← MES Jr. Grade
Past Questions — 2022
Pick the correct statement regarding Delta and Star connections:
A)
Delta connection draws same current as star connection
B)
Delta connection draws 3 times as much current as star connection
C)
Delta connection draws 1/3 times as much current as star connection
D)
Delta connection draws 1/3 times as much current as star connection
To determine the relationship between the line currents of Star (Y) and Delta (Δ) connections for the same load impedance (Z) and supply voltage (V_L), we analyze the current flow in both configurations:
1. **Star Connection (Y):**
- The voltage across each phase (phase voltage, V_ph) is V_L / √3.
- The phase current (I_ph) is V_ph / Z = V_L / (√3 * Z).
- In a star connection, the line current (I_L_star) is equal to the phase current: I_L_star = V_L / (√3 * Z).
2. **Delta Connection (Δ):**
- The voltage across each phase is equal to the line voltage: V_ph = V_L.
- The phase current (I_ph) is V_ph / Z = V_L / Z.
- In a delta connection, the line current (I_L_delta) is √3 times the phase current: I_L_delta = √3 * (V_L / Z).
3. **Comparison:**
- Ratio = I_L_delta / I_L_star = [√3 * (V_L / Z)] / [V_L / (√3 * Z)]
- Ratio = √3 * √3 = 3.
Therefore, for the same load impedance and supply voltage, a Delta connection draws 3 times as much line current as a Star connection. This is why motors are often started in Star and then switched to Delta to reduce the starting current.
A uniform plane wave has components :
A)
E and H existing only in direction perpendicular to direction of propagation
B)
in the direction of propagation, E existing while H is zero
C)
in the perpendicular direction, E existing while H is zero
D)
both E and H existing in all directions
A uniform plane wave is a specific type of electromagnetic wave where the electric field (E) and magnetic field (H) are constant over any plane perpendicular to the direction of propagation. According to Maxwell's equations and the properties of electromagnetic waves in free space or linear media: 1. Transversality: EM waves are transverse waves, meaning the electric field vector (E) and the magnetic field vector (H) are both perpendicular to the direction of wave propagation (the wave vector k). 2. Orthogonality: E and H are also perpendicular to each other. Therefore, neither E nor H has a component in the direction of propagation. Option (a) correctly describes this transverse nature. Option (b) is incorrect because E cannot exist in the direction of propagation for a plane wave. Option (c) is incorrect because a plane wave must have both E and H components to propagate. Option (d) is incorrect because the fields are restricted to the plane perpendicular to the propagation direction, not all directions.
Consider the following statements in connection with boundary relations of electric field: i. In a single medium electric field is continuous. ii. The tangential components are the same on both sides of a boundary between two dielectrics. iii. The tangential electric field at the boundary of a dielectric and a current carrying conductor with finite conductivity is zero. iv. Normal component of the flux density is continuous across the charge-free boundary between two dielectrics. Which of these statements is/are correct?
A)
i only
B)
i, ii and iii
C)
i, ii and iv
D)
iii and iv only
Let's analyze each statement based on the laws of electrostatics and boundary conditions:
1. **Statement i: In a single medium electric field is continuous.**
This is correct. In a homogeneous medium, there are no abrupt changes in permittivity or surface charge densities, so the electric field $\mathbf{E}$ varies continuously.
2. **Statement ii: The tangential components are the same on both sides of a boundary between two dielectrics.**
This is correct. According to the boundary condition derived from Faraday's Law (or the line integral of $\mathbf{E}$), the tangential component of the electric field is continuous across any boundary: $E_{t1} = E_{t2}$.
3. **Statement iii: The tangential electric field at the boundary of a dielectric and a current carrying conductor with finite conductivity is zero.**
This is incorrect. For a perfect conductor ($\sigma \to \infty$), the internal electric field is zero, making the tangential component zero. However, for a conductor with **finite conductivity**, Ohm's law $\mathbf{J} = \sigma \mathbf{E}$ applies. To maintain a current flow $\mathbf{J}$ within the conductor, a non-zero electric field $\mathbf{E}$ must exist. Therefore, the tangential component is not necessarily zero.
4. **Statement iv: Normal component of the flux density is continuous across the charge-free boundary between two dielectrics.**
This is correct. Gauss's Law states that $D_{n1} - D_{n2} = \rho_s$. If the boundary is charge-free ($\rho_s = 0$), then $D_{n1} = D_{n2}$, meaning the normal component of the electric flux density $\mathbf{D}$ is continuous.
Conclusion: Statements i, ii, and iv are correct. Statement iii is false because finite conductivity allows for a non-zero tangential field to drive current.
Explain the concept of superconductivity. Also draw the magnetic field Vs Temperature characteristics.
What is a semiconductor? How does it differ from conductor and insulator? Write some properties of semiconductors.
The impurity atoms in semiconductors :
A)
reduce the energy gap
B)
increase the kinetic energy of valence electrons
C)
inject more charge carriers
D)
increase the energy release
The process of adding impurity atoms to an intrinsic (pure) semiconductor is called 'doping'. The primary purpose of doping is to increase the electrical conductivity of the material.
1. **Mechanism**: In an n-type semiconductor, pentavalent impurities (like Phosphorus) are added, which provide extra electrons (negative charge carriers). In a p-type semiconductor, trivalent impurities (like Boron) are added, which create 'holes' (positive charge carriers) by accepting electrons from the valence band.
2. **Effect**: By introducing these impurities, the concentration of free charge carriers (either electrons or holes) is significantly increased compared to the intrinsic state, thereby 'injecting' more carriers into the crystal lattice.
Why other options are incorrect:
- **Option a**: Impurity atoms create new energy levels (donor or acceptor levels) within the existing energy gap, but they do not reduce the fundamental band gap (Eg) of the semiconductor material itself.
- **Option b**: Kinetic energy of electrons is primarily a function of temperature and the electric field applied, not the presence of impurity atoms.
- **Option d**: 'Energy release' is not a standard term describing the effect of doping on the electrical properties of a semiconductor.
When impurities tend to remain in the liquid rather than the solid, the impurities are :
A)
impossible to remove by normal freezing
B)
impossible to remove by zone refining
C)
more soluble in the solid than in the liquid
D)
more soluble in the liquid than in the solid
This question relates to the principle of 'Zone Refining' (or fractional crystallization), a technique used to purify semiconductors like Silicon and Germanium.
1. **The Principle of Solubility**: The effectiveness of zone refining depends on the distribution coefficient (k), which is the ratio of the solubility of the impurity in the solid phase to its solubility in the liquid phase (k = C_solid / C_liquid).
2. **Analyzing the Statement**: The question states that 'impurities tend to remain in the liquid rather than the solid'. This means that as the material freezes (solidifies), the impurity atoms are 'pushed' ahead of the solidification front and stay in the molten (liquid) zone.
3. **Conclusion**: If an impurity prefers the liquid state over the solid state, it is by definition 'more soluble in the liquid than in the solid'. This property allows the impurities to be swept to one end of the ingot as the molten zone moves, enabling high purification.
4. **Evaluating other options**:
- Option (a) and (b) are incorrect because this specific property (preferring the liquid phase) is exactly what makes zone refining and normal freezing *possible* and effective for purification.
- Option (c) is the opposite of the given condition; if impurities were more soluble in the solid, they would be trapped in the crystal lattice during freezing, making purification difficult.
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.
For the circuit shown below, find the complete expression for the current when the switch is closed at t=0.
What is Hall Effect? Briefly explain any one of the applications of a Hall Effect transducer with a neat diagram.
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.
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.
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.
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.
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.
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.
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'.
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.
Define electric field intensity. Derive the equation for electric field intensity at a distance ‘r’ from a point charge of Q coulombs.