalanced. We must use a star-delta transformation to solve it.

Let's convert the upper delta formed by nodes A, C, and D (resistors 10Ω, 30Ω, and 60Ω) into an equivalent star network with a new center point O.

Step 1: Delta to Star Conversion (Delta ACD -> Star O)

The delta resistors are \(R_{AC} = 10\Omega\), \(R_{AD} = 30\Omega\), and \(R_{CD} = 60\Omega\).

• Resistor from A to O (\(R_{AO}\)): \[ R_{AO} = \frac{R_{AC} \times R_{AD}}{R_{AC} + R_{AD} + R_{CD}} = \frac{10 \times 30}{10 + 30 + 60} = \frac{300}{100} = 3 \Omega \]
• Resistor from C to O (\(R_{CO}\)): \[ R_{CO} = \frac{R_{AC} \times R_{CD}}{100} = \frac{10 \times 60}{100} = \frac{600}{100} = 6 \Omega \]
• Resistor from D to O (\(R_{DO}\)): \[ R_{DO} = \frac{R_{AD} \times R_{CD}}{100} = \frac{30 \times 60}{100} = \frac{1800}{100} = 18 \Omega \]

Step 2: Simplify the new circuit.

After the conversion, the circuit is simplified as follows:

• The resistor \(R_{CO}\) (6Ω) is now in series with \(R_{CB}\) (40Ω).
  Total resistance of this branch: \(R_{COB} = 6 + 40 = 46 \Omega\).
• The resistor \(R_{DO}\) (18Ω) is in series with \(R_{DB}\) (20Ω).
  Total resistance of this branch: \(R_{DOB} = 18 + 20 = 38 \Omega\).
• These two branches (COB and DOB) are in parallel.

Step 3: Calculate the parallel combination.

The equivalent resistance of the parallel branches is:

\[ R_{parallel} = \frac{R_{COB} \times R_{DOB}}{R_{COB} + R_{DOB}} = \frac{46 \times 38}{46 + 38} = \frac{1748}{84} \approx 20.81 \Omega \]

Step 4: Calculate the total equivalent resistance.

This parallel combination is in series with the resistor \(R_{AO}\) (3Ω).

\[ R_{AB} = R_{AO} + R_{parallel} = 3 + 20.81 = 23.81 \Omega \]

The equivalent resistance between terminals A and B is 23.81 Ω.

3. (a) Derive the RMS value of a sinusoidal alternating quantity.

Solution:

Let the sinusoidal alternating voltage be represented by the equation:

\[ v(t) = V_m \sin(\omega t) \] where \(V_m\) is the maximum or peak value, and \(\omega\) is the angular frequency.

The RMS (Root Mean Square) value is found by performing three operations: squaring the quantity, taking the mean (average) over one cycle, and finally taking the square root.

Step 1: Square the instantaneous value.

\[ v(t)^2 = (V_m \sin(\omega t))^2 = V_m^2 \sin^2(\omega t) \] Using the trigonometric identity \(\sin^2(\theta) = \frac{1 - \cos(2\theta)}{2}\), we get: \[ v(t)^2 = V_m^2 \left( \frac{1 - \cos(2\omega t)}{2} \right) \]

Step 2: Find the mean (average) of the squared value over one complete cycle (from 0 to T).

The mean value is found by integrating over one period (T) and dividing by the period. \[ \text{Mean Square Value} = \frac{1}{T} \int_0^T v(t)^2 \,dt = \frac{1}{T} \int_0^T V_m^2 \left( \frac{1 - \cos(2\omega t)}{2} \right) \,dt \] \[ = \frac{V_m^2}{2T} \int_0^T (1 - \cos(2\omega t)) \,dt \] \[ = \frac{V_m^2}{2T} \left[ t - \frac{\sin(2\omega t)}{2\omega} \right]_0^T \] Substitute the limits of integration: \[ = \frac{V_m^2}{2T} \left[ \left(T - \frac{\sin(2\omega T)}{2\omega}\right) - \left(0 - \frac{\sin(0)}{2\omega}\right) \right] \] Since \(\omega = \frac{2\pi}{T}\), we have \(2\omega T = 2 \left(\frac{2\pi}{T}\right) T = 4\pi\). And we know \(\sin(4\pi) = 0\) and \(\sin(0) = 0\). \[ = \frac{V_m^2}{2T} \left[ (T - 0) - (0 - 0) \right] = \frac{V_m^2}{2T} \cdot T = \frac{V_m^2}{2} \]

Step 3: Take the square root of the mean square value.

\[ V_{rms} = \sqrt{\text{Mean Square Value}} = \sqrt{\frac{V_m^2}{2}} \] \[ V_{rms} = \frac{V_m}{\sqrt{2}} \]

Thus, the RMS value of a sinusoidal alternating quantity is its peak value divided by the square root of 2, which is approximately \(0.707\) times the peak value.

OR

3. (b) A series RLC circuit consists of R=100Ω, L=0.15H, and C=150µF across a 230V, 50Hz supply. Calculate: i) \(X_L\), ii) \(X_C\), iii) Z, iv) Current, and v) Power factor.

Solution:

Given parameters:

• Resistance (R) = 100 Ω
• Inductance (L) = 0.15 H
• Capacitance (C) = 150 µF = \(150 \times 10^{-6}\) F
• Voltage (V) = 230 V
• Frequency (f) = 50 Hz

i) Inductive Reactance (\(X_L\)):

\[ X_L = 2\pi f L = 2\pi \times 50 \times 0.15 = 100\pi \times 0.15 = 15\pi \approx 47.12 \, \Omega \]

ii) Capacitive Reactance (\(X_C\)):

\[ X_C = \frac{1}{2\pi f C} = \frac{1}{2\pi \times 50 \times 150 \times 10^{-6}} = \frac{1}{100\pi \times 150 \times 10^{-6}} = \frac{10^6}{15000\pi} \approx 21.22 \, \Omega \]

iii) Total Impedance (Z):

\[ Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{100^2 + (47.12 - 21.22)^2} \] \[ Z = \sqrt{10000 + (25.9)^2} = \sqrt{10000 + 670.81} = \sqrt{10670.81} \approx 103.3 \, \Omega \]

iv) Current (I):

\[ I = \frac{V}{Z} = \frac{230}{103.3} \approx 2.226 \, \text{A} \]

v) Power Factor (cos φ):

\[ \cos(\phi) = \frac{R}{Z} = \frac{100}{103.3} \approx 0.968 \] Since \(X_L > X_C\), the circuit is inductive, and the current lags the voltage. Therefore, the power factor is 0.968 lagging.

4. (a) With a neat diagram, explain the constructional features of a DC machine.

Solution:

A DC machine (generator or motor) consists of two main parts: a stationary part called the stator and a rotating part called the rotor (or armature).

The main constructional features are:

1. Yoke (or Frame): The outer cylindrical frame of the machine. It serves two purposes:
   • It provides mechanical support for the poles and acts as a protective cover.
   • It forms a part of the magnetic circuit, carrying the magnetic flux produced by the field windings. It is typically made of cast iron or cast steel.
2. Field Poles and Pole Shoes: The poles are fixed to the inner periphery of the yoke. Each pole has a pole core and a pole shoe.
   • The pole shoe has a larger cross-sectional area to spread the magnetic flux more uniformly in the air gap and to support the field windings.
3. Field Winding (or Exciting Winding): These are coils wound around the pole cores. When direct current flows through these windings, they produce the main magnetic flux.
4. Armature Core: A cylindrical core mounted on the shaft. It is made of thin, high-grade silicon steel laminations to minimize eddy current losses. The outer surface has slots to house the armature conductors.
5. Armature Winding: A set of insulated conductors housed in the armature slots. This is where the EMF is induced (in a generator) or where the force is produced (in a motor). The windings can be of two types: lap winding or wave winding.
6. Commutator: A cylindrical structure made of wedge-shaped copper segments insulated from each other by thin layers of mica. Its function is to convert the alternating EMF induced in the armature conductors into a unidirectional (DC) EMF at the output terminals. It acts as a mechanical rectifier.
7. Brushes and Brush Gear: Stationary carbon or graphite blocks held against the rotating commutator by springs. Their function is to collect the current from the armature conductors (in a generator) or feed the current to them (in a motor).

OR

4. (b) A 4-pole lap wound DC shunt generator has a useful flux per pole of 0.07 Wb. The armature winding consists of 220 turns, each of 0.004Ω resistance. Calculate the terminal voltage when running at 900 rpm if the armature current is 50A.

Solution:

Given data:

• Number of poles (P) = 4
• Winding type: Lap wound, so Number of parallel paths (A) = P = 4
• Flux per pole (\(\Phi\)) = 0.07 Wb
• Number of turns = 220
• Speed (N) = 900 rpm
• Armature current (\(I_a\)) = 50 A
• Resistance per turn = 0.004 Ω

Step 1: Calculate the total number of armature conductors (Z).

Each turn has two conductors.

\[ Z = \text{Number of turns} \times 2 = 220 \times 2 = 440 \text{ conductors} \]

Step 2: Calculate the generated EMF (\(E_g\)).

The formula for generated EMF in a DC generator is:

\[ E_g = \frac{\Phi Z N P}{60 A} \] For a lap wound machine, P = A, so the formula simplifies to: \[ E_g = \frac{\Phi Z N}{60} = \frac{0.07 \times 440 \times 900}{60} = 0.07 \times 440 \times 15 = 462 \text{ V} \]

Step 3: Calculate the total armature resistance (\(R_a\)).

• Total resistance of all turns if in series = \(220 \text{ turns} \times 0.004 \, \Omega/\text{turn} = 0.88 \, \Omega\).
• Since there are 4 parallel paths in a lap winding, the turns are distributed among these paths.
• Turns per path = \(220 / 4 = 55\) turns.
• Resistance per path = \(55 \text{ turns} \times 0.004 \, \Omega/\text{turn} = 0.22 \, \Omega\).
• The total armature resistance (\(R_a\)) is the equivalent resistance of these 4 paths in parallel: \[ R_a = \frac{\text{Resistance per path}}{\text{Number of paths}} = \frac{0.22}{4} = 0.055 \, \Omega \]

Step 4: Calculate the terminal voltage (V).

For a DC generator, the terminal voltage is the generated EMF minus the armature voltage drop.

\[ V = E_g - I_a R_a \] \[ V = 462 - (50 \times 0.055) = 462 - 2.75 = 459.25 \text{ V} \]

The terminal voltage of the generator is 459.25 V.

5. (a) Explain the open circuit and short circuit tests on a single-phase transformer with neat circuit diagrams.

Solution:

The open circuit (O.C.) and short circuit (S.C.) tests are performed on a transformer to determine its equivalent circuit parameters, efficiency, and voltage regulation without actually loading it.

1. Open Circuit (O.C.) Test

Purpose: To determine the core losses (iron losses) and the no-load circuit parameters (\(R_0\) and \(X_0\)).

Procedure:

• The high voltage (HV) winding is kept open-circuited.
• Measuring instruments (voltmeter, ammeter, wattmeter) are connected to the low voltage (LV) winding.
• Rated voltage at rated frequency is applied to the LV winding.

Observations and Calculations:

• The ammeter reads the no-load current \(I_0\), which is very small (2-5% of full load current).
• The voltmeter reads the rated LV voltage \(V_1\).
• The wattmeter reads the input power \(W_0\), which is almost entirely the core loss (\(P_i\)) since copper losses (\(I_0^2 R_1\)) are negligible. So, \(W_0 = P_i\).
• From the readings, we can find:
  • No-load power factor: \(\cos(\phi_0) = \frac{W_0}{V_1 I_0}\)
  • Core loss component of current: \(I_w = I_0 \cos(\phi_0)\)
  • Magnetizing component of current: \(I_m = \sqrt{I_0^2 - I_w^2}\)
  • Core loss resistance: \(R_0 = \frac{V_1}{I_w}\)
  • Magnetizing reactance: \(X_0 = \frac{V_1}{I_m}\)

2. Short Circuit (S.C.) Test

Purpose: To determine the full-load copper losses and the equivalent series parameters (\(R_{e}\) and \(X_{e}\)).

Procedure:

• The low voltage (LV) winding is short-circuited by a thick conductor.
• Instruments are connected to the high voltage (HV) winding.
• A low voltage (typically 5-10% of rated voltage) is applied to the HV winding and gradually increased until the ammeter reads the rated full-load current.

Observations and Calculations:

• The ammeter reads the full-load current \(I_{sc}\).
• The voltmeter reads the applied voltage \(V_{sc}\).
• The wattmeter reads the input power \(W_{sc}\). Since the applied voltage is very low, the core losses are negligible. Thus, \(W_{sc}\) represents the full-load copper loss (\(P_{cu}\)).
• From the readings, we can find the equivalent parameters referred to the HV side:
  • Equivalent impedance: \(Z_{e1} = \frac{V_{sc}}{I_{sc}}\)
  • Equivalent resistance: \(W_{sc} = I_{sc}^2 R_{e1} \implies R_{e1} = \frac{W_{sc}}{I_{sc}^2}\)
  • Equivalent reactance: \(X_{e1} = \sqrt{Z_{e1}^2 - R_{e1}^2}\)

OR

5. (b) A 30 kVA, 2400/120 V, 50 Hz transformer has a high voltage winding resistance of 0.1Ω and a leakage reactance of 0.22Ω. The low voltage winding resistance is 0.035Ω and the leakage reactance is 0.012Ω. Find the equivalent circuit parameters referred to the i) High voltage side and ii) Low voltage side.

Solution:

Given data:

• High Voltage (HV) side (Primary, side 1): \(V_1 = 2400\) V
  • Resistance \(R_1 = 0.1 \, \Omega\)
  • Reactance \(X_1 = 0.22 \, \Omega\)
• Low Voltage (LV) side (Secondary, side 2): \(V_2 = 120\) V
  • Resistance \(R_2 = 0.035 \, \Omega\)
  • Reactance \(X_2 = 0.012 \, \Omega\)

Step 1: Calculate the transformation ratio (K).

\[ K = \frac{V_2}{V_1} = \frac{120}{2400} = \frac{1}{20} = 0.05 \]

i) Equivalent Circuit Parameters Referred to the High Voltage (HV) Side

To refer the secondary parameters to the primary (HV) side, we divide them by \(K^2\).

• Equivalent Resistance (\(R_{01}\)): \[ R_{01} = R_1 + \frac{R_2}{K^2} = 0.1 + \frac{0.035}{(0.05)^2} = 0.1 + \frac{0.035}{0.0025} = 0.1 + 14 = 14.1 \, \Omega \]
• Equivalent Reactance (\(X_{01}\)): \[ X_{01} = X_1 + \frac{X_2}{K^2} = 0.22 + \frac{0.012}{(0.05)^2} = 0.22 + \frac{0.012}{0.0025} = 0.22 + 4.8 = 5.02 \, \Omega \]
• Equivalent Impedance (\(Z_{01}\)): \[ Z_{01} = \sqrt{R_{01}^2 + X_{01}^2} = \sqrt{(14.1)^2 + (5.02)^2} = \sqrt{198.81 + 25.20} = \sqrt{224.01} \approx 14.97 \, \Omega \]

ii) Equivalent Circuit Parameters Referred to the Low Voltage (LV) Side

To refer the primary parameters to the secondary (LV) side, we multiply them by \(K^2\).

• Equivalent Resistance (\(R_{02}\)): \[ R_{02} = R_2 + R_1 K^2 = 0.035 + 0.1 \times (0.05)^2 = 0.035 + 0.1 \times 0.0025 = 0.035 + 0.00025 = 0.03525 \, \Omega \]
• Equivalent Reactance (\(X_{02}\)): \[ X_{02} = X_2 + X_1 K^2 = 0.012 + 0.22 \times (0.05)^2 = 0.012 + 0.22 \times 0.0025 = 0.012 + 0.00055 = 0.01255 \, \Omega \]
• Equivalent Impedance (\(Z_{02}\)): \[ Z_{02} = \sqrt{R_{02}^2 + X_{02}^2} = \sqrt{(0.03525)^2 + (0.01255)^2} = \sqrt{0.001242 + 0.0001575} = \sqrt{0.001399} \approx 0.0374 \, \Omega \]

6. (a) Explain the principle of operation of a three-phase induction motor.

Solution:

The operation of a three-phase induction motor is based on the principles of electromagnetism and the production of a rotating magnetic field (RMF).

The process can be broken down into the following steps:

1. Production of Rotating Magnetic Field (RMF):
   • The stator of the motor has a three-phase winding distributed in slots.
   • When a three-phase AC supply is connected to this winding, it produces a magnetic field that is constant in magnitude but rotates in space at a constant speed.
   • This speed is called the synchronous speed (\(N_s\)) and is determined by the supply frequency (\(f\)) and the number of stator poles (\(P\)): \( N_s = \frac{120f}{P} \).
2. Induction of EMF in Rotor:
   • The rotor consists of a set of conductors (either squirrel cage bars or wound coils) which are stationary at the start.
   • The rotating magnetic field (RMF) produced by the stator sweeps across the stationary rotor conductors.
   • According to Faraday's Law of Electromagnetic Induction, this relative motion between the flux and the conductors induces an EMF in the rotor conductors.
3. Production of Rotor Current:
   • The rotor conductors are short-circuited either by end rings (in a squirrel cage motor) or through external resistance (in a slip-ring motor).
   • The induced EMF drives a large current through the rotor conductors because the rotor impedance is low.
4. Production of Torque:
   • Now we have current-carrying rotor conductors situated within the stator's magnetic field.
   • These conductors experience a mechanical force (Lorentz force). The direction of this force is given by the motor rule.
   • The sum of the forces on all rotor conductors produces a torque, which acts on the rotor.
5. Rotor Rotation and Slip:
   • This torque causes the rotor to rotate in the same direction as the RMF.
   • The rotor tries to "catch up" with the RMF, but it can never reach synchronous speed. If it did, there would be no relative motion between the RMF and the rotor conductors, no induced EMF, no rotor current, and hence no torque.
   • Therefore, the rotor always runs at a speed (\(N_r\)) slightly less than the synchronous speed (\(N_s\)). The difference between these speeds is called slip (s). \[ s = \frac{N_s - N_r}{N_s} \]
   • This slip is essential for producing the torque required to drive the load.

OR

6. (b) Explain the construction and working principle of a permanent magnet moving coil instrument with a neat diagram.

Solution:

Principle of Operation

The Permanent Magnet Moving Coil (PMMC) instrument works on the principle that a current-carrying conductor placed in a magnetic field experiences a mechanical force, which produces a deflecting torque. This is often called the motor principle. The magnitude of the force (and hence the torque) is directly proportional to the current flowing through the coil.

Construction

The main components of a PMMC instrument are:

1. Permanent Magnet: A strong U-shaped permanent magnet made of materials like Alnico provides a strong and uniform magnetic field.
2. Soft Iron Core: A cylindrical soft iron core is placed between the magnetic poles. It concentrates the magnetic flux, making the field radial and uniform in the air gap where the coil moves.
3. Moving Coil: A lightweight rectangular coil of fine, insulated copper wire is wound on a light aluminum former. This coil is free to rotate in the air gap between the magnet's poles and the iron core.
4. Control System (Springs): Two hairsprings, usually made of phosphor bronze, are attached to the coil spindle. They serve two purposes:
   • They provide the controlling torque (\(T_c\)) which opposes the deflecting torque and brings the pointer to rest at a specific position.
   • They act as leads to carry current into and out of the moving coil.
5. Damping System: Damping is provided by eddy currents induced in the aluminum former as it moves through the magnetic field. This prevents the pointer from oscillating and helps it settle quickly at its final position.
6. Pointer and Scale: A lightweight pointer is attached to the coil spindle and moves over a calibrated, linear scale to indicate the measured quantity.

Working

1. When the instrument is connected in a circuit, the current to be measured (I) flows through the moving coil.
2. Since the coil is in a magnetic field (B), it experiences a deflecting torque (\(T_d\)). This torque is given by: \[ T_d = N \cdot B \cdot A \cdot I \] where N is the number of turns, B is the magnetic flux density, and A is the area of the coil. Since N, B, and A are constant for a given instrument, the deflecting torque is directly proportional to the current: \[ T_d \propto I \]
3. The control springs produce a restoring or controlling torque (\(T_c\)) that is proportional to the angle of deflection (\(\theta\)): \[ T_c \propto \theta \]
4. The pointer comes to rest when the deflecting torque equals the controlling torque (\(T_d = T_c\)). \[ I \propto \theta \]
5. Since the deflection angle (\(\theta\)) is directly proportional to the current (I), the PMMC instrument has a linear (uniformly spaced) scale.
6. Limitation: Because the direction of the torque depends on the direction of the current, PMMC instruments can only be used to measure DC quantities. If AC is applied, the torque would reverse every half-cycle, and the pointer would just vibrate around the zero position.