← MES Jr. Grade
Past Questions — 2022
Series capacitor are used for improving line :
A)
capacitive reactance
B)
inductive reactance effect
C)
voltage
D)
regulation
In power transmission lines, the line impedance is primarily composed of resistance (R) and inductive reactance (XL). The inductive reactance is the dominant factor that limits the power transfer capability of the line and causes voltage drops. Series capacitors are installed in series with the transmission line to provide a capacitive reactance (XC) that opposes and partially cancels out the inductive reactance (XL). The net reactance becomes X_net = XL - XC. By reducing the overall inductive reactance effect, the power transfer capability (P = (V1*V2/X)*sin(delta)) is increased, and the stability of the system is improved.
Why other options are incorrect:
- (a) Capacitive reactance: Series capacitors *add* capacitive reactance to the line; they do not 'improve' it in the sense of reducing it.
- (c) Voltage: While reducing reactance can indirectly help with voltage drop, the primary physical purpose of a series capacitor is the compensation of the inductive reactance. Voltage support is primarily the role of shunt capacitors.
- (d) Regulation: Voltage regulation is a result of the overall impedance; however, the specific mechanism of a series capacitor is the compensation of the inductive reactance effect.
The signal flow graph of a feedback control system is shown below. Determine its closed loop transfer function.
If the back e.m.f of a d.c motor is absent, then :
A)
motor will run at very high speed
B)
motor will run at very low speed
C)
motor will not run at all
D)
motor will burn
The current flowing through the armature of a DC motor is governed by the equation: I_a = (V - E_b) / R_a, where V is the supply voltage, E_b is the back e.m.f., and R_a is the armature resistance.
1. Role of Back e.m.f.: The back e.m.f. (E_b) acts as a self-regulating mechanism that opposes the supply voltage. It ensures that the armature current remains at a safe level during normal operation.
2. Scenario without Back e.m.f.: If the back e.m.f. is absent (E_b = 0), the equation becomes I_a = V / R_a.
3. Impact of Low Resistance: Since the armature resistance (R_a) is designed to be very small to minimize copper losses, the resulting current (I_a) becomes extremely high (essentially a short-circuit condition).
4. Result: This excessive current leads to massive I²R heating in the armature windings, which will quickly melt the insulation and cause the motor to burn out.
Comparing options:
- (a) and (b) are incorrect because speed is a result of rotation; the motor cannot reach a steady speed if it is destroyed by current first.
- (c) is incorrect because the motor will actually attempt to start with a massive surge of current, but it will not 'not run' in a stable sense; it will fail catastrophically.
The air-gap between the yoke and armature in a dc motor is kept small :
A)
to achieve stronger magnetic field
B)
to avoid overheating of the machine
C)
to avoid locking of the armature
D)
to avoid transverse motion
In a DC motor, the air gap is the space between the stationary field poles (attached to the yoke) and the rotating armature. The primary reason for keeping this gap small is based on the concept of magnetic reluctance. Reluctance is the opposition a material offers to the flow of magnetic flux, and it is directly proportional to the length of the path and inversely proportional to the permeability of the material. Since air has a very high reluctance compared to the iron/steel used in the yoke and poles, a larger air gap would significantly increase the total reluctance of the magnetic circuit. By minimizing the air gap, the reluctance is reduced, allowing more magnetic flux to cross from the poles to the armature for a given amount of magnetizing current. A higher flux density results in a stronger magnetic field, which increases the torque production capability of the motor (since Torque ∝ Φ).
Regarding other options:
- (b) Overheating is primarily managed by cooling systems and insulation, not the air gap size.
- (c) Locking of the armature is prevented by precise mechanical bearings and centering, not by the gap size (in fact, a gap that is too small could risk mechanical rubbing).
- (d) Transverse motion is controlled by the mechanical assembly and bearings of the shaft.
A 4-pole induction machine is working as an induction generator. The generator supply frequency is 60 Hz. The rotor current frequency is 5 Hz. The mechanical speed of the rotor in RPM is :
A)
1350
B)
1650
C)
1950
D)
2250
For a 4-pole machine at 60 Hz, synchronous speed is Ns = 120f/P = 120×60/4 = 1800 rpm. In an induction generator, rotor current frequency is fr = |s|f, so |s| = 5/60 = 1/12 = 0.0833. Since it is generating, slip is negative, meaning rotor speed is above synchronous speed: N = Ns(1 - s) = 1800(1 + 0.0833) = 1950 rpm. Hence the correct option is d. The other options are incorrect because 1350 rpm and 1650 rpm are below synchronous speed (motor mode), and 2250 rpm would correspond to a much larger slip than 5 Hz rotor frequency.
If a synchronous motor is switched on to 3-phase supply with its rotor winding short circuited, it will
A)
start
B)
not start
C)
start and continue to run as induction motor
D)
start and continue to run as synchronous motor
1. **Nature of Synchronous Motors**: A synchronous motor is not self-starting because the stator's rotating magnetic field rotates at synchronous speed (Ns) immediately upon startup. The rotor, due to its inertia, cannot instantly accelerate to this speed to lock into step with the field.
2. **The Role of Damper Windings**: To overcome this, synchronous motors are equipped with damper windings (Amortisseur windings) embedded in the rotor poles. These are essentially short-circuited bars, similar to those found in a squirrel-cage induction motor.
3. **Starting Process**: When the motor is switched on and the rotor winding is short-circuited (or the damper windings are present), the rotating stator field induces currents in these short-circuited rotor paths. This creates a torque based on the principle of an induction motor.
4. **Operation**: The motor will accelerate as an induction motor. However, it can never reach synchronous speed (Ns) using induction torque alone, as the slip would become zero and torque would vanish. It will continue to run at a speed slightly below synchronous speed (slip speed) as an induction motor unless an external DC excitation is applied to the rotor to pull it into synchronism.
5. **Analysis of Options**:
- (a) 'start' is too vague; it specifies how it starts and runs.
- (b) 'not start' is incorrect because the short-circuited winding provides induction torque.
- (d) 'run as synchronous motor' is incorrect because synchronism requires DC excitation to lock the rotor to the stator field; short-circuiting alone only provides induction action.
The restriking voltage is measured in :
A)
RMS value
B)
Peak value
C)
Instantaneous value
D)
Average value
Restriking voltage is the transient voltage that appears across the contacts of a circuit breaker at the instant an arc is extinguished. In an AC system, the arc is typically extinguished at a current zero. Immediately after the arc is extinguished, the system voltage rises toward its peak value. Because the restriking voltage is a transient phenomenon that can lead to the re-ignition of the arc if it exceeds the dielectric strength of the gap, it is critical to monitor the maximum voltage reached. Therefore, restriking voltage is always expressed and measured as a peak value.
- RMS value (a) is used for steady-state power calculations but does not represent the maximum stress on the insulation during a transient.
- Instantaneous value (c) refers to the voltage at any specific moment in time, whereas 'restriking voltage' specifically refers to the peak of the transient oscillation.
- Average value (d) is not applicable for analyzing transient dielectric breakdown in switchgear.
In transformers, which of the following statements is valid?
A)
In an open circuit test, copper losses are obtained while in short circuit test, core losses are obtained
B)
In an open circuit test, current is drawn at high power factor
C)
In a short circuit test, current is drawn at zero power factor
D)
In an open circuit test, current is drawn at low power factor
To determine the correct statement, let's analyze the characteristics of Transformer tests:
1. **Open Circuit (OC) Test**: This test is performed to find core losses (iron losses) and magnetizing parameters. The transformer is supplied at rated voltage and frequency on the Low Voltage (LV) side while the High Voltage (HV) side is left open. Because the secondary is open, the current drawn (no-load current $I_0$) is very small (typically 2-5% of rated current). This current is primarily used to magnetize the core and supply losses. Since the core is highly inductive, the no-load current lags the voltage by a large angle (typically 75° to 85°), resulting in a very **low power factor** (cos φ ≈ 0.1 to 0.2).
2. **Short Circuit (SC) Test**: This test is performed to find copper losses and equivalent impedance. The HV side is supplied at a reduced voltage while the LV side is short-circuited. The current drawn is the rated full-load current. Since the transformer is essentially a large inductor at this stage, the current lags the voltage by nearly 90°, meaning it is drawn at a **very low power factor** (close to zero), but not exactly zero.
Evaluating the options:
- **Option a**: Incorrect. In an OC test, core losses are obtained; in an SC test, copper losses are obtained. The option states the reverse.
- **Option b**: Incorrect. In an OC test, the current is drawn at a very low power factor, not high.
- **Option c**: Incorrect. While the power factor in an SC test is very low, it is not exactly zero (as there are resistive losses in the windings).
- **Option d**: Correct. In an OC test, the no-load current is primarily magnetizing current, which lags the voltage significantly, leading to a low power factor.
If any two phases of the 3-phase supply are interchanged, then the motor will :
A)
still run in the same direction
B)
stop running
C)
run in the reverse direction
D)
draw high current
In a 3-phase induction motor, the direction of rotation is determined by the sequence of the phases (e.g., R-Y-B). This sequence creates a Rotating Magnetic Field (RMF) that rotates in a specific direction around the stator. According to the principles of electromagnetism, if any two of the three supply phases are interchanged (for example, changing R-Y-B to Y-R-B), the direction of the Rotating Magnetic Field is reversed. Since the rotor is dragged along by this magnetic field, the motor will run in the reverse direction. This is the standard method used in industry to reverse the direction of a 3-phase motor.
Why other options are incorrect:
- (a) The motor cannot run in the same direction because the phase sequence determines the RMF direction.
- (b) The motor will not stop; it will still receive power and generate torque, just in the opposite direction.
- (d) Interchanging phases does not inherently cause the motor to draw high current, as the electrical characteristics of the windings remain the same.
A 220V d.c shunt motor runs at 500 r.p.m when the armature current is 50A. Calculate the speed if the torque is double. Given that Ra = 0.2Ω.
If the supply frequency increases, the skin effect is :
A)
decreased
B)
increased
C)
remain same
D)
flactuate
The skin effect is the tendency of an alternating electric current (AC) to distribute itself within a conductor such that the current density is largest near the surface of the conductor and decreases with greater depths. The depth to which the current penetrates is characterized by the 'skin depth' (δ), which is mathematically defined as: δ = 1 / √(π f μ σ), where 'f' is the supply frequency, 'μ' is the magnetic permeability of the conductor, and 'σ' is the electrical conductivity. From this formula, it is evident that the skin depth (δ) is inversely proportional to the square root of the frequency (f). Therefore, as the supply frequency increases, the skin depth decreases, meaning the current is pushed further toward the surface (the 'skin') of the conductor. This intensification of the current distribution at the surface is described as an increase in the skin effect. Consequently, higher frequencies lead to a more pronounced skin effect, which increases the effective resistance of the conductor.
Bulk power transmission over long HVDC lines are preferred, on account of :
A)
low cost of HVDC terminals
B)
no harmonics problems
C)
minimum line power losses
D)
simple protection
Bulk power transmission over long distances is preferred using HVDC (High Voltage Direct Current) for several technical and economic reasons:
1. **Minimum Line Power Losses (Correct):** In HVDC, there is no skin effect (which increases effective resistance in AC lines) and no reactive power flow (which causes $I^2X$ losses). For a given power rating and distance, HVDC lines have lower conductor losses compared to HVAC lines.
2. **Cost of Terminals (Incorrect):** HVDC terminals (converter stations) are significantly more expensive than AC substations due to the complex power electronics required for rectification and inversion. This is why HVDC is only economical for very long distances (the 'break-even distance').
3. **Harmonics (Incorrect):** HVDC systems actually introduce significant harmonic distortion into the AC system due to the switching of thyristors/IGBTs in the converters, requiring expensive AC and DC filters to mitigate.
4. **Protection (Incorrect):** Protection of HVDC lines is more complex than AC lines. AC systems benefit from well-established distance and differential protection; HVDC requires specialized control systems to handle faults and converter stability.
Power transmission lines are transposed to reduce :
A)
Skin effect
B)
Ferranti effect
C)
Transmission loss
D)
Interference with neighbouring communication lines
Transposition is the process of swapping the positions of the three phases of a three-phase transmission line at regular intervals along the length of the line.
1. Why it is done: In an asymmetrical transmission line (where conductors are not equidistant), the inductance and capacitance of each phase are different. This leads to unbalanced voltage drops and unbalanced currents in the neutral/ground.
2. Impact on Communication Lines: These unbalanced currents create an asymmetrical magnetic field around the transmission line. This magnetic field induces an electromotive force (EMF) in nearby communication lines (telephone lines), causing electromagnetic interference (EMI) or 'noise'.
3. How Transposition helps: By rotating the positions of the conductors, the average inductance and capacitance for all three phases are equalized over the total length of the line. This balances the phase voltages and cancels out the net induced voltage in neighboring communication lines.
Analysis of other options:
- Skin effect (a): This is a property of the conductor's material and frequency; it is reduced by using stranded conductors or hollow conductors, not transposition.
- Ferranti effect (b): This is the rise in voltage at the receiving end of a long line under no-load or light-load conditions; it is mitigated by using shunt reactors.
- Transmission loss (c): Losses are primarily $I^2R$ losses; transposition does not significantly reduce the resistance or the current magnitude to lower these losses.
A 0.2 H inductor with an initial current of 4 A is in parallel with a resistor of 100 W. The current at 0.8ms is :
A)
4e^-0.4 A
B)
4e^-16x10^-6 A
C)
4e^-0.4x10^-3 A
D)
4e^-16x10^-3 A
1. **Identify the Circuit Type**: This is a source-free RL circuit where an inductor (L) is in parallel with a resistor (R). In such a configuration, the inductor discharges its stored energy through the resistor.
2. **Given Parameters**:
- Inductance (L) = 0.2 H
- Resistance (R) = 100 Ω (Note: The question says '100 W', which is a typo for 100 Ω)
- Initial Current (I₀) = 4 A
- Time (t) = 0.8 ms = 0.8 × 10⁻³ s
3. **Formula for Current Decay**: The current through an inductor in an RL circuit decays exponentially according to the formula:
i(t) = I₀ * e^(-t / τ)
where τ (time constant) = L / R
4. **Calculate the Time Constant (τ)**:
τ = L / R = 0.2 / 100 = 0.002 s = 2 ms
5. **Calculate the Current at t = 0.8 ms**:
i(0.8ms) = 4 * e^(-0.8 × 10⁻³ / 0.002)
i(0.8ms) = 4 * e^(-0.4)
6. **Conclusion**: The current at 0.8 ms is 4e⁻⁰.⁴ A. Comparing this to the options, option (a) matches exactly. Options (b), (c), and (d) result from incorrect calculations of the exponent or incorrect handling of the time units.
Determine the Thevenin’s equivalent circuit across terminals AB of the circuit shown below.
Which one of the following is the main cause of creeping in the induction type energy meters?
A)
Friction compensation
B)
Lag/lead compensation
C)
Overload compensation
D)
Braking torque producing system
Creeping in an induction-type energy meter is a phenomenon where the disc rotates slowly even when there is no load connected (current coil is not energized), provided the voltage coil is energized. This happens because of small stray magnetic fields, over-compensation of friction, or the effect of the voltage coil's magnetic field on the disc. To prevent creeping, a 'braking torque producing system' (specifically a permanent magnet brake) is used. The permanent magnet is adjusted to provide a small amount of braking torque that opposes the motion of the disc when the driving torque is zero or very low. If the braking system is insufficient or improperly adjusted, the disc may 'creep'. Therefore, the braking torque system is the primary mechanism designed to counteract and control creeping; its absence or inadequacy is the cause of the phenomenon.
Regarding other options:
- Friction compensation (a) is used to ensure accuracy at low loads, but over-compensation can contribute to creeping; however, the 'braking system' is the specific engineering solution to stop it.
- Lag/lead compensation (b) is used to correct phase errors to ensure the meter reads energy accurately regardless of the load power factor.
- Overload compensation (c) is used to maintain accuracy when the meter is operated beyond its rated current.
The relative permeability of paramagnetic materials is :
A)
very high
B)
slightly more than one
C)
equal to one
D)
less than one
To determine the relative permeability of paramagnetic materials, we must look at the definition of magnetic permeability and the behavior of different materials in a magnetic field:
1. **Relative Permeability (μr)**: It is defined as the ratio of the permeability of a material (μ) to the permeability of free space (μ₀), i.e., μr = μ / μ₀. It can also be expressed as μr = 1 + χ, where χ is the magnetic susceptibility.
2. **Paramagnetic Materials**: These materials have a small, positive magnetic susceptibility (χ > 0). When placed in an external magnetic field, they develop a weak magnetization in the same direction as the field. Because χ is a small positive value, μr = 1 + (small positive value), which means μr is slightly greater than 1.
3. **Analysis of Options**:
- **Option (a) 'very high'**: This describes **Ferromagnetic** materials (like iron), where μr is typically in the hundreds or thousands.
- **Option (b) 'slightly more than one'**: This correctly describes **Paramagnetic** materials (like aluminum or platinum).
- **Option (c) 'equal to one'**: This describes a **Vacuum** or non-magnetic materials.
- **Option (d) 'less than one'**: This describes **Diamagnetic** materials (like copper or bismuth), where χ is negative, making μr < 1.
The Maxwell equation VxH = J + ∂D/∂t is based on :
A)
Ampere’s law
B)
Gauss’s law
C)
Faraday’s law
D)
Coulomb’s law
The given equation $\nabla \times \mathbf{H} = \mathbf{J} + \frac{\partial \mathbf{D}}{\partial t}$ is the Maxwell-Ampère Law.
1. **Analysis of the Equation**:
- $\nabla \times \mathbf{H}$ represents the curl of the magnetic field intensity.
- $\mathbf{J}$ represents the conduction current density.
- $\frac{\partial \mathbf{D}}{\partial t}$ represents the displacement current density (introduced by James Clerk Maxwell).
2. **Why Option (a) is correct**: Ampere's original law stated that the line integral of the magnetic field around a closed loop is proportional to the electric current passing through the loop ($\nabla \times \mathbf{H} = \mathbf{J}$). Maxwell extended this by adding the term $\frac{\partial \mathbf{D}}{\partial t}$ to account for time-varying electric fields (such as those between capacitor plates), making the equation consistent with the conservation of charge. Thus, this is the generalized Ampere's Law.
3. **Why other options are incorrect**:
- **Gauss's Law (b)**: Relates the divergence of the electric field to the charge density ($\nabla \cdot \mathbf{D} = \rho$).
- **Faraday's Law (c)**: Relates the curl of the electric field to the time-varying magnetic field ($\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}$).
- **Coulomb's Law (d)**: Describes the electrostatic force between two stationary point charges, not a differential field equation.
Fields are said to be circularly polarized if their magnitudes are :
A)
Equal and they are in phase
B)
Equal and they differ in phase by ± 90°
C)
Unequal and they differ in phase by ± 90°
D)
Unequal and they are in phase
Circular polarization occurs when a wave is composed of two orthogonal linear components (e.g., along the x and y axes) that satisfy two specific conditions: 1) The magnitudes of the two components must be equal. 2) The phase difference between the two components must be exactly 90 degrees (π/2 radians) or 270 degrees (-90 degrees). Under these conditions, the resultant electric field vector maintains a constant magnitude but rotates in a circle as the wave propagates.
Analysis of other options:
- Option (a): If magnitudes are equal and they are in phase, the result is a linearly polarized wave oriented at 45 degrees.
- Option (c): If magnitudes are unequal but the phase difference is 90°, the result is elliptical polarization.
- Option (d): If magnitudes are unequal and they are in phase, the result is a linearly polarized wave at an angle determined by the ratio of the magnitudes.
Suppose that resistors R1 and R2 are connected in parallel to give an equivalent resistor R. If resistors R1 and R2 have tolerance of 1% each, the equivalent resistor R for resistors R1 =300 W and R2 =200 W will have tolerance of :
A)
0.5%
B)
1%
C)
1.2%
D)
2%
To find the tolerance of the equivalent resistance R for two resistors R1 and R2 in parallel, we use the formula for parallel resistance: 1/R = 1/R1 + 1/R2, or R = (R1 * R2) / (R1 + R2).
Given:
- R1 = 300 Ω, Tolerance (dR1/R1) = 1% = 0.01
- R2 = 200 Ω, Tolerance (dR2/R2) = 1% = 0.01
Using the method of partial derivatives for error propagation, the relative error (tolerance) of R is given by:
dR/R = |(∂R/∂R1) * (dR1/R)| + |(∂R/∂R2) * (dR2/R)|
Alternatively, for a product/quotient relationship, we can use the logarithmic differentiation method:
ln(R) = ln(R1) + ln(R2) - ln(R1 + R2)
Taking the differential:
dR/R = dR1/R1 + dR2/R2 - d(R1+R2)/(R1+R2)
Let's calculate the specific weights:
1. dR/R = (R2 / (R1 + R2)) * (dR1/R1) + (R1 / (R1 + R2)) * (dR2/R2)
2. Substitute the values:
dR/R = (200 / (300 + 200)) * (1%) + (300 / (300 + 200)) * (1%)
dR/R = (200 / 500) * 1% + (300 / 500) * 1%
dR/R = 0.4 * 1% + 0.6 * 1%
dR/R = 0.4% + 0.6% = 1%
Therefore, the tolerance of the equivalent resistor R is 1%. This makes sense intuitively: if both components have the same percentage tolerance, the combined parallel (or series) resistance will also maintain that same percentage tolerance.
The current in the RLC series circuit at resonance is :
A)
maximum
B)
minimum
C)
infinity
D)
zero
In a series RLC circuit, the total impedance (Z) is given by the formula: Z = √[R² + (X_L - X_C)²], where R is the resistance, X_L is the inductive reactance (ωL), and X_C is the capacitive reactance (1/ωC).
1. Resonance occurs when the inductive reactance equals the capacitive reactance (X_L = X_C).
2. At this condition, the term (X_L - X_C) becomes zero, and the impedance Z reduces to its minimum possible value, which is equal to the resistance R (Z = R).
3. According to Ohm's Law for AC circuits, Current (I) = V / Z.
4. Since the impedance Z is at its minimum at resonance, the current I reaches its maximum value for a given applied voltage V.
Why other options are incorrect:
- b. minimum: This occurs in a parallel RLC circuit at resonance, not a series circuit.
- c. infinity: This would require impedance to be zero, which is impossible in a real circuit due to the presence of resistance R.
- d. zero: This would require impedance to be infinite, which does not happen at resonance in a series circuit.
The depth of penetration of wave in a lossy dielectric increases with increasing :
A)
Conductivity
B)
Permeability
C)
Wavelength
D)
Permittivity
The depth of penetration, also known as the skin depth (δ), is the distance over which the amplitude of an electromagnetic wave decreases to 1/e (approximately 37%) of its original value as it penetrates a lossy medium. For a good conductor or a lossy dielectric, the formula for skin depth is given by: δ = 1 / α, where α is the attenuation constant. For a lossy medium, α is approximately proportional to the square root of the frequency (f) and the conductivity (σ), and inversely proportional to the square root of the permeability (μ) and permittivity (ε). Specifically, δ ≈ 1 / sqrt(π f μ σ). Since frequency (f) is inversely proportional to wavelength (λ) (f = c/λ), the skin depth δ is proportional to the square root of the wavelength (δ ∝ √λ). Therefore, as the wavelength increases, the depth of penetration also increases. Conversely, increasing conductivity (σ), permeability (μ), or permittivity (ε) generally leads to a decrease in the depth of penetration because they increase the attenuation of the wave within the medium.
Soft iron is used to manufacture electromagnets because it has :
A)
High retentivity
B)
High coercive field
C)
Low retentivity
D)
Low coercive field
An electromagnet is a temporary magnet that must be magnetized when current flows through the coil and demagnetized immediately when the current is switched off. To achieve this, the core material must be a 'soft' magnetic material.
1. **Coercivity (Coercive Field):** This is the intensity of the external magnetic field required to reduce the magnetization of a material to zero after it has been magnetized. A 'low coercive field' means the material loses its magnetism very easily once the external current is removed. This is essential for electromagnets to prevent them from remaining magnetized (becoming permanent magnets).
2. **Retentivity:** This is the ability of a material to retain some magnetism after the external field is removed. While low retentivity is also desirable for electromagnets, the defining characteristic that allows for the rapid switching of the magnetic state is the low coercivity.
Comparing the options:
- **High retentivity (a):** Characteristic of permanent magnets (hard magnets), not electromagnets.
- **High coercive field (b):** Characteristic of permanent magnets; it would make the electromagnet very difficult to turn off.
- **Low retentivity (c):** While soft iron has relatively low retentivity compared to steel, the primary requirement for the 'switching' action of an electromagnet is the low coercive force.
- **Low coercive field (d):** This is the most accurate technical requirement for a soft magnetic material used in electromagnets to ensure the magnetic field disappears quickly when the power is cut.
Wattmeter measures :
A)
apparent power
B)
true power
C)
volt ampere
D)
volt ampere reactive
A wattmeter is an instrument used to measure the active power (also known as real power or true power) in an electrical circuit.
1. **True Power (Active Power):** This is the actual power consumed by the load to perform useful work (e.g., heating or rotating a motor). It is measured in Watts (W) and is calculated as P = V * I * cos(φ), where cos(φ) is the power factor.
2. **Apparent Power (Option a):** This is the product of RMS voltage and RMS current (V * I) without considering the phase angle. It is measured in Volt-Amperes (VA). This is measured by a Voltmeter and Ammeter combined, not a wattmeter.
3. **Volt-Ampere (Option c):** This is the unit of Apparent Power, not a type of power measured by a wattmeter.
4. **Volt-Ampere Reactive (Option d):** This refers to Reactive Power (Q), which is the power that oscillates between the source and the load (stored in inductors and capacitors). It is measured in VAR and requires a VAR-meter or a power analyzer, not a standard wattmeter.
Therefore, the wattmeter specifically measures the true power.
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.