← MES Jr. Grade
Past Questions
The air-gap between the yoke and armature in a dc motor is kept small :
A)
to achieve stronger magnetic field
B)
to avoid overheating of the machine
C)
to avoid locking of the armature
D)
to avoid transverse motion
In a DC motor, the air gap is the space between the stationary field poles (attached to the yoke) and the rotating armature. The primary reason for keeping this gap small is based on the concept of magnetic reluctance. Reluctance is the opposition a material offers to the flow of magnetic flux, and it is directly proportional to the length of the path and inversely proportional to the permeability of the material. Since air has a very high reluctance compared to the iron/steel used in the yoke and poles, a larger air gap would significantly increase the total reluctance of the magnetic circuit. By minimizing the air gap, the reluctance is reduced, allowing more magnetic flux to cross from the poles to the armature for a given amount of magnetizing current. A higher flux density results in a stronger magnetic field, which increases the torque production capability of the motor (since Torque ∝ Φ).
Regarding other options:
- (b) Overheating is primarily managed by cooling systems and insulation, not the air gap size.
- (c) Locking of the armature is prevented by precise mechanical bearings and centering, not by the gap size (in fact, a gap that is too small could risk mechanical rubbing).
- (d) Transverse motion is controlled by the mechanical assembly and bearings of the shaft.
Which one of the following relays has the capability of anticipating the possible major fault in a transformer?
A)
Overcurrent relay
B)
Differential relay
C)
Buchholz relay
D)
Overfluxing relay
The Buchholz relay is a gas-actuated relay specifically designed for oil-immersed transformers. It is installed in the pipe connecting the main transformer tank to the conservator tank. Its primary function is to detect incipient (developing) faults—such as insulation breakdown, core heating, or minor sparking—which produce small amounts of gas. These gases accumulate in the relay, triggering an alarm to warn operators of a potential major fault before it becomes catastrophic. If a severe fault occurs (like a short circuit), the resulting surge of oil triggers a trip signal to disconnect the transformer immediately.
Why other options are incorrect:
- a. Overcurrent relay: This is a protective relay that trips when the current exceeds a predefined limit; it reacts to the fault itself, not the 'anticipation' of one.
- b. Differential relay: This is used for internal fault protection by comparing currents entering and leaving the transformer; it operates once a fault has already occurred.
- d. Overfluxing relay: This protects the transformer from saturation caused by overvoltage or under-frequency conditions, which is a specific operating condition rather than a general anticipation of internal faults.
The signal flow graph of a feedback control system is shown below. Determine its closed loop transfer function.
A 220V d.c shunt motor runs at 500 r.p.m when the armature current is 50A. Calculate the speed if the torque is double. Given that Ra = 0.2Ω.
Fermi energy is the amount of energy which :
A)
a valence electron can have at room temperature
B)
must be given to an electron to move it to conduction band
C)
must be given to a hole to move it to valence band
D)
a hole can have at room temperature
The Fermi energy level (Ef) represents the highest energy level that an electron can occupy at absolute zero temperature (0 K). In the context of semiconductors, the Fermi level is a theoretical energy level that defines the probability of occupancy. For an electron in the valence band to contribute to electrical conduction, it must gain enough energy to cross the energy gap (band gap) and enter the conduction band. While the Fermi level is a statistical measure, in many simplified exam contexts, the energy required to move an electron from the valence band (or the Fermi level in intrinsic semiconductors) to the conduction band is associated with the concept of Fermi energy/activation energy.
Analyzing the options:
- Option (a) is incorrect because valence electrons have a range of energies, not a single fixed 'Fermi energy'.
- Option (b) is the most appropriate answer in this context, as it describes the energy transition required for conduction.
- Option (c) is incorrect because holes are already in the valence band; they do not move 'to' the valence band.
- Option (d) is incorrect because holes are the absence of electrons and their energy state is defined relative to the valence band edge, not by the Fermi energy itself.
Magnetisation is nonlinearly related to the applied field in case of :
A)
diamagnetic material
B)
paramagnetic material
C)
ferromagnetic material
D)
diamagnetic & paramagnetic material
The relationship between Magnetization (M) and the applied Magnetic Field Intensity (H) is defined by the equation M = χH, where χ is the magnetic susceptibility.
1. Diamagnetic Materials: These have a small, negative, and constant susceptibility. The relationship between M and H is linear (M ∝ H) and negative.
2. Paramagnetic Materials: These have a small, positive, and constant susceptibility (at a constant temperature). The relationship between M and H is linear (M ∝ H).
3. Ferromagnetic Materials: These materials exhibit a complex relationship. Due to the presence of magnetic domains, the magnetization increases rapidly at first and then levels off as the domains become fully aligned (saturation). This relationship is represented by the 'Hysteresis Loop' (B-H curve), which is inherently non-linear. The susceptibility χ is not constant and depends on the strength of the applied field.
Therefore, only ferromagnetic materials show a non-linear relationship between magnetization and the applied field.
Ceramic loses its insulating properties above a temperature called :
A)
Melting point
B)
Solidification point
C)
Curie point
D)
Flash point
The Curie point (or Curie temperature) is the critical temperature above which certain materials lose their permanent magnetic properties. In the context of ceramic insulating materials (specifically ferroelectric ceramics like Barium Titanate), the Curie point is the temperature at which the material undergoes a phase transition from a ferroelectric state to a paraelectric state. At this point, the dielectric constant changes significantly, and the material's specific insulating/dielectric properties are altered.
Why other options are incorrect:
- Melting point: This is the temperature at which a solid becomes a liquid; while it would lose insulating properties, it is a physical state change, not a specific property threshold for insulation.
- Solidification point: This is the temperature at which a liquid becomes a solid (freezing point).
- Flash point: This is the lowest temperature at which a liquid gives off enough vapor to ignite in air, which is relevant to flammable liquids, not ceramic insulators.
In two wattmeter method of measuring 3-phase power, power factor is 0.5, then one of the wattmeter will read :
A)
W
B)
Zero
C)
2 W
D)
W/2
In the two-wattmeter method for measuring three-phase power, the readings of the two wattmeters (W1 and W2) are given by the formulas:
W1 = V_L I_L cos(30° + φ)
W2 = V_L I_L cos(30° - φ)
Where:
- V_L is the line voltage
- I_L is the line current
- φ is the phase angle (the angle by which the current lags the voltage).
The power factor (pf) is given as 0.5.
Therefore, cos(φ) = 0.5, which means φ = 60°.
Now, let's substitute φ = 60° into the wattmeter equations:
For W1:
W1 = V_L I_L cos(30° + 60°) = V_L I_L cos(90°)
Since cos(90°) = 0, W1 = 0.
For W2:
W2 = V_L I_L cos(30° - 60°) = V_L I_L cos(-30°)
Since cos(-30°) = cos(30°) = √3/2, W2 = V_L I_L (√3/2).
Since one of the wattmeters (W1) reads zero when the power factor is 0.5, the correct option is 'Zero'.
For the circuit shown below, find the complete expression for the current when the switch is closed at t=0.
Which one of the following transducers is an active transducer?
A)
Piezo-electric pressure transducer
B)
Metallic strain gauge
C)
Semi-conductor strain gauge
D)
Platinum resistance thermometer
To solve this question, we must distinguish between Active and Passive transducers:
1. **Active Transducers (Self-generating):** These transducers generate their own electrical output (voltage or current) from the energy of the physical quantity being measured. They do not require an external power source to operate.
- A **Piezo-electric transducer** works on the piezoelectric effect, where applying mechanical pressure to certain crystals (like quartz) generates an electric charge. Therefore, it is an active transducer.
2. **Passive Transducers (Externally powered):** These transducers do not generate electricity. Instead, they change a passive electrical property (like resistance, capacitance, or inductance) in response to a physical stimulus. They require an external power source (excitation voltage) to produce a measurable output signal.
- **Metallic strain gauge (Option b):** Changes its resistance when stretched. It requires an external bridge circuit/power source. (Passive)
- **Semi-conductor strain gauge (Option c):** Similar to metallic gauges, it changes resistance based on strain. (Passive)
- **Platinum resistance thermometer (Option d):** Also known as a RTD, it changes resistance based on temperature. It requires an external current source to measure the voltage drop. (Passive)
Conclusion: Only the Piezo-electric pressure transducer is self-generating, making it the active transducer.
Define electric field intensity. Derive the equation for electric field intensity at a distance ‘r’ from a point charge of Q coulombs.
Pick up false statement regarding measurement accuracy:
A)
The degree of closeness with which a given value can be repeatedly measured is called reproducibility
B)
When for a given input the measured value does not vary with time, it is said to have zero drift
C)
Any signal conveying no useful information is called noise
D)
The nature of repeatability is deterministic
To identify the false statement, let's analyze each option based on measurement science (metrology):
- **Option (a):** Reproducibility refers to the closeness of agreement between measurements of the same measurand when carried out under changed conditions (different observers, different instruments, or different locations). While the wording in the option is slightly simplified, it generally aligns with the concept of consistency across varied trials.
- **Option (b):** Drift is the gradual change in the output of a measuring instrument over time while the input remains constant. Therefore, if the measured value does not vary with time for a given input, the instrument has zero drift. This is a true statement.
- **Option (c):** In measurement and signal processing, noise is defined as any unwanted signal or interference that obscures the useful information of the primary signal. This is a true statement.
- **Option (d):** Repeatability is the closeness of agreement between independent test results obtained with the same method on identical test items in the same laboratory. The variations in repeatability are caused by random errors (stochastic processes), not deterministic ones. Deterministic errors are systematic and predictable (like offset or scale factor errors), whereas repeatability deals with the random dispersion of measurements. Therefore, stating that the nature of repeatability is 'deterministic' is false.
Since the question asks for the false statement, option (d) is the correct choice.
Two semi-infinite dielectric regions are separated by a plane boundary at y = 0. The dielectric constants of region 1 (y < 0) and region 2 (y > 0) are 2 and 5, respectively. Region 1 has uniform electric field E = 3ax + 4ay + 2az. The electric field in region 2 is :
A)
3ax + 1.6ay + 2az
B)
1.2ax + 4ay + 2az
C)
1.2ax + 4ay + 0.8az
D)
3ax + 10ay + 0.8az
To find the electric field in region 2, we apply the boundary conditions for electromagnetic fields at the interface between two dielectrics.
1. **Identify the Boundary and Normal Vector**: The boundary is the plane y = 0. The normal vector to this boundary is in the y-direction (ay).
2. **Tangential Component Condition**: The tangential component of the electric field (E_t) is continuous across the boundary.
E1 = 3ax + 4ay + 2az
The tangential components are those in the x and z directions: E1t = 3ax + 2az.
Therefore, E2t = E1t = 3ax + 2az.
3. **Normal Component Condition**: Assuming there is no free surface charge at the interface, the normal component of the electric flux density (D_n) is continuous across the boundary.
D1n = ε1 * E1n = ε0 * εr1 * E1y
D2n = ε2 * E2n = ε0 * εr2 * E2y
Setting D1n = D2n:
εr1 * E1y = εr2 * E2y
Given εr1 = 2, εr2 = 5, and E1y = 4:
2 * 4 = 5 * E2y
8 = 5 * E2y
E2y = 8 / 5 = 1.6
4. **Combine the Components**:
E2 = E2t + E2n
E2 = (3ax + 2az) + 1.6ay
E2 = 3ax + 1.6ay + 2az
Comparing this result to the options:
Option (a) matches exactly. Options (b), (c), and (d) incorrectly modify the tangential components or the normal component calculation.
A moving iron ammeter produces a full scale torque of 240 µNm with a deflection of 120° at a current of 10 A. The rate of change of self inductance (µH/radian) of the instrument at full scale is :
A)
2.0 µH/radian
B)
4.8 µH/radian
C)
12.0 µH/radian
D)
114.6µH/radian
For a moving iron instrument, the deflecting torque is
T = \tfrac{1}{2} I^2 \frac{dL}{d\theta}.
Given full-scale torque T = 240 \u00b5N m = 240 \times 10^{-6} N m and current I = 10 A,
\frac{dL}{d\theta} = \frac{2T}{I^2} = \frac{2 \times 240 \times 10^{-6}}{10^2} = 4.8 \times 10^{-6} H/radian = 4.8 \u00b5H/radian.
So the correct option is b.
Why the others are incorrect:
- a (2.0 \u00b5H/radian): too small; it would give only 100 \u00b5N m torque at 10 A.
- c (12.0 \u00b5H/radian): too large; it would imply 600 \u00b5N m torque, not 240 \u00b5N m.
- d (114.6 \u00b5H/radian): far too large and not consistent with the torque-current relation.
Note: The given 120\u00b0 deflection is not needed for this calculation because the torque and current are already specified at full scale.
In the circuit shown below, when V =0, I=3A. When V =12V, the value of I becomes:
A)
3A
B)
6A
C)
7A
D)
9A
The circuit is a linear network, so the current varies linearly with the applied voltage. Given that when V = 0, the current is I = 3 A, this means the circuit has a 3 A offset current. When V is increased to 12 V, the current increases by another 3 A to become 6 A. Hence, the correct answer is 6 A.
Why the others are incorrect:
- a) 3 A: This is only the current at V = 0, not at V = 12 V.
- c) 7 A: This does not match the linear change implied by the given condition.
- d) 9 A: This would require a larger increase than what the circuit relation indicates.
Magnetic susceptibility of an ideal type I superconductor in the superconductivity state is :
A)
-∞
B)
-1
C)
Between 0 and -1
D)
Zero
1. **The Meissner Effect**: An ideal Type I superconductor exhibits the Meissner effect, which is the complete expulsion of magnetic flux from its interior when it is cooled below its critical temperature (Tc).
2. **Magnetic Induction Equation**: The magnetic induction inside a material is given by the relation: B = μ₀(H + M), where B is the magnetic flux density, H is the external magnetic field intensity, and M is the magnetization of the material.
3. **Condition for Superconductivity**: For an ideal superconductor, the internal magnetic field B must be zero (B = 0). Therefore, 0 = μ₀(H + M), which simplifies to M = -H.
4. **Magnetic Susceptibility (χ)**: Magnetic susceptibility is defined as the ratio of magnetization (M) to the magnetic field intensity (H): χ = M / H.
5. **Calculation**: Substituting M = -H into the formula, we get χ = -H / H = -1.
6. **Conclusion**: A susceptibility of -1 indicates a 'perfect diamagnet'.
Comparing with options:
- Option (a) -∞ is incorrect; this does not represent physical susceptibility.
- Option (b) -1 is correct as it represents perfect diamagnetism.
- Option (c) Between 0 and -1 describes ordinary diamagnets, not ideal superconductors.
- Option (d) Zero describes a non-magnetic material.
If a capacitor is charged by a square wave current source, the voltage across the capacitor is :
A)
a square wave
B)
triangular wave
C)
step function
D)
zero
The relationship between the current (i) flowing through a capacitor and the voltage (v) across it is given by the fundamental equation: i(t) = C * (dv/dt). Rearranging this to find the voltage, we get: v(t) = (1/C) * ∫ i(t) dt.
1. A square wave current source provides a constant current (+I) for one half-cycle and a constant current (-I) for the next half-cycle.
2. The integral of a constant is a linear function (a ramp).
3. During the first half-cycle (positive constant current), the voltage increases linearly: v(t) = (I/C)t.
4. During the second half-cycle (negative constant current), the voltage decreases linearly: v(t) = v_peak - (I/C)t.
5. This sequence of linear increases and decreases results in a triangular waveform.
Comparing with other options:
- (a) A square wave voltage would require an impulsive (delta function) current source.
- (c) A step function voltage would require a single pulse of current.
- (d) Zero voltage would require zero current.
Inductance is measured by :
A)
Wien bridge
B)
Schering bridge
C)
Maxwell’s bridge
D)
Hay bridge
To determine the correct bridge for measuring inductance, we analyze the purpose of each bridge listed in the options:
1. **Maxwell's Bridge (Maxwell-Wien Bridge):** This is specifically designed to measure an unknown inductance by comparing it with a known standard capacitance. It is primarily used for measuring medium-Q coils (Q factor between 1 and 10).
2. **Hay's Bridge:** While Hay's bridge is also used to measure inductance, it is specifically optimized for high-Q coils (Q > 10). However, in general textbook contexts for 'Inductance measurement' without specifying the Q factor, Maxwell's bridge is the fundamental reference. (Note: In some contexts, both c and d are used for inductance, but Maxwell's is the primary general-purpose bridge for this category).
3. **Wien Bridge:** This bridge is used for the measurement of frequency and is commonly used in audio frequency oscillators.
4. **Schering Bridge:** This bridge is used for the measurement of capacitance and the dissipation factor (dielectric loss) of a capacitor.
Therefore, Maxwell's bridge is the standard answer for measuring inductance among the given choices.
Two wattmeters are connected to measure the input power to a three phase balanced load in two-wattmeter method. If the two wattmeters read 8kW and 4kW then power factor is
A)
0.5
B)
0.866
C)
1
D)
0.7
In the two-wattmeter method for measuring power in a three-phase balanced load, the readings of the two wattmeters (W1 and W2) are related to the phase angle (φ) of the load.
1. Given values:
W1 = 8 kW
W2 = 4 kW
2. The formula to find the phase angle (φ) is:
tan(φ) = sqrt(3) * [(W1 - W2) / (W1 + W2)]
3. Substituting the given values:
tan(φ) = sqrt(3) * [(8 - 4) / (8 + 4)]
tan(φ) = sqrt(3) * [4 / 12]
tan(φ) = sqrt(3) * (1/3)
tan(φ) = 1 / sqrt(3)
4. Finding the angle φ:
φ = arctan(1 / sqrt(3)) = 30°
5. The power factor (pf) is the cosine of the phase angle:
pf = cos(φ) = cos(30°)
pf = sqrt(3) / 2 ≈ 0.866
Therefore, the power factor of the load is 0.866. Option (b) is correct.
The attenuation in wave guide near the cut off frequency is :
A)
high
B)
low
C)
zero
D)
negative
In a waveguide, the attenuation constant (α) is composed of two main parts: conductor loss (due to the finite conductivity of the walls) and dielectric loss. The attenuation due to conductor loss is inversely proportional to the guide wavelength (λg) and the group velocity. As the operating frequency (f) approaches the cut-off frequency (fc), the guide wavelength λg tends toward infinity (λg = λ / sqrt(1 - (fc/f)^2)). More importantly, the group velocity (vg) approaches zero as f approaches fc. Since the power flow (group velocity) becomes very slow near cut-off, the electromagnetic wave spends more time interacting with the waveguide walls per unit length, leading to significantly higher ohmic losses. Consequently, the attenuation increases sharply and becomes very high near the cut-off frequency. Conversely, far above the cut-off frequency, the attenuation decreases and stabilizes. Therefore, option (a) is correct, while (b), (c), and (d) are physically incorrect for a passive waveguide.
In ceramic insulators, glazes are used to improve :
A)
mechanical properties
B)
chemical properties
C)
electrical properties
D)
physical properties
Ceramic insulators (typically made of porcelain) are porous materials. Glazing is the process of applying a thin layer of glass-like material to the surface of the ceramic and firing it. The primary purpose of this glaze is to seal the pores of the ceramic body. By sealing these pores, the glaze prevents the absorption of moisture and protects the insulator from chemical attacks caused by environmental pollutants, acids, and salts. This significantly improves the 'chemical properties' (specifically chemical resistance and stability) of the insulator. While it may marginally affect physical appearance or surface conductivity (electrical), its fundamental engineering purpose is to prevent chemical degradation and moisture ingress. Mechanical properties are determined by the bulk material and the firing process, not the surface glaze.
What is Hall Effect? Briefly explain any one of the applications of a Hall Effect transducer with a neat diagram.
Using Schering bridge show how capacitance and dissipation factor of unknown capacitor is measured.
Draw and explain the operation of a 2 bit flash type analog-to-digital converter.
Discuss the attenuation of waveguides.