MES Jr. Grade • 2025 • Paper-I
Magnetic field inside an ideal solenoid is:
A)
Zero
B)
Uniform
C)
Increasing with radius
D)
Inversely proportional to current
An ideal solenoid is defined as an infinitely long coil of tightly wound wire. According to Ampere's Circuital Law, the magnetic field (B) inside an ideal solenoid is given by the formula B = μ₀nI, where μ₀ is the permeability of free space, n is the number of turns per unit length, and I is the current. Since μ₀, n, and I are constants throughout the interior of the solenoid, the magnetic field is constant in both magnitude and direction, making it 'uniform'.
Why other options are incorrect:
- (a) Zero: The magnetic field is zero outside an ideal solenoid, but it is at its maximum and uniform inside.
- (c) Increasing with radius: The field does not depend on the radial distance from the axis inside the solenoid; it remains constant.
- (d) Inversely proportional to current: According to the formula B = μ₀nI, the magnetic field is directly proportional to the current, not inversely proportional.
MES Jr. Grade • 2025 • Paper-I
Which material is commonly used for making heating elements?
A)
Copper
B)
Aluminum
C)
Nichrome
D)
Iron
To function as a heating element, a material must possess three primary characteristics: high electrical resistivity (to generate heat via Joule heating), a high melting point (to withstand extreme temperatures), and resistance to oxidation at high temperatures.
1. **Nichrome (Nickel-Chromium alloy)**: This is the ideal material because it has high resistivity and forms a protective layer of chromium oxide when heated, which prevents further oxidation and degradation. This makes it stable and durable for use in electric kettles, toasters, and heaters.
2. **Copper**: It has very low electrical resistivity, meaning it conducts electricity too efficiently to generate significant heat unless the wire is extremely thin, and it would oxidize rapidly at high temperatures.
3. **Aluminum**: Like copper, it has low resistivity and is not suitable for generating heat in a compact element.
4. **Iron**: While it has higher resistivity than copper, it oxidizes (rusts) very quickly when heated in the presence of air, leading to the rapid failure of the element.
MES Jr. Grade • 2025 • Paper-I
Magnetic moment is:
A)
Pole strength
B)
Universal constant
C)
Scalar quantity
D)
Vector quantity
The magnetic moment (m) of a magnetic dipole is defined as the product of the pole strength (q_m) and the distance (d) between the two poles. Mathematically, it is expressed as m = q_m × d. Because it has both a magnitude and a specific direction (directed from the south pole to the north pole), it is classified as a vector quantity.
Analysis of other options:
- Option (a) Pole strength: This is a component used to calculate the magnetic moment, not the magnetic moment itself.
- Option (b) Universal constant: Magnetic moment varies depending on the material, the current, and the geometry of the loop; it is not a constant of nature.
- Option (c) Scalar quantity: Since magnetic moment has a defined direction in space, it cannot be a scalar.
MES Jr. Grade • 2025 • Paper-I
For a given conductor, if length is doubled and cross-sectional area is halved, the resistance becomes:
A)
Doubled
B)
Quadrupled
C)
Same
D)
Halved
The resistance (R) of a conductor is given by the formula: R = ρ(L/A), where ρ is the resistivity of the material, L is the length of the conductor, and A is the cross-sectional area.
1. Let the initial resistance be R1 = ρ(L/A).
2. According to the problem, the new length (L') is doubled: L' = 2L.
3. The new cross-sectional area (A') is halved: A' = A/2.
4. Substitute these new values into the resistance formula to find the new resistance (R2):
R2 = ρ(L' / A')
R2 = ρ(2L / (A/2))
R2 = ρ(2L * 2 / A)
R2 = 4 * ρ(L/A)
5. Since R1 = ρ(L/A), we have R2 = 4 * R1.
Therefore, the resistance becomes four times the original value, which is described as 'quadrupled'.
- Option (a) is incorrect because doubling only the length would double the resistance.
- Option (c) is incorrect because the changes in length and area both act to increase the resistance.
- Option (d) is incorrect because the resistance increases, not decreases.
MES Jr. Grade • 2025 • Paper-I
The poorest conductor of electricity is:
A)
Carbon
B)
Steel
C)
Silver
D)
Aluminium
To determine the poorest conductor of electricity among the given options, we must compare their electrical conductivities.
1. **Silver (c):** Silver is the most conductive element known to man. It has the highest electrical conductivity of all metals.
2. **Aluminium (d):** Aluminium is an excellent conductor, though less so than silver or copper. It is widely used in power transmission lines due to its conductivity and light weight.
3. **Steel (b):** Steel is an alloy primarily composed of iron. While it is a conductor, its conductivity is significantly lower than that of pure metals like silver or aluminium due to the presence of alloying elements (like carbon) which scatter electrons.
4. **Carbon (a):** Carbon exists in several allotropes. While graphite is a conductor, carbon in general (especially in forms like diamond or amorphous carbon) is a poor conductor compared to metals. Even graphite's conductivity is orders of magnitude lower than that of silver, aluminium, or steel.
Comparing the four, carbon has the highest electrical resistivity (and thus the lowest conductivity) among the listed materials.
MES Jr. Grade • 2025 • Paper-I
Super conductivity can be destroyed by:
A)
Adding impurities
B)
Reducing temperatures
C)
Application of magnetic field
D)
None of the above
Superconductivity is a quantum mechanical phenomenon where a material exhibits zero electrical resistance and the expulsion of magnetic fields (Meissner effect). However, this state is fragile and can be destroyed by three primary factors: 1. Temperature: If the temperature rises above the Critical Temperature (Tc), the material returns to a normal conducting state. 2. Magnetic Field: If an external magnetic field exceeds the Critical Magnetic Field (Hc), the superconducting state is destroyed. 3. Current Density: If the current flowing through the superconductor exceeds the Critical Current Density (Jc), it reverts to a normal state.
Analyzing the options:
- Option (a): Adding impurities can affect the critical temperature or the type of superconductor (Type I vs Type II), but it does not inherently 'destroy' the phenomenon of superconductivity across the board.
- Option (b): Reducing temperatures actually helps a material enter the superconducting state; it does not destroy it.
- Option (c): Application of a magnetic field (above the critical threshold) directly destroys the superconducting state by penetrating the material and breaking the Cooper pairs. Therefore, this is the correct answer.
MES Jr. Grade • 2025 • Paper-I
Displacement current density is:
A)
Jd = ε∂E/∂t
B)
Jd = σE
C)
Jd = μ∂H/∂t
D)
Jd = εμ∂E/∂t
1. Definition: Displacement current density (Jd) is a term introduced by James Clerk Maxwell to account for the change in electric flux density over time, ensuring that Ampere's Law is consistent with the conservation of charge (continuity equation).
2. Mathematical Derivation: According to Maxwell's equations, the displacement current density is defined as the time rate of change of the electric flux density (D).
Jd = ∂D/∂t
3. Relation to Electric Field: In a linear, isotropic medium, the electric flux density D is related to the electric field E by the permittivity (ε) of the medium: D = εE.
4. Substitution: Substituting D = εE into the displacement current formula gives:
Jd = ∂(εE)/∂t
Assuming the medium is not time-varying (ε is constant), we get:
Jd = ε(∂E/∂t).
Analysis of other options:
- Option b (Jd = σE) represents the conduction current density (Jc) according to Ohm's Law in point form.
- Option c (Jd = μ∂H/∂t) is incorrect; the time derivative of the magnetic field is related to the induced electric field (Faraday's Law), not displacement current.
- Option d (Jd = εμ∂E/∂t) is dimensionally incorrect as it includes permeability (μ) in a way that does not align with the definition of current density.
MES Jr. Grade • 2025 • Paper-I
Copper is preferred over aluminum for electrical conductors because:
A)
It is cheaper
B)
It has higher conductivity and better mechanical strength
C)
It has higher resistivity
D)
It is lighter in weight
Step-by-step reasoning:
1. **Electrical Conductivity**: Copper has a higher electrical conductivity than aluminum. In the International Annealed Copper Standard (IACS), pure copper is the benchmark (100%), while aluminum is approximately 61% as conductive as copper. This means copper can carry more current for the same cross-sectional area.
2. **Mechanical Strength**: Copper possesses superior tensile strength and ductility compared to aluminum. It is more resistant to mechanical wear and fatigue, making it more durable for wiring in buildings and electronic devices.
3. **Evaluating Options**:
- **Option a (Cheaper)**: Incorrect. Aluminum is generally cheaper than copper.
- **Option b (Higher conductivity and better mechanical strength)**: Correct. This aligns with the physical and electrical properties of copper.
- **Option c (Higher resistivity)**: Incorrect. Higher resistivity means lower conductivity. Copper has lower resistivity than aluminum.
- **Option d (Lighter in weight)**: Incorrect. Aluminum is significantly lighter (lower density) than copper, which is why aluminum is preferred for long-distance overhead power lines where weight is a critical factor.
MES Jr. Grade • 2025 • Paper-I
Which of the following materials is a soft magnetic material?
A)
Alnico
B)
Silicon steel
C)
Ferrite
D)
Both (b) and (c)
Magnetic materials are broadly classified into two categories based on their hysteresis loop and coercivity: Hard and Soft magnetic materials.
1. **Soft Magnetic Materials**: These materials have low coercivity, low retentivity, and a narrow hysteresis loop. They are easily magnetized and demagnetized, making them ideal for applications where magnetic fields must change rapidly (e.g., transformer cores, inductors).
- **Silicon Steel**: An alloy of iron and silicon. The addition of silicon increases electrical resistivity and reduces hysteresis loss, making it a classic soft magnetic material used in transformer cores.
- **Ferrites**: Ceramic-like compounds of iron oxide mixed with other metals. They have high electrical resistivity and low coercivity, classifying them as soft magnetic materials (specifically soft ferrites).
2. **Hard Magnetic Materials**: These have high coercivity and high retentivity, meaning they retain their magnetism even after the external field is removed. They are used to make permanent magnets.
- **Alnico**: An alloy consisting of Aluminum, Nickel, and Cobalt. It is one of the most well-known hard magnetic materials used for permanent magnets.
Since both Silicon steel and Ferrite are soft magnetic materials, option (d) is the correct choice.
MES Jr. Grade • 2025 • Paper-I
In a PMMC instrument, the deflecting torque is produced by
A)
Eddy currents
B)
Electromagnetic interaction
C)
Spring control
D)
Air friction
A Permanent Magnet Moving Coil (PMMC) instrument operates on the principle of the Lorentz Force Law. When a current-carrying conductor (the moving coil) is placed within a magnetic field produced by a permanent magnet, it experiences a mechanical force. The deflecting torque (Td) is given by the formula Td = BINA, where B is the magnetic flux density, I is the current, N is the number of turns, and A is the area of the coil. This force is a direct result of the electromagnetic interaction between the magnetic field of the permanent magnet and the magnetic field generated by the current flowing through the coil.
Why other options are incorrect:
- a. Eddy currents: These are circulating currents induced in conductors by changing magnetic fields, usually causing losses or used in specific types of meters (like induction meters), but not for deflection in PMMC.
- c. Spring control: This provides the 'controlling torque' (Tc) to oppose the deflecting torque and return the pointer to zero; it does not produce the deflection.
- d. Air friction: This is a form of 'damping torque' used to prevent the pointer from oscillating, not to produce the deflection.
MES Jr. Grade • 2025 • Paper-I
The burden of a CT is expressed in:
A)
Power rating of the resistor
B)
VA rating
C)
Current rating of secondary winding
D)
Current and voltage rating of secondary
In the context of Current Transformers (CTs), the 'burden' refers to the total load connected to the secondary winding of the CT. This load includes the impedance of the connected measuring instruments (like ammeters), relays, and the connecting leads.
1. **Definition**: Burden is the total impedance of the secondary circuit. Since the secondary of a CT is designed to provide a specific current (usually 1A or 5A), the burden is expressed as the apparent power that the CT can deliver to the load while maintaining its specified accuracy class.
2. **Mathematical Expression**: Apparent Power (S) = I² × Z, where I is the secondary current and Z is the burden impedance. The unit for apparent power is Volt-Ampere (VA).
3. **Why other options are incorrect**:
- (a) Power rating of the resistor: This refers to the heat dissipation capacity (Watts), not the load impedance seen by the CT.
- (c) Current rating: This is the rated secondary current (e.g., 5A), not the load/burden.
- (d) Current and voltage rating: While VA is derived from these, the standard industry term for 'burden' is specifically the VA rating.
MES Jr. Grade • 2025 • Paper-I
What is the total reactance of a series RLC circuit at resonance?
A)
Equal to XL
B)
Equal to XC
C)
Equal to R
D)
Zero
In a series RLC circuit, the total impedance (Z) is given by the formula: Z = √[R² + (XL - XC)²], where R is the resistance, XL is the inductive reactance, and XC is the capacitive reactance. The 'total reactance' (X) of the circuit is the difference between the inductive and capacitive reactances: X = XL - XC. Resonance is defined as the condition where the inductive reactance and capacitive reactance are equal in magnitude but opposite in phase (XL = XC). Therefore, at resonance: Total Reactance = XL - XC = 0. This results in the circuit becoming purely resistive, meaning the impedance is at its minimum (Z = R) and the current is at its maximum.
MES Jr. Grade • 2025 • Paper-I
The Ampere circuit law is used to find:
A)
Magnetic field
B)
Electric field
C)
Force
D)
Velocity
Ampere's Circuital Law states that the line integral of the magnetic field (B) around any closed loop (Amperian loop) is equal to $\mu_0$ times the total current (I) passing through the surface enclosed by the loop. Mathematically, it is expressed as $\oint B \cdot dl = \mu_0 I$. This law is primarily used to calculate the magnetic field intensity produced by symmetric current distributions, such as long straight wires, solenoids, and toroids.
Why other options are incorrect:
- b. Electric field: The electric field is typically found using Gauss's Law for electricity.
- c. Force: Magnetic force is calculated using the Lorentz force law ($F = q(E + v \times B)$) or the force between currents ($F = I L \times B$).
- d. Velocity: Velocity is a kinematic property and is not the primary target of Ampere's Law.
MES Jr. Grade • 2025 • Paper-I
The electric flux through a closed surface in free space is:
A)
Zero always
B)
Equal to enclosed charge divided by ε0
C)
Equal to electric field times surface area always
D)
Equal to potential difference across the surface
This question is a direct application of Gauss's Law in electromagnetics. Gauss's Law states that the total electric flux (Φ) passing through any closed surface is equal to the total net charge (Q_enclosed) enclosed by that surface divided by the permittivity of free space (ε0). Mathematically, this is expressed as: Φ = ∮ E · dA = Q_enclosed / ε0.
Analysis of options:
- Option (a) is incorrect because the flux is only zero if the net enclosed charge is zero.
- Option (b) is correct as it is the formal definition of Gauss's Law.
- Option (c) is incorrect because 'Electric field times surface area' (E * A) only calculates flux if the electric field is uniform and perpendicular to the surface; it is not a general rule for all closed surfaces.
- Option (d) is incorrect because potential difference is a measure of energy per unit charge between two points, not a measure of electric flux through a surface.
MES Jr. Grade • 2025 • Paper-I
The electric field inside a uniformly charged hollow sphere is:
A)
Zero
B)
Constant
C)
Increasing linearly with r
D)
Decreasing as 1/r²
To find the electric field inside a uniformly charged hollow sphere (spherical shell), we use Gauss's Law. Gauss's Law states that the total electric flux through a closed surface is equal to the total enclosed charge divided by the permittivity of free space (ε₀): ∮ E · dA = Q_enclosed / ε₀.
1. Consider a Gaussian surface (an imaginary sphere) with radius 'r' located inside the hollow sphere (where r < R, and R is the radius of the shell).
2. Since the charge is distributed uniformly on the surface of the hollow sphere, there is no charge contained within the volume of the Gaussian surface (Q_enclosed = 0).
3. Substituting this into Gauss's Law: E(4πr²) = 0 / ε₀, which simplifies to E = 0.
Therefore, the electric field at any point inside a uniformly charged hollow sphere is zero.
Analysis of other options:
- Option b (Constant): Incorrect, as the field is specifically zero, not a non-zero constant.
- Option c (Increasing linearly): This occurs inside a uniformly charged solid sphere, not a hollow one.
- Option d (Decreasing as 1/r²): This describes the electric field outside the sphere (r > R), where the sphere behaves like a point charge located at the center.
MES Jr. Grade • 2025 • Paper-I
For a two port network to be reciprocal:
A)
Z11=Z22
B)
Y21 = Y22
C)
h21= - h12
D)
AD-BC=0
A two-port network is reciprocal if the ratio of the response at one port to the excitation at the other port remains the same when the positions of the excitation and response are interchanged. In terms of network parameters, the conditions for reciprocity are:
1. Z-parameters: Z12 = Z21
2. Y-parameters: Y12 = Y21
3. ABCD-parameters: AD - BC = 1
4. h-parameters: h12 = -h21 (which is equivalent to h21 = -h12)
Evaluating the options:
- Option (a) Z11 = Z22: This is the condition for symmetry, not reciprocity.
- Option (b) Y21 = Y22: This is not a standard condition for reciprocity or symmetry.
- Option (c) h21 = -h12: This is the correct condition for a reciprocal network in terms of hybrid parameters.
- Option (d) AD - BC = 0: For a reciprocal network, AD - BC must equal 1. A value of 0 would imply a non-reciprocal or singular network.
Therefore, option (c) is the correct condition.
MES Jr. Grade • 2025 • Paper-I
Which among the following is an active transducer?
A)
Photo emissive cell
B)
Selsyn
C)
Strain gauge
D)
Photovoltaic cell
In instrumentation and measurements, transducers are classified into two main categories based on their power requirements:
1. **Active Transducers (Self-Generating):** These transducers generate an electrical signal (voltage or current) directly from the physical quantity being measured without requiring an external power source. Examples include thermocouples, piezoelectric crystals, and photovoltaic cells.
2. **Passive Transducers (Externally Powered):** These transducers require an external power source (excitation voltage) to operate. They typically work by changing a passive electrical property (like resistance, capacitance, or inductance) in response to a physical stimulus. Examples include strain gauges, LVDTs, and RTDs.
Analysis of the options:
- **a. Photo emissive cell:** These typically require an external bias to emit electrons or operate as part of a circuit; they are generally not categorized as primary self-generating active transducers in this context.
- **b. Selsyn:** A Selsyn (self-synchronous) is an electromagnetic transducer used for angle measurement. It requires an AC power supply to operate, making it a passive transducer.
- **c. Strain gauge:** A strain gauge works by changing its electrical resistance when stretched or compressed. It requires an external voltage source (usually via a Wheatstone bridge) to measure this change. Therefore, it is a passive transducer.
- **d. Photovoltaic cell:** A photovoltaic cell converts light energy directly into electrical energy (voltage) via the photovoltaic effect. It generates its own power without needing an external source, making it a classic example of an active transducer.
MES Jr. Grade • 2025 • Paper-I
Which of the following laws do not form a Maxwell equation?
A)
Planck’s law
B)
Gauss’s Law
C)
Faraday’s law
D)
Ampere’s Law
Maxwell's equations are a set of four fundamental partial differential equations that form the foundation of classical electromagnetism. They are:
1. Gauss's Law for Electricity: Relates electric fields to charge densities.
2. Gauss's Law for Magnetism: States that there are no magnetic monopoles.
3. Faraday's Law of Induction: Describes how a changing magnetic field creates an electric field.
4. Ampere's Law (with Maxwell's addition of displacement current): Describes how electric currents and changing electric fields create magnetic fields.
Comparing this to the options:
- Option b (Gauss's Law), Option c (Faraday's Law), and Option d (Ampere's Law) are all integral components of Maxwell's equations.
- Option a (Planck's Law) describes the spectral density of electromagnetic radiation emitted by a black body in thermal equilibrium. It is a fundamental law of Quantum Mechanics, not classical electromagnetism, and is not part of Maxwell's equations.
MES Jr. Grade • 2025 • Paper-I
Hysteresis loss in magnetic materials is proportional to:
A)
Area of hysteresis loop
B)
Square of flux density
C)
Square of current
D)
Magnetizing force
Hysteresis loss occurs in ferromagnetic materials when they are subjected to a cycle of magnetization and demagnetization. This energy loss is caused by the internal friction encountered by magnetic domains as they reorient themselves. Mathematically, the work done per unit volume per cycle to magnetize the material is equal to the integral of H (magnetizing force) with respect to B (magnetic flux density), which is represented by the area enclosed by the B-H hysteresis loop. Therefore, the energy lost as heat per cycle is directly proportional to the area of the hysteresis loop.
Why other options are incorrect:
- b. Square of flux density: This is not a general rule for hysteresis loss; flux density is a variable within the loop, not the determinant of the total loss per cycle.
- c. Square of current: This typically relates to copper losses (I²R losses) in electrical conductors, not magnetic hysteresis losses.
- d. Magnetizing force: While the magnetizing force (H) is used to create the loop, the loss is not simply proportional to H, but to the integral of H over the full cycle of B.
MES Jr. Grade • 2025 • Paper-I
The reciprocity theorem is valid for:
A)
Linear bilateral networks
B)
Non-linear bilateral networks
C)
Linear unilateral networks
D)
Non-linear unilateral networks
The Reciprocity Theorem states that in any linear, bilateral network, the ratio of an excitation (voltage or current) at one point to the response (current or voltage) at another point remains the same if the positions of the excitation and response are interchanged.
1. **Linearity**: The theorem relies on the principle of superposition and the fact that the relationship between voltage and current is a linear function (V = IZ). If the network were non-linear (e.g., containing diodes or transistors), the proportionality would change with the magnitude of the signal, making the theorem invalid.
2. **Bilateralism**: A bilateral element is one whose properties are the same regardless of the direction of current flow (e.g., resistors, inductors, capacitors). If a network contains unilateral elements (like diodes), the current cannot flow in the reverse direction with the same characteristics, thus violating the reciprocity condition.
Therefore:
- Option (b) is incorrect because non-linear networks do not satisfy the proportionality required.
- Option (c) is incorrect because unilateral networks change behavior based on direction.
- Option (d) is incorrect because it combines both non-linearity and unilateralism.
MES Jr. Grade • 2025 • Paper-I
The intrinsic impedance of free space is:
A)
120 V
B)
377 Ω
C)
50 V
D)
100 V
The intrinsic impedance of a medium (denoted as η) is the ratio of the electric field (E) to the magnetic field (H) for a plane wave. For free space (vacuum), it is calculated using the formula: η₀ = √(μ₀ / ε₀), where μ₀ is the permeability of free space (4π × 10⁻⁷ H/m) and ε₀ is the permittivity of free space (approximately 8.854 × 10⁻¹² F/m). Plugging in these values: η₀ = √( (4π × 10⁻⁷) / (8.854 × 10⁻¹²) ) ≈ 376.73 Ω, which is commonly rounded to 377 Ω. Options a, c, and d are incorrect because they provide values in Volts (V), which is a unit of electric potential, not impedance (which is measured in Ohms Ω).
MES Jr. Grade • 2025 • Paper-I
The main advantage of an electronic voltmeter over a PMMC voltmeter is:
A)
Higher input impedance
B)
Lower cost
C)
Requires no power supply
D)
Better temperature stability only
1. PMMC (Permanent Magnet Moving Coil) voltmeters are passive devices. To measure voltage, they require a large series resistance (multiplier) to limit current. However, this resistance is finite, meaning the PMMC meter draws some current from the circuit being measured, which causes a 'loading effect' and introduces measurement error.
2. Electronic voltmeters (such as those based on FETs or Op-Amps) utilize high-impedance input stages. This allows them to have an extremely high input impedance (often in the Mega-ohms or Giga-ohms range).
3. A higher input impedance ensures that the voltmeter draws negligible current from the circuit under test, thereby minimizing the loading effect and providing a much more accurate reading of the actual voltage.
4. Regarding other options: (b) Electronic voltmeters are generally more complex and expensive than simple PMMC meters; (c) Electronic voltmeters require an external or internal power supply to operate the active components, whereas PMMC meters are self-powered by the circuit; (d) While they may have different stability characteristics, the primary and most significant advantage in measurement theory is the input impedance.
MES Jr. Grade • 2025 • Paper-I
Superposition theorem is not valid for:
A)
Voltage responses
B)
Current responses
C)
Power responses
D)
All the three
The Superposition Theorem states that in a linear bilateral network containing multiple independent sources, the current through (or voltage across) any element is the algebraic sum of the currents (or voltages) produced by each source acting alone, while all other independent sources are replaced by their internal impedances.
1. **Linearity**: The theorem relies on the principle of linearity. Voltage (V) and Current (I) are linear responses in a linear circuit. Therefore, options (a) and (b) are valid applications of the theorem.
2. **Non-linearity of Power**: Power (P) is calculated as P = I²R or P = V²/R or P = VI. Because power depends on the square of the current or voltage, it is a non-linear (quadratic) function.
3. **Mathematical Proof**: If two sources produce currents I1 and I2 respectively, the total current is I = I1 + I2. The total power is P_total = (I1 + I2)²R = (I1² + I2² + 2I1I2)R. However, the sum of individual powers would be P1 + P2 = I1²R + I2²R. Since (I1 + I2)²R ≠ I1²R + I2²R (unless one current is zero), the superposition theorem cannot be applied to power.
MES Jr. Grade • 2025 • Paper-I
A digital voltmeter has a voltage controlled generator which provides a width of 5 μ S/V of unit signal, uses a clock of 20 MHz. The pulse count corresponding to an input signal of 12 V would be:
A)
600
B)
1200
C)
1800
D)
2400
Step-by-step solution:
1. **Identify the given parameters:**
- Input Voltage ($V_{in}$) = $12\text{ V}$
- Pulse width per unit voltage ($\text{width/V}$) = $5\text{ μs/V}$
- Clock Frequency ($f$) = $20\text{ MHz} = 20 \times 10^6\text{ Hz}$
2. **Calculate the total pulse width ($T$):**
The voltage-controlled generator produces a pulse width proportional to the input voltage.
$T = V_{in} \times (\text{width/V})$
$T = 12\text{ V} \times 5\text{ μs/V} = 60\text{ μs} = 60 \times 10^{-6}\text{ s}$
3. **Calculate the pulse count ($N$):**
The pulse count is the number of clock cycles that occur during the duration of the generated pulse width.
$N = T \times f$
$N = (60 \times 10^{-6}\text{ s}) \times (20 \times 10^6\text{ Hz})$
$N = 60 \times 20 = 1200$
Conclusion: The pulse count corresponding to an input signal of $12\text{ V}$ is $1200$. Therefore, option (b) is correct.
MES Jr. Grade • 2025 • Paper-I
Which of the following Maxwell’s equations introduces the concept of displacement current?
A)
Gauss’s Law
B)
Faraday’s Law
C)
Ampere–Maxwell Law
D)
Gauss’s Law for Magnetism
The correct answer is **Ampere–Maxwell Law**. Maxwell modified Ampere’s law by adding the **displacement current** term, \(\varepsilon_0 \frac{d\Phi_E}{dt}\), to account for time-varying electric fields and to make the law consistent with charge conservation. This is the equation that introduces the concept of displacement current.
Why the others are incorrect:
- **Gauss’s Law for Magnetism**: states that the net magnetic flux through a closed surface is zero, meaning there are no magnetic monopoles. It does not involve displacement current.
- **Faraday’s Law**: relates a changing magnetic field to an induced electric field. It describes electromagnetic induction, not displacement current.
- **Gauss’s Law**: relates electric flux to enclosed charge. It does not introduce displacement current.