← MES Jr. Grade
Past Questions
Explain boundary conditions for Electric field and Magnetic field.
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.
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.
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.
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.
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.
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.
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).
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.
How the range of DC ammeter and DC voltmeter can be extended. Derive the expression to find the shunt resistance and multiplier resistance?
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).
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.
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.
For intrinsic GaAs, the room-temperature electrical conductivity is 10-6 (ohm-m)-1, the electron and the hole mobilities are, respectively, 0.85 and 0.04 m2/V-s. What is the intrinsic carrier concentration ni at the room temperature?
A)
10^21 m-3
B)
10-20 m-3
C)
7.0 x 10+12 m-3
D)
7.0 x 10-20 m-3
For an intrinsic semiconductor, the conductivity is
σ = q ni (μe + μh)
Given:
σ = 10^-6 (ohm-m)^-1 = 10^-6 S/m
q = 1.6 × 10^-19 C
μe = 0.85 m^2/V-s
μh = 0.04 m^2/V-s
So,
ni = σ / [q(μe + μh)]
= 10^-6 / [1.6 × 10^-19 × (0.85 + 0.04)]
= 10^-6 / [1.6 × 10^-19 × 0.89]
≈ 10^-6 / 1.424 × 10^-19
≈ 7.0 × 10^12 m^-3
Hence the correct option is c.
Why the others are incorrect:
- a (10^21 m^-3) is far too large; it would give a conductivity many orders of magnitude higher than 10^-6 S/m.
- b (10^-20 m^-3) is absurdly small and physically impossible for an intrinsic semiconductor at room temperature.
- d (7.0 × 10^-20 m^-3) is also far too small and inconsistent with the given conductivity.
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.
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.
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.
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.
What is meant by creeping? What are the causes of creeping and how it can be eliminated?
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.
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.