← MES Jr. Grade
Past Questions — 2022
High permittivity ceramic is used for capacitors of
A)
a few pF to a few hundred pF
B)
a few µF to a few hundred µF
C)
A few nF to a few hundred nF
D)
A few mF to a few hundred mF
The capacitance of a capacitor is given by the formula C = (ε₀ εᵣ A) / d, where εᵣ is the relative permittivity (dielectric constant) of the material. Ceramic capacitors are categorized based on their dielectric materials: 1. Low permittivity ceramics (Class 1) are used for very small, stable values (typically in the pF range). 2. High permittivity ceramics (Class 2, such as Barium Titanate) allow for much higher capacitance values in a small physical volume. These high-permittivity ceramics are typically used to achieve capacitance values ranging from a few nanofarads (nF) to a few hundred nanofarads (nF), and sometimes into the low microfarad (µF) range. Option (a) refers to low-permittivity ceramics. Options (b) and (d) represent values (µF to mF) that are typically achieved using electrolytic or tantalum capacitors, as ceramic materials cannot practically reach those magnitudes without becoming excessively large or losing stability.
A capacitor with initial charge q at t=0+ acts as a :
A)
short circuit
B)
open circuit
C)
current source
D)
voltage source
Step-by-step reasoning:
1. **Capacitor Fundamental Equation**: The voltage across a capacitor is given by V = q/C, where 'q' is the charge and 'C' is the capacitance.
2. **Initial Condition**: At t = 0+, the capacitor is given to have an initial charge 'q'. This means it has an initial voltage V(0+) = q/C.
3. **Behavior at t = 0+**: In transient analysis, a capacitor cannot change its voltage instantaneously. Therefore, at the instant t = 0+, the capacitor maintains the voltage V = q/C regardless of the current flowing through it (for an infinitesimal duration).
4. **Circuit Equivalent**: An element that maintains a specific voltage across its terminals regardless of the current is defined as a voltage source. Specifically, a capacitor with an initial charge acts as a voltage source of value V = q/C.
Why other options are incorrect:
- **a. short circuit**: A capacitor acts as a short circuit at t = 0+ ONLY if its initial charge/voltage is zero (q = 0).
- **b. open circuit**: A capacitor acts as an open circuit as t approaches infinity (steady state) in a DC circuit, not at t = 0+.
- **c. current source**: A capacitor does not maintain a constant current; its current depends on the external circuit and the rate of change of voltage (i = C dv/dt).
Draw the block diagram of generalized digital data acquisition system and describe the function of each component.
Select the equation which is not Maxwell :
A)
D = εE
B)
E = D/ε
C)
J = σ E
D)
B = μ H
Maxwell's equations describe how electric and magnetic fields are generated and altered by each other and by charges and currents. While Maxwell's equations are typically presented as four fundamental partial differential equations, they are supported by 'constitutive relations' that describe the properties of the medium.
1. Options (a) D = εE, (b) E = D/ε, and (d) B = μH are constitutive relations for linear, isotropic media. These relate the field intensities (E, H) to the flux densities (D, B) and are integral to the application of Maxwell's equations in matter.
2. Option (c) J = σE is known as Ohm's Law in point form (or the conduction current density equation). While it is used alongside Maxwell's equations to solve boundary value problems in conducting media, it is a phenomenological law describing the behavior of charge carriers in a conductor, not a fundamental Maxwell equation or a constitutive relation defining the medium's permittivity or permeability.
Therefore, J = σE is the equation that is not a Maxwell equation (nor a constitutive relation of the field itself).
When a charge ‘q’ moves with velocity ‘v’ in an electric field ‘E’ and magnetic field ‘B’ the Lorentz force ‘F’ is given by :
A)
F = qE
B)
F = vBq
C)
F = qE + vBq
D)
F = Zero
The Lorentz force is the total force exerted on a charged particle moving through both an electric field and a magnetic field.
1. **Electric Force (Fe):** A charge 'q' in an electric field 'E' experiences a force given by Fe = qE. This force acts in the direction of the field (for a positive charge).
2. **Magnetic Force (Fm):** A charge 'q' moving with velocity 'v' in a magnetic field 'B' experiences a force given by Fm = q(v × B). The magnitude of this force is qvB sin(θ), where θ is the angle between velocity and the magnetic field. In simplified notation often used in multiple-choice questions, this is represented as vBq.
3. **Total Lorentz Force (F):** The total force is the vector sum of the electric and magnetic forces: F = Fe + Fm = qE + q(v × B).
Comparing this to the options:
- Option (a) only considers the electric field.
- Option (b) only considers the magnetic field.
- Option (c) combines both components (F = qE + vBq), which correctly represents the Lorentz force.
- Option (d) is incorrect as force is only zero in specific cases (like a velocity selector), not as a general definition.
A popular method of increasing the range of an AC instrument is the use of :
A)
shunt
B)
multiplier
C)
AC potentiometers
D)
instrument transformer
To increase the range of an AC instrument (such as an ammeter or voltmeter) to measure high currents or high voltages, 'Instrument Transformers' are used. These consist of Current Transformers (CTs) for extending the range of ammeters and Potential Transformers (PTs) for extending the range of voltmeters. They provide two main advantages: they step down the high values to a safe, standardized level (e.g., 5A or 110V) for the meter, and they provide electrical isolation between the high-voltage circuit and the measuring instrument.
Why other options are incorrect:
- a. Shunt: A shunt is a low-resistance resistor used to extend the range of a DC ammeter. While it can be used in AC, it is not the 'popular' or standard method for high-range AC measurements due to safety and burden issues.
- b. Multiplier: A multiplier is a high-resistance resistor used to extend the range of a DC voltmeter. Similar to shunts, it is not the preferred method for high-range AC measurements.
- c. AC potentiometers: These are used for precise measurement of voltage and calibration, not for extending the range of standard AC instruments.
In the given circuit, current I in the 3 ohm resistor is given by :
A)
-2A
B)
2A
C)
4A
D)
6A
Using circuit reduction and Ohm’s law, the 3 Ω resistor carries a current of 2 A in the indicated direction, so the correct answer is 2 A. Hence option (b) is correct. Option (a) is incorrect because the current is not opposite to the assumed direction. Options (c) and (d) are too large and do not satisfy the circuit’s equivalent resistance/voltage relation.
The circuit given below is in steady state for a long time with switch S open. The switch is closed at t=0. The current through R at t=0 will be (R=10Ω):
A)
1/3 A
B)
2/3 A
C)
1 A
D)
2 A
With the switch S open for a long time, the circuit is in DC steady state, so the inductor behaves like a short circuit and the capacitor behaves like an open circuit. Hence the current through the resistor R just before switching is set by the DC source and the resistive path. At the instant t=0+ after the switch is closed, the inductor current cannot change abruptly, so the current through the branch containing R remains the same as at t=0−. Using the steady-state equivalent, the current through R is 2/3 A. Therefore option (b) is correct.
Why others are wrong: 1 A and 2 A would require a larger instantaneous change in inductor current, which is impossible at t=0+. 1/3 A would correspond to a different steady-state division of current and does not match the DC equivalent of the circuit before switching.
For the network shown below, find the voltage reading on voltmeter, if voltmeter sensitivity is 1kW/Volt. If the voltmeter is replaced by another voltmeter having sensitivity 25kW/volt, find the new reading. Comment on the answer.
For increasing the range of an ammeter, connect :
A)
a high value resistance in series with the ammeter coil
B)
a high value resistance in parallel with the ammeter coil
C)
a low value resistance in parallel with the ammeter coil
D)
a low value resistance in series with the ammeter coil
An ammeter is essentially a galvanometer with a shunt resistance. To increase the range of an ammeter (meaning it can measure higher currents without damaging the meter movement), we must limit the amount of current that actually flows through the sensitive ammeter coil. This is achieved by adding a 'multiplier' or 'series' resistance.
1. **Mechanism**: By connecting a high-value resistance in series with the coil, the total resistance of the ammeter increases. According to Ohm's Law (V=IR), for a given voltage, a higher resistance results in a lower current flowing through the meter movement, allowing the device to be used in circuits with higher total currents.
2. **Calculation**: The required series resistance (Rs) is calculated as Rs = (I - Im) * Rm, where I is the new desired full-scale range, Im is the original full-scale current of the coil, and Rm is the internal resistance of the coil. As the desired range (I) increases, the required series resistance (Rs) must also increase.
**Why other options are incorrect:**
- **Option b & c**: Connecting resistance in parallel (shunting) is used to increase the range of a *galvanometer* to make it an ammeter, but to increase the range of an *existing* ammeter to measure even higher currents, the current must be diverted or limited. However, the standard method to extend the range of a meter's scale in this context refers to adding a series resistor to protect the coil from overcurrent.
- **Option d**: A low value resistance in series would not provide enough voltage drop to significantly protect the coil from high currents, failing to effectively increase the range.
The equivalent resistance R will be :
A)
120 ohm
B)
150 ohm
C)
25 ohm
D)
30 ohm
The equivalent resistance is 25 ohm, so option (c) is correct. The other options (30 ohm, 120 ohm, and 150 ohm) do not match the resistance obtained from the circuit’s series/parallel reduction.
In series as well as parallel resonant circuit, increasing the value of resistance would lead to :
A)
Increase in band-width of both the circuits
B)
Decreasing in the band-width of both the circuits
C)
Increase in band-width in series circuit and decrease in parallel circuit
D)
decrease in band-width in series circuit and increase in parallel circuit
To determine the effect of resistance on the bandwidth of resonant circuits, we analyze the formulas for both series and parallel resonance:
1. **Series Resonant Circuit**:
The bandwidth (BW) is defined as the range of frequencies over which the power delivered to the circuit is at least half of the maximum power. The formula for bandwidth in a series RLC circuit is:
BW = ω₀ / Q = R / L
Where ω₀ is the resonant frequency, Q is the quality factor, R is the resistance, and L is the inductance.
From the formula BW = R/L, it is evident that if the resistance (R) increases, the bandwidth (BW) increases linearly.
2. **Parallel Resonant Circuit**:
For an ideal parallel resonant circuit (or a practical one with resistance in series with the inductor), the quality factor Q is inversely proportional to the resistance (Q = ω₀L/R or Q = R_parallel / ω₀L).
The bandwidth is given by:
BW = ω₀ / Q
In a parallel circuit, the 'effective' resistance that limits the Q-factor is the internal resistance of the coil. As the resistance increases, the Q-factor decreases. Since BW = ω₀ / Q, a decrease in Q leads to an increase in bandwidth.
**Conclusion**: In both series and parallel resonant circuits, an increase in resistance leads to a decrease in the Quality Factor (Q), which consequently results in an increase in the bandwidth (BW). Therefore, option (a) is correct.
Explain boundary conditions for Electric field and Magnetic field.
An electron is not deflected while passing through a certain region. Can we be sure that there is no magnetic field in the region? Explain.
Discuss the attenuation of waveguides.
How the range of DC ammeter and DC voltmeter can be extended. Derive the expression to find the shunt resistance and multiplier resistance?
Two port networks are connected in cascade. The combination is to be represented as a single two-port network. The parameters of the network are obtained by multiplying the individual
A)
z-parameter matrix
B)
h-parameter matrix
C)
y-parameter matrix
D)
ABCD parameter matrix
In network analysis, when two-port networks are connected in cascade (meaning the output of the first network is connected directly to the input of the second network), the overall relationship between the input voltage/current and the output voltage/current is determined by the product of their transmission matrices.
1. **ABCD Parameters (Transmission Parameters):** These are defined such that the input variables (V1, I1) are expressed in terms of the output variables (V2, I2). For a single network: [V1; I1] = [A B; C D] * [V2; -I2]. When two networks are cascaded, the output of the first is the input of the second. Mathematically, this results in the matrix multiplication of the individual ABCD matrices: [T_total] = [T1] * [T2].
2. **Why other options are incorrect:**
- **z-parameters (Impedance):** These are used for networks connected in series.
- **y-parameters (Admittance):** These are used for networks connected in parallel.
- **h-parameters (Hybrid):** These are typically used for networks connected in a specific hybrid configuration (series-parallel), but not for simple cascading.
What is meant by creeping? What are the causes of creeping and how it can be eliminated?
The accuracy of moving coil instruments as compare to moving iron instruments is :
A)
high
B)
low
C)
same
D)
reasonable accurate
Permanent Magnet Moving Coil (PMMC) instruments are significantly more accurate than Moving Iron (MI) instruments due to several fundamental design factors:
1. Linear Scale: PMMC instruments have a linear scale because the deflecting torque is directly proportional to the current (Td ∝ I). In contrast, MI instruments have a non-linear (squared) scale (Td ∝ I²), which leads to lower precision, especially at the lower end of the scale.
2. Hysteresis and Eddy Currents: MI instruments use ferromagnetic materials that are subject to hysteresis and eddy current losses, which introduce errors. PMMC instruments use a permanent magnet and a coil, avoiding these specific magnetic losses in the moving part.
3. Sensitivity: PMMC instruments are much more sensitive and can detect very small changes in current compared to MI instruments.
4. External Field Interference: PMMC instruments are typically shielded from stray magnetic fields, whereas MI instruments are more susceptible to external magnetic interference.
Therefore, when comparing the two, the accuracy of moving coil instruments is high.
Regarding Thevenin equivalent, which of the following is not correct?
A)
The voltage source in the Thevenin equivalent circuit is the open circuit voltage of the network when load is disconnected.
B)
The Thevenin equivalent resistance (impedance) is the resistance (impedance) of the network when all voltage sources are short circuited.
C)
Thevenin equivalent resistance is calculated when all voltage sources are open circuited.
D)
Thevenin equivalent is the voltage equivalent of the network.
Thevenin's Theorem states that any linear electrical network with voltage and current sources and resistances can be replaced at its terminals by an equivalent voltage source (Vth) in series with an equivalent resistance (Rth).
Let's analyze the options:
- Option (a) is CORRECT: The Thevenin voltage (Vth) is indeed the open-circuit voltage measured across the load terminals when the load is removed.
- Option (b) is CORRECT: To find the Thevenin resistance (Rth), all independent sources must be deactivated. Deactivating a voltage source means replacing it with its internal resistance, which for an ideal voltage source is a short circuit (0 ohms).
- Option (c) is INCORRECT: This is the opposite of the correct procedure. Voltage sources are short-circuited, not open-circuited, when calculating the equivalent resistance. Current sources are the ones that are open-circuited.
- Option (d) is CORRECT: Thevenin's theorem provides a voltage-source equivalent, whereas Norton's theorem provides a current-source equivalent.
Since the question asks which statement is NOT correct, option (c) is the right answer.
The electric field on equipotential surface is :
A)
Unity
B)
Always perpendicular to the surface
C)
Always parallel to the surface
D)
Zero
An equipotential surface is defined as a surface where the electric potential (V) is constant at every point. The relationship between the electric field (E) and the electric potential is given by the gradient: E = -∇V. Mathematically, the work done (W) in moving a charge (q) along an equipotential surface is W = qΔV. Since ΔV = 0 on an equipotential surface, the work done is zero. Work is also defined as W = ∫ E · dl cos(θ). For the work to be zero for any displacement (dl) along the surface, the cosine of the angle (θ) between the electric field vector and the displacement vector must be zero, meaning θ = 90°. Therefore, the electric field must always be perpendicular to the equipotential surface.
Why other options are incorrect:
- Option (a) Unity: The magnitude of the electric field depends on the charge distribution and distance, not a constant value of 1.
- Option (c) Always parallel: If the field were parallel, there would be a potential difference between points on the surface, contradicting the definition of an equipotential surface.
- Option (d) Zero: The electric field is not necessarily zero; it is simply perpendicular to the surface. A zero field would imply no charge source nearby.
The electric field at the centre of a circular loop of radius ‘r’ and carrying current ‘I’ is given by the expression :
A)
I/r
B)
I/2r
C)
I^2/2r
D)
I^2/2εr
The question asks for the 'electric field' at the center of a circular loop carrying a current 'I'. In a steady-state current-carrying conductor, the charges are distributed uniformly. For a circular loop of radius 'r' carrying a constant current 'I', the charge distribution is symmetric. According to the principle of superposition, the electric field contributions from opposite elements of the loop cancel each other out exactly at the center. Therefore, the net electric field (E) at the center of a current-carrying circular loop is zero. It is important to distinguish between the Electric Field (E) and the Magnetic Field (B). The Magnetic Field at the center of such a loop is given by B = μ₀I / 2r, but the Electric Field is zero. None of the provided options (a, b, c, d) are correct as they all suggest a non-zero value. However, if the question intended to ask for the Magnetic Field, the form would be similar to option 'b' but with the permeability constant μ₀. Given the literal text 'electric field', the answer is 0.
What do you mean by insulating materials? Explain its classification based on temperature.
The current i (in ampere) in the 2 W resistor of the given network is :
A)
0 A
B)
1.2 A
C)
2 A
D)
4.5 A
The 2 Ω resistor is connected between two points that are at the same potential (the bridge is balanced/symmetrical), so the voltage across it is zero. Hence the current through the 2 Ω resistor is i = V/R = 0/2 = 0 A. Therefore option (a) is correct. The other options (1.2 A, 2 A, 4.5 A) would require a nonzero potential difference across the 2 Ω resistor, which does not exist in this network.
A digital –to-analog converter with a full-scale output voltage of 3.5V has a resolution close to 14mV. Its bit size is :
A)
4
B)
8
C)
16
D)
32
Step-by-step calculation:
1. **Understand the Formula**: The resolution of a Digital-to-Analog Converter (DAC) is the smallest change in output voltage that can be produced. It is defined as the Full-Scale Voltage (Vfs) divided by the total number of steps. For an n-bit DAC, there are 2^n levels, meaning there are (2^n - 1) intervals.
Formula: Resolution = Vfs / (2^n - 1) or approximately Vfs / 2^n.
2. **Given Data**:
- Full-scale output voltage (Vfs) = 3.5 V
- Resolution = 14 mV = 0.014 V
3. **Set up the Equation**:
0.014 = 3.5 / (2^n)
(Note: For larger n, 2^n - 1 is approximately 2^n)
4. **Solve for 2^n**:
2^n = 3.5 / 0.014
2^n = 3500 / 14
2^n = 250
5. **Determine n**:
We look for the power of 2 closest to 250.
- 2^4 = 16
- 2^7 = 128
- 2^8 = 256
- 2^9 = 512
Since 256 is very close to 250, the bit size (n) is 8.
Verification: If n = 8, Resolution = 3.5 / (256 - 1) = 3.5 / 255 ≈ 0.0137 V ≈ 13.7 mV, which is close to 14 mV. Therefore, option (b) is correct.