PSC SOLVED QUESTIONS

diploma queation and answers 3

Determine the generated emf in a lap wound, 4 pole dc generator having useful flux per pole 0.07Wb, 220 armature turns, and runs at 900 rpm

Problem Statement

Determine the generated EMF in a lap-wound, 4-pole DC generator having a useful flux per pole of $0.07\text{ Wb}$, $220$ armature turns, and running at $900\text{ rpm}$.

Given Data

Number of poles ($P$) = $4$
Useful flux per pole ($\Phi$) = $0.07\text{ Wb}$
Number of armature turns = $220$
Speed of armature ($N$) = $900\text{ rpm}$
Type of winding = Lap wound

Formula

The general EMF equation of a DC generator is:

$$E_g = \frac{\Phi \cdot Z \cdot N \cdot P}{60 \cdot A}$$

where:

$\Phi$ = Flux per pole in Webers ($\text{Wb}$)
$Z$ = Total number of armature conductors
$N$ = Speed in revolutions per minute ($\text{rpm}$)
$P$ = Number of poles
$A$ = Number of parallel paths

Step-by-Step Solution

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

Each turn consists of two active conductors (sides):

$$Z = 2 \times \text{Number of Turns} = 2 \times 220 = 440\text{ conductors}$$

Step 2: Determine number of parallel paths ($A$)

For a simplex lap-wound armature, the number of parallel paths equals the number of poles:

$$A = P = 4$$

Step 3: Calculate the generated EMF ($E_g$)

Substitute the given values into the EMF equation:

$$E_g = \frac{0.07 \times 440 \times 900 \times 4}{60 \times 4}$$

Cancelling $P$ and $A$ since $P = A = 4$:

$$E_g = \frac{0.07 \times 440 \times 900}{60}$$

$$E_g = \frac{27,720}{60} = \mathbf{462\text{ V}}$$

Final Answer:
The generated EMF in the lap-wound DC generator is 462 V.

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Summarize the Use of Dummy Coils in a DC Generator

1. What is a Dummy Coil?

A dummy coil (also called an idle coil) is a coil placed in the armature slots of a DC machine that is physically identical to the active armature coils but is never connected electrically to the commutator or the rest of the winding circuit. Both ends of a dummy coil are taped and insulated, leaving it entirely disconnected.

2. Why are Dummy Coils Needed?

Dummy coils are exclusively used in Wave-Wound DC Armatures when standard commercially manufactured armature cores are utilized.

The Mathematical Condition in Wave Winding:

In a simplex wave winding, the commutator pitch ($Y_c$) must strictly be an integer:

$$Y_c = \frac{C \pm 1}{P/2}$$

where:

$C$ = Total number of commutator segments (and coils)
$P$ = Number of poles

Standard mass-produced armature punchings (stampings) come with fixed numbers of slots. In many cases, the total number of slots available on a standard core produces a coil count $C$ that does not yield an integer value for $Y_c$.

To make wave winding feasible without creating custom, expensive punchings:

One coil is left out of the electrical circuit so that the effective number of connected coils satisfies the wave winding equation.
This disconnected coil is placed inside the vacant slot to serve as a dummy coil.

3. Primary Uses and Functions

Mechanical Balance of the Armature Rotor: If a slot were left completely empty, the rotor would suffer from an asymmetrical weight distribution. At high operating speeds, this uneven mass causes severe dynamic unbalance, leading to heavy rotor vibrations, noisy running, and premature bearing wear. Placing a dummy coil ensures perfect dynamic mechanical balance.
Uniform Slot Filling: It keeps all armature slots mechanically full and uniform, preventing the adjacent conductors from shifting or vibrating inside loose slots during operation.
Standardization & Cost Reduction: Manufacturers can use standard, off-the-shelf laminated cores for different voltage/pole specifications without the heavy expense of designing custom core stampings.

4. Summary Comparison

Aspect	Active Armature Coil	Dummy Coil
Electrical Connection	Connected to commutator segments	Isolated; ends taped and disconnected
EMF & Current	Carries load current & contributes to induced EMF	No current flow; zero contribution to generated EMF
Winding Type	Used in both Lap and Wave windings	Used only in Wave winding
Main Purpose	Electromechanical power conversion	Mechanical rotor balancing and slot filling

Exam Summary (4 Marks Key Point):
Dummy coils are electrically isolated coils used exclusively in wave-wound armatures to satisfy the wave winding pitch condition while maintaining complete mechanical dynamic balance of the rotating armature core.

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Derive the emf equation of dc generator

1. Notations Used

Let:

$\Phi$ = Magnetic flux per pole in Webers ($\text{Wb}$)
$P$ = Total number of field poles
$Z$ = Total number of armature conductors
$N$ = Rotational speed of the armature in revolutions per minute ($\text{rpm}$)
$A$ = Number of parallel paths in the armature winding
$E_g$ = Generated EMF across the armature terminals (in Volts)

2. Step-by-Step Derivation

Step 1: Flux cut by one conductor in one full revolution ($d\Phi$)

When an armature conductor completes one full revolution ($360^\circ$), it passes under all $P$ poles. The total magnetic flux cut by that single conductor is:

$$d\Phi = \Phi \times P \quad \text{Webers}$$

Step 2: Time taken to complete one revolution ($dt$)

Since the armature rotates at $N$ revolutions per minute:

$$\text{Revolutions per second} = \frac{N}{60}$$

Therefore, the time taken for one single revolution is:

$$dt = \frac{60}{N} \quad \text{seconds}$$

Step 3: Average EMF induced in a single conductor ($e$)

According to Faraday’s Law of Electromagnetic Induction:

$$e = \frac{d\Phi}{dt} = \frac{\Phi \cdot P}{\left(\frac{60}{N}\right)} = \frac{\Phi \cdot P \cdot N}{60} \quad \text{Volts}$$

Step 4: Number of conductors connected in series per parallel path

The total $Z$ conductors are distributed evenly into $A$ parallel paths. Therefore, the number of conductors connected in series in each path is:

$$\text{Conductors in series per path} = \frac{Z}{A}$$

Step 5: Total Generated EMF ($E_g$)

The terminal EMF of the generator equals the total EMF generated across any one parallel path:

$$E_g = (\text{EMF per conductor}) \times (\text{Number of conductors in series per path})$$

$$E_g = \left( \frac{\Phi \cdot P \cdot N}{60} \right) \times \left( \frac{Z}{A} \right)$$

$$E_g = \frac{\Phi \cdot Z \cdot N \cdot P}{60 \cdot A} \quad \text{Volts}$$

3. Value of Parallel Paths ($A$) for Different Windings

For Simplex Lap Winding: The number of parallel paths equals the number of poles:
$$A = P \implies E_g = \frac{\Phi \cdot Z \cdot N}{60} \quad \text{Volts}$$
(Used in high-current, low-voltage machines)

For Simplex Wave Winding: The number of parallel paths is always $2$, irrespective of the number of poles:
$$A = 2 \implies E_g = \frac{\Phi \cdot Z \cdot N \cdot P}{120} \quad \text{Volts}$$
(Used in high-voltage, low-current machines)

Key Takeaway: For a given manufactured machine, $Z$, $P$, and $A$ are fixed constants. Hence, the generated EMF is directly proportional to the product of flux and speed:

$$E_g \propto \Phi \cdot N$$

shyamlal

diploma queation and answers 2

Explain the electrical and mechanical characteristics of dc shunt motor

Electrical and Mechanical Characteristics of a DC Shunt Motor

In a DC shunt motor, the field winding is connected in parallel (shunt) with the armature winding across the constant supply voltage ($V$). Therefore, the shunt field current remains essentially constant:

$$I_{sh} = \frac{V}{R_{sh}} \approx \text{Constant}$$

Consequently, the magnetic flux per pole ($\Phi$) remains practically constant from no-load to full-load conditions (neglecting minor demagnetizing effects of armature reaction).

Characteristic Curves

1. Electrical Characteristics

The electrical behavior of a DC motor relates internal electrical quantities: armature current ($I_a$), torque ($T_a$), and speed ($N$).

A. Torque vs. Armature Current ($T_a$ vs. $I_a$)

The general torque equation of a DC motor is:
$$T_a \propto \Phi \cdot I_a$$
Since flux ($\Phi$) is constant in a shunt motor:
$$T_a \propto I_a$$
Curve Nature: The characteristic is a straight line passing through the origin. Torque increases linearly with armature current.
Effect of Heavy Load: At very high loads, the demagnetizing effect of armature reaction slightly weakens the main field flux, causing the actual torque curve to bend slightly downward from the theoretical straight line.

B. Speed vs. Armature Current ($N$ vs. $I_a$)

The speed equation of a DC motor is:
$$N \propto \frac{E_b}{\Phi} = \frac{V - I_a R_a}{\Phi}$$
Because $\Phi$ is constant:
$$N \propto (V - I_a R_a)$$
Curve Nature: As the load current ($I_a$) increases from no-load to full-load, the armature resistance drop ($I_a R_a$) increases slightly. Because $R_a$ is very small, this drop is marginal (only about 5% to 8% of rated speed from no-load to full-load).
Hence, the curve is a nearly horizontal line drooping very slightly downwards. For all practical purposes, the DC shunt motor is regarded as a constant-speed motor.

2. Mechanical Characteristic ($N$ vs. $T_a$)

The mechanical characteristic shows the relationship between mechanical output torque ($T_a$) and operating speed ($N$).

Since $T_a \propto I_a$, substituting $I_a \propto T_a$ into the speed equation gives:
$$N = \frac{V}{\Phi} - \left(\frac{R_a}{k \Phi^2}\right) T_a = N_0 - m T_a$$
where $N_0$ is the ideal no-load speed and $m$ is a constant representing the slope.
Curve Nature: The speed-torque characteristic is an almost straight, slightly drooping line. As the mechanical load (torque demand) increases, the speed drops only marginally.
Starting on No-Load: Unlike a DC series motor, a DC shunt motor can safely be started on no-load because its no-load speed is strictly finite and well-defined by $\frac{V - I_{a0} R_a}{k \Phi}$.

3. Summary Table

Characteristic	Mathematical Relation	Nature of Curve
Torque vs. Current ($T_a$ vs. $I_a$)	$T_a \propto I_a$	Straight line through origin (slight droop at severe overloads).
Speed vs. Current ($N$ vs. $I_a$)	$N \propto (V - I_a R_a)$	Nearly horizontal line with a slight downward droop (5–8%).
Speed vs. Torque ($N$ vs. $T_a$)	$N = N_0 - m T_a$	Slightly drooping straight line (mechanical characteristic).

Typical Applications: Lathe machines, centrifugal pumps, fans, conveyors, printing presses, and machine tools where approximately constant speed is required across variable loads.

-----------------------------------------------------------------------------------------------------------------------------Explain series parallel starting of dc traction motor

Series-Parallel Starting of DC Traction Motors

1. Introduction & Objective

In electric traction applications (such as electric trains, trams, and locomotives), two or more identical DC series motors are coupled to different driving axles.

Starting traction motors solely using external series resistors causes enormous $I^2 R$ power losses dissipated as heat. The Series-Parallel Starting Method significantly reduces these energy losses (saving up to 50% of rheostatic energy loss during starting) while providing smooth acceleration and two economical operating running speeds.

2. Circuit Connections & Operating Stages

3. Step-by-Step Working Principle

Step 1: Series Starting Period (0 to 50% Speed)

The two traction motors are connected in series with external starting resistance ($R_{\text{start}}$) across the overhead line voltage ($V$).
Because the motors are in series, the same current flows through both armatures. The supply voltage divides equally across the two machines:
$$V_{\text{motor}} = \frac{V - I_a R_{\text{start}}}{2}$$
As the train gains momentum, back EMF accumulates, and the controller cuts out starting resistance in steps.
Once all resistance is eliminated, each motor operates directly across $\frac{V}{2}$ without any external rheostatic loss. This corresponds to the first economical running speed ($\approx \frac{1}{2}$ rated base speed).

Step 2: Transition Period

To accelerate further, the circuit switches from series to parallel configuration. Three transition methods are used in traction systems:

Open Circuit Transition: Power to one or both motors is disconnected during switching. (Causes sudden tractive effort loss and mechanical jerks; rarely used now).
Shunt Transition: One motor remains powered while the second is momentarily short-circuited through a shunting resistor, disconnected, and reconnected in parallel. (Maintains continuous torque on one axle).
Bridge Transition (Most Preferred): Utilizes a bridge circuit of contactors and resistors. The tractive effort is never interrupted during transition, producing exceptionally smooth acceleration without current surges.

Step 3: Parallel Running Period (50% to 100% Speed)

Both motors are connected in parallel across the full line voltage ($V$), with individual resistor banks inserted into each branch.
The branch resistance is cut out in sequential controller notches until both motors run directly across the full line voltage ($V$).
At this stage, each motor develops rated back EMF and operates at full rated speed ($N$), forming the second economical running speed.

4. Energy Saving Derivation

Consider two identical motors accelerated to full base speed $N$ drawing a constant starting current $I$ during acceleration:

Parameter	Plain Rheostatic Starting	Series-Parallel Starting
Total Energy Drawn from Line	$$E_{\text{in}} = 2 V \cdot I \cdot t$$	$$E_{\text{in}} = \frac{3}{2} V \cdot I \cdot t$$
Useful Energy Converted	$$E_{\text{useful}} = V \cdot I \cdot t$$	$$E_{\text{useful}} = V \cdot I \cdot t$$
Starting Energy Wasted in Resistors	$$E_{\text{loss}} = V \cdot I \cdot t \quad (50\%)$$	$$E_{\text{loss}} = \frac{1}{2} V \cdot I \cdot t \quad (25\%)$$
Rheostatic Loss Reduction	Baseline (0% saved)	50% reduction in rheostatic loss

Summary of Benefits:

Cuts starting energy dissipation in half compared to plain rheostatic starting.
Offers two stable, zero-resistance running speeds ($\frac{N}{2}$ and $N$).
Reduces the physical size and thermal rating needed for starting resistance banks.

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Explain Speed Control of DC Shunt Motor by Field (Flux) Control Method

1. Working Principle

The speed ($N$) of a DC motor is governed by the fundamental speed equation:

$$N = \frac{E_b}{k\Phi} = \frac{V - I_a R_a}{k\Phi}$$

In a DC shunt motor operating on a fixed terminal voltage ($V$), the voltage drop across the armature resistance ($I_a R_a$) is relatively small compared to $V$. Thus, back EMF remains nearly constant ($E_b \approx V$), leading to the relation:

$$N \propto \frac{1}{\Phi}$$

This indicates that the speed of a DC motor is inversely proportional to the magnetic flux per pole ($\Phi$). By weakening the magnetic flux, the motor speed can be increased above its normal rated (base) speed.

2. Circuit Arrangement

A variable external resistor known as a Field Rheostat ($R_{\text{ext}}$) is connected in series with the shunt field winding across the DC supply mains.

3. Step-by-Step Operation

Normal Condition ($R_{\text{ext}} = 0$): When the external field rheostat resistance is zero, maximum rated field current flows through the field winding:
$$I_{sh} = \frac{V}{R_{sh}}$$
At this point, the magnetic flux ($\Phi$) is at its rated maximum, and the motor runs at its **normal rated speed (base speed, $N_b$)**.
Increasing Resistance ($R_{\text{ext}} > 0$): When resistance is added to the field circuit by adjusting the rheostat, the total circuit resistance increases to $(R_{sh} + R_{\text{ext}})$.
Drop in Field Current and Flux: The shunt field current decreases:
$$I_{sh} = \frac{V}{R_{sh} + R_{\text{ext}}} \quad \implies \quad \Phi \text{ decreases}$$
Speed Increase: Because back EMF cannot change instantaneously, the sudden drop in $\Phi$ causes the armature current to surge temporarily, generating excess accelerating torque. The motor accelerates to a higher equilibrium speed where the back EMF recovers to balance the applied voltage. As a result:
$$N > N_b$$

4. Constant Power Drive Characteristic

In this method of speed control:

The armature current ($I_a$) and terminal voltage ($V$) remain at their rated values. Thus, the power rating of the motor remains constant:
$$P = V \times I_a = \text{Constant}$$
Since torque is given by $T \propto \Phi I_a$, and flux $\Phi$ is reduced to gain speed, the developed mechanical **torque decreases inversely as speed increases** ($T \propto \frac{1}{N}$).
Therefore, the flux control method is classified as a Constant Power and Variable Torque Drive.

5. Advantages

High Efficiency: Shunt field current is small (only 2% to 5% of total line current). Hence, the $I_{sh}^2 R_{\text{ext}}$ power loss in the field rheostat is negligible.
Compact & Inexpensive Rheostat: Because it carries only a small current, the rheostat is physically small, light, and low-cost.
Smooth & Stepless Speed Regulation: Speed can be adjusted smoothly across the entire control range above base speed.

6. Limitations

Only Above-Base Speeds Possible: Flux cannot be increased beyond its rated value without saturating the magnetic iron core and overheating the field coils. Therefore, this method cannot achieve speeds below normal base speed.
Poor Commutation at Very High Speeds: At very high speeds (weakened main field), the demagnetizing armature reaction becomes dominant, shifting the magnetic neutral axis significantly and causing severe brush sparking. Interpoles are necessary to enable wider speed ranges (usually up to 2:1 or 3:1).
Mechanical Speed Limits: Centrifugal forces acting on the armature conductors and commutator bars limit the maximum allowable operating speed.

Summary: The flux control method is the most efficient and practical method for obtaining speeds above the base speed in DC shunt motors, operating as a constant power drive.

shyamlal

diploma queation and answers

Efficiency Calculation of DC Shunt Motor by Swinburne’s Test

Question:
A 200-V, 14.92 kW DC shunt motor when tested by the Swinburne method gave the following results:
Running light: armature current was 6.5 A and field current 2.2 A.
With the armature locked, the current was 70 A when a potential difference of 3 V was applied to the brushes.
Estimate the efficiency of the motor when working under full-load conditions.

Given Data:

Supply Voltage ($V$) = $200\text{ V}$
Full-load shaft output ($P_{\text{out}}$) = $14.92\text{ kW} = 14,920\text{ W}$
No-load armature current ($I_{a0}$) = $6.5\text{ A}$
Shunt field current ($I_{sh}$) = $2.2\text{ A}$
Locked rotor voltage across brushes ($V_{br}$) = $3\text{ V}$
Locked rotor armature current ($I_{br}$) = $70\text{ A}$

Step 1: Armature Resistance ($R_a$)

Locked armature conditionil brush voltage-um current-um upayogichu armature resistance kandethunnu:

$$R_a = \frac{V_{br}}{I_{br}} = \frac{3}{70} \approx 0.04286\ \Omega$$

Step 2: Constant Losses ($W_c$)

No-load (running light) conditionil ulla losses:

No-load armature input:
$$P_{a0} = V \times I_{a0} = 200 \times 6.5 = 1300\text{ W}$$
No-load armature copper loss:
$$P_{cu0} = I_{a0}^2 \times R_a = (6.5)^2 \times 0.04286 \approx 1.81\text{ W}$$
Iron, friction and windage loss (Stray losses, $W_m$):
$$W_m = P_{a0} - P_{cu0} = 1300 - 1.81 = 1298.19\text{ W}$$
Shunt field copper loss ($W_{sh}$):
$$W_{sh} = V \times I_{sh} = 200 \times 2.2 = 440\text{ W}$$
Total constant losses ($W_c$):
$$W_c = W_m + W_{sh} = 1298.19 + 440 = 1738.19\text{ W}$$

Step 3: Full-Load Calculations and Efficiency

Method 1: Standard Approximate Method (Common Exam Method)

Full-load input power, output power-inu thulyamaanu ennu karuthiyaal:

$$I_{FL} \approx \frac{P_{\text{out}}}{V} = \frac{14,920}{200} = 74.6\text{ A}$$

Full-load armature current:

$$I_a = I_{FL} - I_{sh} = 74.6 - 2.2 = 72.4\text{ A}$$

Full-load armature copper loss:

$$P_{cu,FL} = I_a^2 \times R_a = (72.4)^2 \times 0.04286 \approx 224.66\text{ W}$$

Total losses at full load:

$$W_{\text{total}} = W_c + P_{cu,FL} = 1738.19 + 224.66 = 1962.85\text{ W}$$

Full-load electrical input power:

$$P_{\text{in}} = P_{\text{out}} + W_{\text{total}} = 14,920 + 1962.85 = 16,882.85\text{ W}$$

Full-load efficiency ($\eta$):

$$\eta = \left( \frac{P_{\text{out}}}{P_{\text{in}}} \right) \times 100 = \left( \frac{14,920}{16,882.85} \right) \times 100 \approx \mathbf{88.37\%}$$

Method 2: Exact Calculation

Armature-il develop cheytha mechanical power:

$$E_b I_a = P_{\text{out}} + W_m$$

$$(V - I_a R_a) I_a = 14,920 + 1298.19 = 16,218.19$$

$$0.04286 I_a^2 - 200 I_a + 16,218.19 = 0$$

Solving for $I_a$:

$$I_a \approx 82.55\text{ A}$$

Full-load armature copper loss:

$$P_{cu,FL} = (82.55)^2 \times 0.04286 \approx 292.09\text{ W}$$

Total losses:

$$W_{\text{total}} = 1738.19 + 292.09 = 2030.28\text{ W}$$

Total power input:

$$P_{\text{in}} = 14,920 + 2030.28 = 16,950.28\text{ W}$$

Full-load efficiency ($\eta$):

$$\eta = \left( \frac{14,920}{16,950.28} \right) \times 100 \approx \mathbf{88.02\%}$$

Answer: The estimated full-load efficiency of the DC shunt motor is approximately 88.37% (or 88.02% by exact method).

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Three-Point Starter for DC Shunt Motor

1. Need for a Starter

At starting, the motor armature is stationary, so the back EMF ($E_b$) is zero:

$$I_a = \frac{V - E_b}{R_a} = \frac{V - 0}{R_a} = \frac{V}{R_a}$$

Since the armature resistance ($R_a$) is very low, a dangerously high starting current flows through the armature. This can blow fuses, damage the commutator, and burn out the windings. A Three-Point Starter inserts variable external resistance into the armature circuit to safely restrict this starting current.

2. Circuit Diagram

3. Three Main Terminals

L (Line Terminal): Connected to the incoming positive supply via the Overload Release (OLR) coil.
A (Armature Terminal): Connected directly to the armature winding of the DC shunt motor.
F (Field Terminal): Connected to the shunt field winding through the No-Volt Coil (NVC).

4. Construction & Working Operation

Starting Position (Stud 1): To start the motor, the handle is gently pulled clockwise to touch Stud 1. At this point:
- The entire starting resistance is connected in series with the armature winding, effectively limiting the inrush starting current.
- The brass arc establishes a circuit directly feeding full supply voltage to the shunt field winding through the NVC, securing maximum flux ($\Phi$) and maximum starting torque.
Gradual Acceleration (Studs 2 to 4): As the motor picks up speed, back EMF ($E_b$) builds up. The starting handle is moved progressively across studs 2, 3, and 4, gradually cutting out the external starting resistance.
Normal Running Position (Stud 5 / RUN): In the final 'RUN' position, all starting resistance is completely cut out of the armature circuit, and the motor runs at rated speed.

5. Protective Devices in 3-Point Starter

No-Volt Coil (NVC) / Under-Voltage Protection: The NVC is an electromagnet connected in series with the shunt field. In the 'RUN' position, it magnetically attracts the soft-iron keeper on the starting handle, holding it firmly against the tension of the spiral return spring. If the power supply fails or voltage drops below a safe limit, the NVC loses its magnetism, and the return spring pulls the handle back to the 'OFF' position.
Overload Release (OLR) / Overload Protection: The OLR is an electromagnet connected in series with the main line terminal ($L$). If the motor draws excessive current due to overloading, the magnetic pull of the OLR lifts its movable iron plunger. This bridges the tripping contacts, short-circuiting the NVC coil. As a result, the NVC demagnetizes immediately, releasing the handle back to the 'OFF' position and cutting off the motor.

Limitation of 3-Point Starter: When controlling the speed above rated speed by weakening the shunt field (field rheostat control), the field current decreases. If it becomes too weak, the NVC may accidentally drop the handle back to the OFF position. (This drawback is rectified using a Four-Point Starter).

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Define torque in dc motor and compare armature torque and shaft
torque

Torque in DC Motor: Armature Torque vs. Shaft Torque

1. Definition of Torque

Torque is defined as the turning or twisting moment of a force about an axis of rotation. In a DC motor, it is the rotational force produced on the armature conductors when current-carrying conductors interact with the magnetic field of the stator poles.

Mathematically, torque ($T$) is expressed as:

$$T = F \times r \quad (\text{N}\cdot\text{m})$$

where $F$ is the tangential force exerted on the armature conductors (in Newtons), and $r$ is the radius of the armature core (in meters).

2. Types of Torque in a DC Motor

A. Armature Torque ($T_a$)

Armature Torque (Gross Torque): The total gross electromagnetic torque developed inside the armature winding due to electromechanical energy conversion.

The standard expression for armature torque is:

$$T_a = \frac{1}{2\pi} \left(\frac{P Z}{A}\right) \Phi I_a = 0.159 \times \Phi I_a \times \left(\frac{P Z}{A}\right) \quad (\text{N}\cdot\text{m})$$

Also, in terms of mechanical power developed:

$$E_b I_a = \omega T_a = \left(\frac{2\pi N}{60}\right) T_a \implies T_a = \frac{E_b I_a}{\left(\frac{2\pi N}{60}\right)} = 9.55 \times \frac{E_b I_a}{N} \quad (\text{N}\cdot\text{m})$$

B. Shaft Torque ($T_{sh}$)

Shaft Torque (Net Output Torque): The useful net torque available at the motor shaft for driving the mechanical load. It is always less than the armature torque because a part of the gross torque is consumed to overcome internal iron losses (hysteresis and eddy current) and mechanical losses (friction and windage).

$$T_{sh} = T_a - T_f$$

where $T_f$ is the lost torque due to mechanical and core friction losses.

In terms of useful output power ($P_{\text{out}}$):

$$P_{\text{out}} = \omega T_{sh} = \left(\frac{2\pi N}{60}\right) T_{sh} \implies T_{sh} = \frac{P_{\text{out}}}{\left(\frac{2\pi N}{60}\right)} = 9.55 \times \frac{P_{\text{out}}}{N} \quad (\text{N}\cdot\text{m})$$

3. Comparison Between Armature Torque and Shaft Torque

Feature	Armature Torque ($T_a$)	Shaft Torque ($T_{sh}$)
Definition	Gross torque developed internally in the armature.	Net useful torque delivered at the output shaft.
Magnitude	Higher ($T_a > T_{sh}$).	Lower due to rotational losses.
Power Association	Corresponds to gross electrical power converted ($E_b I_a$).	Corresponds to net mechanical shaft output power ($P_{\text{out}}$).
Loss Consideration	Does not account for iron, friction, and windage losses.	Derived after subtracting mechanical & iron losses ($T_a - T_f$).
Formula	$$T_a = 9.55 \times \frac{E_b I_a}{N}$$	$$T_{sh} = 9.55 \times \frac{P_{\text{out}}}{N}$$
Practical Significance	Indicates electromagnetic performance and torque generation capability.	Determines the actual load-driving capacity of the motor.

Relationship:
$$\text{Lost Torque } (T_f) = T_a - T_{sh} = 9.55 \times \frac{\text{Mechanical \& Iron Losses}}{N}$$

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Explain Three point starter with the help of a figure.

Three-Point Starter for DC Shunt Motor

1. Need for a Starter

When a DC motor is at rest, the armature is stationary, meaning the back EMF ($E_b$) is zero. The armature current is governed by Ohm's law:

$$I_a = \frac{V - E_b}{R_a} = \frac{V - 0}{R_a} = \frac{V}{R_a}$$

Because the armature winding resistance ($R_a$) is extremely small, connecting the motor directly across the full supply voltage would cause an enormous inrush current (typically 10 to 20 times the rated full-load current). This excessive current can:

Blow out fuses or trip line breakers.
Damage the commutator surface and produce severe brush sparking.
Burn out the armature winding insulation due to excessive $I^2 R$ heating.

To prevent this, a Three-Point Starter inserts an external variable starting resistance into the armature circuit at standstill and gradually cuts it out as the motor gains speed and establishes back EMF.

2. Schematic Diagram

3. Main Terminals

L (Line Terminal): Connected to the positive DC power supply line through the Overload Release (OLR) coil.
A (Armature Terminal): Connected directly to the motor's armature terminal.
F (Field Terminal): Connected to the motor's shunt field winding in series with the No-Volt Coil (NVC).

4. Construction and Operation

Starting Step (Stud 1): The operator pulls the handle clockwise from the 'OFF' position until it touches Stud 1. At this point:
- The full bank of starting resistance is placed directly in series with the armature winding, suppressing the heavy starting inrush current.
- The handle simultaneously makes contact with the continuous brass arc, which feeds the full line voltage directly to the shunt field winding via the NVC coil. This establishes maximum field flux ($\Phi$) right away, producing high starting torque.
Intermediate Steps (Studs 2 to 4): As the rotor begins turning, it produces a proportional back EMF ($E_b$). This opposing voltage naturally curbs the armature current. The operator advances the lever smoothly across studs 2, 3, and 4, cutting out sections of the external starting resistance in steps.
Normal Running Position (Stud 5 / 'RUN'): Once the handle reaches Stud 5:
- All external starting resistance is completely removed from the armature circuit, allowing the motor to run at full rated speed.
- The soft-iron piece attached to the handle is held firmly in place by the magnetic pull of the energised No-Volt Coil (NVC), working against the tension of the spiral return spring.

5. Built-in Protective Mechanisms

No-Volt Release (NVC) — Low/Zero Voltage Protection: If the incoming power line fails or line voltage drops below a minimum threshold, the current through the shunt circuit drops. The NVC demagnetizes and loses its grip on the handle. The spiral spring promptly snaps the handle back to the 'OFF' position, preventing sudden, uncontrolled restarts when supply voltage is restored.
Overload Release (OLR) — Over-Current Protection: If the motor experiences an excessive mechanical overload, the resulting heavy line current flows through the series OLR coil. This creates a strong electromagnetic field that pulls up the movable iron plunger. The plunger bridges two fixed tripping contacts beneath it, short-circuiting the NVC coil. Deprived of current, the NVC loses its magnetic hold, allowing the return spring to snap the handle back to 'OFF' and disconnect the motor.

Limitation: When field weakening is used for speed control above base speed, the field current drops. If reduced too far, the magnetic hold of the NVC weakens and can cause the spring to pull the starter arm to 'OFF' during normal running. This drawback is resolved by using a Four-Point Starter, where the NVC circuit is wired independently across the supply.

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Q:What is a Synchronous Generator?

Answer:The synchronous generator is a synchronous machine which converts mechanical power into AC electric power through the process of electromagnetic induction.

Synchronous generators are also referred to as alternators or AC generators. The term "alternator" is used since it produces AC power. It is called synchronous generator because it must be driven at synchronous speed to produce AC power of the desired frequency.

A synchronous generator can be either single-phase or poly-phase generally 3phase.

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Q:Explain the working of AC series motor.

answer:

The AC Series motor is a modified version of the DC series motor adapted to run on alternating current (AC).motor or universal motor works by having its field and armature windings connected in series, ensuring that both currents reverse at the same time when supplied with AC. This simultaneous reversal of the magnetic flux and armature current ensures the torque remains in the same direction, producing continuous rotation despite the alternating supply. The high starting torque and ability to run on both AC and DC power make it suitable for various appliances like drills and vacuum cleaners

Q:Compare salient pole rotor and cylindrical pole rotor.

Answer:

Salient pole rotors have projecting poles suitable for low-speed machines (100-1500 RPM) like hydro-generators, featuring a large diameter, small axial length, many poles, and a non-uniform air gap. In contrast, cylindrical rotors have a smooth, non-projecting cylindrical shape for high-speed (1500-3000 RPM) applications like turbo-generators, characterized by a smaller diameter, longer axial length, few poles (2 or 4), and a uniform air gap

Q:What is a Salient Pole Rotor Synchronous Generator?

Answer:

When the synchronous generator uses a projected pole type rotor, it is known as salient pole rotor synchronous generator or salient pole alternator.

In a salient pole alternator, the rotor poles are made of steel laminations and are fixed to the rotor hub. This type of rotor has rotor poles that are physically separated. Each pole carries a concentrated excitation winding. The salient pole rotor is usually used in alternators having 4 poles or more.

The salient pole alternators are mainly used in the applications where the speed of the prime mover is less because at high speeds the centrifugal forces will be large and may damage the poles of the rotor.

Q:What is an Alternator?

Answer:An electromechanical energy conversion device which converts mechanical energy into AC (alternating current) electrical energy is called an alternator.

An alternator consists of two main parts namely stator and rotor. The stator carries the armature winding whereas the rotor carriers the magnetic field winding or poles. When the rotor rotates, its magnetic field cuts the armature conductors and as a result of it an emf is induced in the armature winding. Since the magnetic poles (N and S) of the rotor alternatively cutting the armature windings, hence they induce an alternating (changing direction alternatively) emf in the armature. In this way, the alternator produces an alternating electricity.

The alternators are used in power generating station and automobiles, etc.

Q:What is a Generator?

Answer:The electrical machine which converts mechanical energy into electrical energy is known as generator or electric generator. The electric generator works on the principle of electromagnetic induction, i.e. when a conductor is moved in a magnetic field, an EMF is induced in it. The electric generators produce electricity from many different sources of mechanical energy such as internal combustion engines, steam turbines, gas turbines, water turbines, etc.

The electric generator is one of the most useful electrical machine used for generating electricity during a power shutdown. However, its maintenance is a big problem because it uses coal, oils, natural gases, etc. as the source of power. But, the generator has one main advantage that it can be operated for longer period of time.

Q:What is an Inverter?

Answer:Inverter is a power electronic device which converts direct current (DC) stored in the batteries into alternating current (AC) is known inverter. The inverter have an extra electronic circuit for controlling the battery charging and load management to enable the use of standard electrical appliances.

Basically, an inverter acts like a power adaptor to power the low power domestic and commercial electrical appliances through the conventional electric wiring based on a battery powered system. The inverters are used to run most of the modern domestic and commercial electrical appliances like lamps, fans, refrigerator, water purifier and other low power devices. The inverters can be operated in stand-alone mode as well as in connected mode to the main power grid.

Q:What is a UPS?

Answer:UPS stands for Uninterrupted Power Supply. As its name suggests, it is a device used to stop the interruption in the electric power supply caused to electrical devices during the cut out of electricity.In actual practice, the UPS is mainly used with the computer and other IT systems to provide the electric power for the sufficient amount of time to save the data and safely shutdown the computer when sudden power cut occurs.The main parts of a typical UPS system are: rectifier, battery, inverter and controller. The rectifier converts the AC supply in DC supply to charge the battery. The battery is connected to the inverter which converts the DC output of battery into AC and supply to the connected device or system. The controller is provided to control the operation of the entire system.A UPS provides a backup of up to 10 to 15 minutes. Therefore, the UPS is mainly used to provide backup power to the electronic devices and IT systems that may get damaged with the sudden power failures.

Q:What is Potential Difference?

Answer:In an electric circuit, the difference in the potential of two points is known as potential difference (P.D.). The potential difference is also known as voltage. The SI unit potential difference is Volt.

In simple terms, the potential difference can be defined as the arithmetical difference of a higher potential and a lower potential in an electric circuit. Basically, the potential difference is the amount of energy required to move a unit charge from one point to another point in an electric circuit. The potential difference between two points in an electric circuit can be established with the help of a source of EMF like cell, battery, etc. Just like the voltage drop, we can use a voltmeter to measure the potential difference in the circuit.

Mathematically, if we require W joules of energy (or work done) to move an electric charge of Q coulombs from one point to another in an electric circuit, then the potential difference between those two points is given by,

$$V= \frac{W}{Q}$$

Potential Difference is the parameter in any electric circuit which is entirely responsible for the flow of current in the circuit. Thus, if there is no potential difference, then no current flow.

Q:What is a DC Generator?

Answer:electric generator which converts input mechanical energy into DC electrical energy is called the DC generator or direct current generator. DC generator is also known as dynamo.

A DC generator consists of a rotating armature and static magnetic field. When the rotating armature moves in the stationary magnetic field, an alternating current is induced in the armature winding and it is converted into direct current by using a commutator (a mechanical rectifier) and supplied to the external circuit.

Q:What is an AC Generator?

Answer:The type of electric generator which converts mechanical energy input into AC electrical energy output is known as AC generator. It is also known as alternator, as it produces alternating current electricity. In practice, the AC generator is a type of generator designed to generate alternating current with a frequency of 50 Hz or 60 Hz.

The AC generator consists of fixed armature and a rotating magnetic field. When rotating magnetic field links with the stationary armature winding, produces alternating current in the winding by the principle of electromagnetic induction.

Q:What is AC Electricity?

Answer:The alternating current electricity is a form of electrical current flow that alternates between positive and negative at regular intervals. This indicates that the voltage and current alternate in direction on a regular basis, resulting in a sinusoidal waveform. The type of current delivered by electrical power grids and used to power most household appliances and devices is alternating current

AC electricity has a frequency of 50 or 60 hertz, which means the current changes direction 50 or 60 times per second. The voltage of alternating current (AC) energy varies based on the electrical power grid; however, it is commonly 120 or 240 volts in residential areas.

One advantage of alternating current power is that it can be easily converted to different voltage levels using transformers. This allows for efficient long-distance transmission of power with minimum power loss. AC electricity is also more efficient for powering some types of motors, such as induction motors found in appliances and industrial equipment

Q:What is DC Electricity?

Answer:The DC electricity (direct current) is a form of electrical current flow that flows in only one direction, from positive to negative. Unlike AC power, which flips direction on a regular basis, DC current flows in a continuous direction. Batteries, electronic gadgets, and some motors all use direct current.

DC energy voltage varies based on the application and the equipment being powered, but it is commonly between 1.5 and 12 volts for household batteries and up to several hundred volts for applications in industry. The current in a direct current circuit can also fluctuate based on the circuit's resistance and the load being powered.

One of the primary advantages of DC electricity is that it is more efficient for powering certain types of motors, such as brushed DC motors. DC motors are widely found in small electronic gadgets and appliances such as fans, drills, and power tools. Another advantage is that DC power transmission often results in less power loss over distance than AC power transfer.

However, DC electricity has a few disadvantages. For example, DC cannot be easily changed to multiple voltage levels using transformers, making long-distance power transmission more complex. In addition, at lower voltages, DC power can be more harmful to people, causing burns and tissue damage.

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PSC SOLVED QUESTIONS

diploma queation and answers 3

Determine the generated emf in a lap wound, 4 pole dc generator having useful flux per pole 0.07Wb, 220 armature turns, and runs at 900 rpm

Problem Statement

Given Data

Formula

Step-by-Step Solution

Summarize the Use of Dummy Coils in a DC Generator

1. What is a Dummy Coil?

2. Why are Dummy Coils Needed?

The Mathematical Condition in Wave Winding:

3. Primary Uses and Functions

4. Summary Comparison

Derive the emf equation of dc generator

1. Notations Used

2. Step-by-Step Derivation

3. Value of Parallel Paths (\(A\)) for Different Windings

diploma queation and answers 2

Electrical and Mechanical Characteristics of a DC Shunt Motor

Characteristic Curves

1. Electrical Characteristics

A. Torque vs. Armature Current (\(T_a\) vs. \(I_a\))

B. Speed vs. Armature Current (\(N\) vs. \(I_a\))

2. Mechanical Characteristic (\(N\) vs. \(T_a\))

3. Summary Table

Series-Parallel Starting of DC Traction Motors

1. Introduction & Objective

2. Circuit Connections & Operating Stages

3. Step-by-Step Working Principle

Step 1: Series Starting Period (0 to 50% Speed)

Step 2: Transition Period

Step 3: Parallel Running Period (50% to 100% Speed)

4. Energy Saving Derivation

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Explain Speed Control of DC Shunt Motor by Field (Flux) Control Method

1. Working Principle

2. Circuit Arrangement

3. Step-by-Step Operation

4. Constant Power Drive Characteristic

5. Advantages

6. Limitations

diploma queation and answers

Efficiency Calculation of DC Shunt Motor by Swinburne’s Test

Given Data:

Step 1: Armature Resistance (\(R_a\))

Step 2: Constant Losses (\(W_c\))

Step 3: Full-Load Calculations and Efficiency

Method 1: Standard Approximate Method (Common Exam Method)

Method 2: Exact Calculation

Three-Point Starter for DC Shunt Motor

1. Need for a Starter

2. Circuit Diagram

3. Three Main Terminals

4. Construction & Working Operation

5. Protective Devices in 3-Point Starter

Define torque in dc motor and compare armature torque and shafttorque

Torque in DC Motor: Armature Torque vs. Shaft Torque

1. Definition of Torque

2. Types of Torque in a DC Motor

A. Armature Torque (\(T_a\))

B. Shaft Torque (\(T_{sh}\))

3. Comparison Between Armature Torque and Shaft Torque

Three-Point Starter for DC Shunt Motor

1. Need for a Starter

2. Schematic Diagram

3. Main Terminals

4. Construction and Operation

5. Built-in Protective Mechanisms

qustion bank

Q:Explain the working of AC series motor.

answer:

Q:What is an Alternator?

Q:What is a Generator?

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