← MES Jr. Grade
Past Questions — 2022
What is frequency response? What is resonant peak of a frequency domain response? Derive the expression for resonant peak and resonant frequency.
The conditions for running two alternators in parallel are :
A)
terminal voltage should be same
B)
frequency should be same
C)
phase sequence should be same
D)
all of these
For two alternators (synchronous generators) to be connected in parallel (synchronized) to a common busbar, certain strict conditions must be met to prevent massive circulating currents and mechanical stress that could damage the machines. These conditions are:
1. **Terminal Voltage (Voltage Magnitude):** The RMS voltage of the incoming alternator must be equal to the busbar voltage. If they differ, a large reactive current will flow between the machines.
2. **Frequency:** The frequency of the incoming alternator must be equal to the frequency of the busbar. If they differ, the machines will experience 'hunting' or power oscillations, and the incoming machine may act as a motor or generator unpredictably.
3. **Phase Sequence:** The phase sequence (e.g., R-Y-B) of the incoming alternator must match that of the busbar. If the phase sequence is reversed, it would be equivalent to a short circuit across the phases, leading to catastrophic failure.
4. **Phase Angle:** Although not listed as a separate option, the phase angle between the voltages must also be zero at the moment of switching.
Since options (a), (b), and (c) are all mandatory requirements for synchronization, the correct answer is 'all of these'.
Conditionally stable system is one which exhibits poor stability at :
A)
increases values of open-loop gain
B)
reduced values of open-loop gain
C)
low frequencies
D)
increas values of close loop
In control systems, stability is often analyzed using the Routh-Hurwitz criterion or Root Locus. A 'conditionally stable' system is one that is stable only for a specific range of the open-loop gain (K). Specifically, it is stable when the gain is between a minimum value (K_min) and a maximum value (K_max). If the open-loop gain is increased beyond the upper limit (K_max), the closed-loop poles move into the right-half of the s-plane, causing the system to become unstable. Therefore, such a system exhibits poor stability (or becomes unstable) as the open-loop gain increases beyond its stable threshold. Option (b) is incorrect because the system is also unstable below K_min, but the standard definition of conditional stability focuses on the 'window' of stability. Option (c) refers to frequency response, which is a different characteristic. Option (d) is phrased incorrectly as 'close loop' values are the result of the gain, not the parameter being adjusted.
During short circuit test, the iron loss of a transformer is negligible because :
A)
the entire input is just sufficient to meet copper loss only
B)
voltage applied across the H.V side is a small fraction of the rated voltage and so is the flux
C)
iron core becomes fully saturated
D)
supply frequency is held constant
In a Short Circuit (SC) test, the secondary winding is short-circuited and a reduced voltage is applied to the primary (usually the High Voltage side) until the rated current flows through the windings.
1. **Iron Loss (Core Loss)** consists of hysteresis and eddy current losses, which are primarily dependent on the magnetic flux density (B) and frequency (f). The formula for core loss is proportional to the square of the applied voltage (V²).
2. **Flux Relationship**: The flux (Φ) in the core is proportional to the applied voltage (V) divided by frequency (f) (V ∝ fΦ).
3. **SC Test Condition**: Since only a very small fraction of the rated voltage (typically 2% to 10%) is applied during the SC test to achieve rated current, the resulting magnetic flux in the core is extremely small compared to the flux at rated voltage.
4. **Conclusion**: Because the flux is so low, the iron losses (which depend on flux) become negligible compared to the copper losses (I²R), which are at their maximum because the rated current is flowing.
Analysis of other options:
- Option (a) is a result of the test condition, not the physical reason why iron loss is negligible.
- Option (c) is incorrect; the core is far from saturation because the voltage is very low.
- Option (d) is true (frequency is constant), but constant frequency does not make iron loss negligible; the low voltage does.
In pump storage hydropower plant, the electrical machine is made to work alternately as generator and motor. The efficiency of the generator working at the same electrical power level is :
A)
greater that that as motor
B)
equal to that as motor
C)
less than that as moor
D)
greater or less than that as motor depending on the type of the machine
In a pumped-storage hydropower plant, a synchronous machine is typically used because it can operate reversibly as both a motor (to pump water to the upper reservoir) and a generator (to produce power when water flows down). The efficiency of an electrical machine is primarily determined by its losses: copper losses (I²R), iron losses (hysteresis and eddy current), and mechanical losses (friction and windage). These losses depend on the current flowing through the windings, the magnetic flux in the core, and the rotational speed of the machine. Since the question specifies that the machine is working at the 'same electrical power level' and the same physical construction is used for both modes, the losses incurred during the motoring phase are virtually identical to those incurred during the generating phase. Therefore, the efficiency (which is the ratio of output power to input power) remains the same regardless of whether the machine is acting as a motor or a generator. Option (a) and (c) are incorrect because there is no inherent physical property that makes one mode more efficient than the other at the same power level. Option (d) is incorrect because the principle of reversibility in synchronous machines ensures consistent efficiency across both modes of operation.
Explain the importance of bus impedance matrix in fault calculation.
Two transformers are operating in parallel. They will share the load depending upon their :
A)
Efficiency
B)
Rating
C)
Leakage reactance
D)
Per-unit impedance
When two transformers operate in parallel, the load is shared between them in inverse proportion to their impedances. Specifically, for the transformers to share the load in proportion to their kVA ratings (which is the ideal operating condition), their per-unit impedances must be identical.
Mathematically, the load shared by a transformer is given by:
Load Share = (Total Load) × [ (Z_total) / (Z_transformer) ]
Since the actual impedance (in ohms) depends on the voltage and rating of the transformer, the 'per-unit impedance' (which is a normalized value) is the critical parameter that determines how the load is distributed relative to the capacity of each unit. If one transformer has a lower per-unit impedance than the other, it will take a disproportionately larger share of the load, potentially leading to overloading even if the total load is within the combined rating of the two transformers.
Why other options are incorrect:
- a. Efficiency: Efficiency is a result of the operation and losses, not a governing factor for load distribution.
- b. Rating: While we want the load to be shared according to the rating, the rating itself does not force the load distribution; the impedance does.
- c. Leakage reactance: While leakage reactance is a component of impedance, the total per-unit impedance (which includes resistance) is the comprehensive factor used for parallel operation calculations.
Explain the build-up process of voltage of a DC generator. Mention the different reasons of failure to building up process of DC shunt generator.
If the dielectric strength of the medium between contacts builds up more rapidly than the re-striking voltage, then the arc will :
A)
be extinguished
B)
not be extinguished
C)
increase
D)
will not change
In switchgear and circuit breaker theory, the process of arc extinction depends on the competition between two voltages: the Restriking Voltage (TRV - Transient Recovery Voltage) and the Dielectric Strength of the medium between the contacts.
1. **Restriking Voltage**: This is the transient voltage that appears across the contacts at the instant of arc extinction. It attempts to re-establish the arc.
2. **Dielectric Strength**: This is the ability of the medium (air, oil, SF6, etc.) between the contacts to withstand voltage without breaking down.
For a successful arc interruption, the dielectric strength of the gap must increase faster than the restriking voltage. If the dielectric strength builds up more rapidly, the medium becomes an insulator before the restriking voltage can reach a magnitude high enough to cause a breakdown (re-ignition). Consequently, the arc is successfully extinguished. If the restriking voltage were to rise faster than the dielectric strength, the arc would be re-struck, and the circuit would not be interrupted.
If the fault current is 2000 amps, the relay setting is 50% and CT ratio is 400/5, the plug setting multiplier will be :
A)
8
B)
10
C)
12
D)
15
To find the Plug Setting Multiplier (PSM), we use the following formula:
PSM = (Fault Current) / (Relay Setting × CT Ratio)
Given data:
1. Fault Current = 2000 A
2. Relay Setting = 50% = 0.5
3. CT Ratio = 400/5 = 80
Step-by-step calculation:
Step 1: Calculate the rated primary current of the CT. The CT ratio is 400/5, meaning 400A on the primary side corresponds to 5A on the secondary side.
Step 2: Calculate the 'Pick-up Current' of the relay. The pick-up current is the current at which the relay is set to operate.
Pick-up Current = Relay Setting × CT Primary Current
Pick-up Current = 0.5 × 400 A = 200 A
Step 3: Calculate the PSM. The PSM is the ratio of the actual fault current to the pick-up current.
PSM = Fault Current / Pick-up Current
PSM = 2000 A / 200 A = 10
Therefore, the Plug Setting Multiplier is 10. Option (b) is correct.
Bandwidth is used as means of specifying performance of a control system related to :
A)
relative stability of the system
B)
the speed of response
C)
the constant gain
D)
time response
In control systems, Bandwidth (BW) is defined as the range of frequencies over which the system gain falls within -3 dB of its low-frequency value. There is a direct relationship between the bandwidth of a system and its speed of response:
1. A larger bandwidth indicates that the system can track higher-frequency inputs, which corresponds to a faster rise time and a quicker response to changes in the input signal.
2. Conversely, a narrow bandwidth indicates a slower response (longer rise time) but provides better noise filtering (attenuation of high-frequency noise).
Evaluating the options:
- (a) Relative stability is typically specified by Gain Margin and Phase Margin.
- (b) The speed of response is directly proportional to the bandwidth. This is the correct answer.
- (c) Constant gain refers to the steady-state gain of the system, not the frequency range.
- (d) Time response is a general term encompassing stability, speed, and accuracy; while bandwidth affects time response, it specifically specifies the 'speed' aspect of it.
The speed of a 50 Hz induction motor under full load conditions is 720 r.p.m. The numbers of poles of the motor are :
A)
4
B)
2
C)
8
D)
none of these
For an induction motor, synchronous speed is given by Ns = 120f / P. At 50 Hz, possible synchronous speeds are: 2 poles → 3000 rpm, 4 poles → 1500 rpm, 6 poles → 1000 rpm, 8 poles → 750 rpm. The full-load speed is 720 rpm, which is slightly less than the synchronous speed due to slip. Since 720 rpm is close to 750 rpm, the motor must be an 8-pole motor. Therefore, the correct option is c. The 2-pole and 4-pole motors would run much faster (around 3000 rpm and 1500 rpm), so they are incorrect. Option d is incorrect because 8 poles is among the given choices.
In Routh-Hurwitz criterion, if there are changes of signs in the elements of the first column, then the number of sign changes indicates :
A)
the number of roots with negative real parts
B)
the number of roots with positive real parts
C)
the number of pair of roots of opposite sign
D)
the number of pairs of roots of same sign
The Routh-Hurwitz stability criterion is a mathematical test used to determine the stability of a linear time-invariant (LTI) system by examining the coefficients of its characteristic equation.
1. **The Criterion**: The Routh array is constructed using the coefficients of the characteristic polynomial. The first column of this array is the critical component for stability analysis.
2. **Stability Condition**: For a system to be stable, all roots of the characteristic equation must lie in the left-half of the s-plane (i.e., they must have negative real parts). This occurs if and only if all elements in the first column of the Routh array have the same sign (usually positive).
3. **Sign Changes**: If there are sign changes in the first column, it indicates that some roots lie in the right-half of the s-plane (RHP). According to the Routh-Hurwitz theorem, the number of sign changes in the first column of the Routh array is exactly equal to the number of roots of the characteristic equation that have positive real parts.
Analysis of options:
- Option (a) is incorrect because roots with negative real parts are indicated by the absence of sign changes (or the remaining roots after subtracting RHP roots from the total).
- Option (b) is correct as per the fundamental theorem of Routh-Hurwitz.
- Option (c) and (d) are incorrect as the criterion specifically identifies the count of roots in the right-half plane, not pairs of opposite or same signs.
A star connected 3-phase, 10MVA, 6.6KW alternator is protected by Merz-price circulating current principle using 1000/5 amperes current transformers. The star point of the alternator is earthed through a resistance of 7.5W. If the minimum operating current for the relay is 0.5A, calculate the percentage of each phase of the stator winding which is unprotected against earth fault when the machine is operating at normal voltage.
Ferranti effect on long overhead lines is experienced when :
A)
the line is lightly loaded
B)
the power factor is unity
C)
the power factor is leading
D)
corona effect is dominant
The Ferranti effect is a phenomenon observed in long transmission lines where the receiving-end voltage (Vr) becomes higher than the sending-end voltage (Vs).
1. **Physical Cause**: Long transmission lines possess a significant shunt capacitance (capacitance between the line and ground). When the line is lightly loaded or open-circuited, the charging current (which flows due to this shunt capacitance) flows through the line's inductance.
2. **Voltage Rise**: This charging current is leading in nature. The voltage drop across the line inductance due to this leading current actually adds to the sending-end voltage at the receiving end, causing the voltage to rise.
3. **Condition**: This effect is most pronounced when the line is 'lightly loaded' or 'no-load' because the inductive voltage drop caused by the load current (which would normally cause a voltage drop) is absent or minimal, allowing the capacitive charging effect to dominate.
Analysis of other options:
- **b. Power factor is unity**: At unity power factor, the load current is in phase with the voltage, which typically results in a standard voltage drop across the line, not a rise.
- **c. Power factor is leading**: While a leading load can exacerbate the voltage rise, the Ferranti effect is a characteristic of the line itself (due to its capacitance) and is specifically identified as occurring during light load or no-load conditions.
- **d. Corona effect**: Corona is a loss phenomenon involving the ionization of air; while it may slightly affect the effective capacitance, it is not the cause of the Ferranti effect.
The concept of an electrically short, medium and long line is primarily based on the :
A)
nominal voltage of the line
B)
physical length of the line
C)
wavelength of the line
D)
power transmitted over the line
In power system analysis, the classification of a transmission line as short, medium, or long is fundamentally determined by the relationship between the physical length of the line (L) and the wavelength (λ) of the operating frequency.
1. Short Line: When the length is small compared to the wavelength (typically L < 0.1λ or < 80 km), the shunt capacitance is negligible and can be ignored.
2. Medium Line: When the length is between 0.1λ and 0.2λ (typically 80 km to 250 km), the shunt capacitance cannot be ignored and is usually represented as lumped parameters (e.g., nominal-π or nominal-T models).
3. Long Line: When the length exceeds 0.2λ (typically > 250 km), the parameters must be treated as distributed along the entire length of the line, requiring the use of wave equations and hyperbolic functions.
While we often use physical distance (km) as a rule of thumb, these distances are derived from the standard power frequency (50Hz or 60Hz). Since wavelength λ = v/f (where v is the velocity of propagation, approximately the speed of light), the classification is physically based on the wavelength.
- Option (a) is incorrect because voltage affects insulation and conductor size, not the classification of line models.
- Option (b) is a common practical shorthand, but the theoretical basis is the ratio of length to wavelength.
- Option (d) is incorrect as power transmission levels do not dictate whether a line is modeled as short, medium, or long.
The critical clearing time of a fault in power system is related to :
A)
Reactive power limit
B)
Short-circuit limit
C)
Steady state stability limit
D)
Transient curves
Step-by-step reasoning: 1. Definition of Critical Clearing Time (CCT): CCT is the maximum time duration for which a fault can be maintained on a power system before the system becomes unstable. If the fault is cleared after this time, the generator will lose synchronism with the rest of the grid. 2. Relation to Stability: CCT is a fundamental parameter of 'Transient Stability'. Transient stability refers to the ability of the power system to maintain synchronism after being subjected to a severe disturbance (like a short circuit). 3. The Role of Transient Curves: To determine CCT, engineers use the 'Equal Area Criterion' or numerical integration of the swing equation. These methods plot the power angle (delta) versus time or the acceleration/deceleration areas on a power-angle curve. These plots are essentially 'transient curves' that describe the dynamic behavior of the rotor during the fault and post-fault periods. 4. Analysis of other options: (a) Reactive power limits relate to voltage stability, not the timing of fault clearing. (b) Short-circuit limits relate to the current-carrying capacity of equipment. (c) Steady state stability limit refers to the maximum power transfer under small, gradual changes, whereas CCT deals with sudden, large disturbances (transients). Therefore, CCT is directly derived from the analysis of transient curves.
A single phase transformer has a maximum efficiency of 90% at full load and unity power factor. Efficiency at half load, at the same power factor is :
A)
86.7%
B)
88.26%
C)
88.9%
D)
87.8%
For a transformer, maximum efficiency occurs when iron loss equals copper loss at that load. Given maximum efficiency = 90% at full load and unity power factor:
Let full-load output = P_out and full-load losses = P_i + P_c, where at maximum efficiency P_i = P_c.
Efficiency at full load:
η = P_out / (P_out + P_i + P_c) = 0.9
Since P_i = P_c, total loss = 2P_i.
So, P_out / (P_out + 2P_i) = 0.9
=> P_out = 18P_i
=> P_i = P_c = P_out/18.
At half load, copper loss becomes (1/2)^2 = 1/4 of full-load copper loss:
P_c(half) = P_c/4 = P_out/72.
Output at half load = P_out/2.
Thus efficiency at half load:
η_half = (P_out/2) / (P_out/2 + P_i + P_out/72)
= 36 / (36 + 4 + 1)
= 36/41
= 0.8780 ≈ 87.8%.
So the correct option is d.
Why others are incorrect:
- a) 86.7% is too low; it would correspond to larger losses than actually present.
- b) 88.26% is slightly higher than the correct value; it does not match the loss calculation.
- c) 88.9% is also too high and exceeds the computed half-load efficiency.
Note: The numerical result is 87.8%, which corresponds to option d.
A single phase line is transmitting 1100KW power to a factory at 11KV and at 0.8 pf lagging. It has a total resistance of 2W and a loop reactance of 3W. Determine the voltage at the sending end and transmission efficiency.
Obtain the transfer function of the system given below using block diagram reduction techniques.
The oil used in the transformer should be free from moisture because moisture will
A)
reduce its density
B)
reduce its dielectric strength
C)
cause its lubricating property to deteriorate
D)
cause the transformer core to rust
In a transformer, the insulating oil serves two primary purposes: cooling the core and windings, and providing electrical insulation (dielectric strength) between the live parts and the grounded tank.
1. **Dielectric Strength**: This refers to the maximum electric field that a material can withstand without breaking down (becoming conductive). Pure transformer oil has a very high dielectric strength.
2. **Effect of Moisture**: Water is a polar molecule and is significantly more conductive than transformer oil. When moisture enters the oil, it forms droplets or dissolves, creating conductive paths. This drastically lowers the breakdown voltage of the oil, meaning the oil can no longer effectively prevent electrical arcs or short circuits between the windings or between the windings and the tank.
3. **Analysis of other options**:
- (a) Density: While moisture might slightly change the density, this is not the critical failure mechanism for transformer operation.
- (c) Lubricating property: Transformer oil is used for insulation and cooling, not primarily for lubrication of moving parts (as there are few to none in a static transformer).
- (d) Core rust: While moisture can cause rust over a very long period, the immediate and most dangerous electrical effect is the reduction of dielectric strength, which leads to catastrophic insulation failure.
The system with the characteristic equation, (s+1)(s+2)(s-3)=0 is :
A)
stable
B)
marginally stable
C)
not necessarily stable
D)
unstable
Step-by-step reasoning:
1. **Identify the Characteristic Equation**: The given characteristic equation is (s + 1)(s + 2)(s - 3) = 0.
2. **Find the Poles**: The poles of the system are the roots of the characteristic equation. Setting each factor to zero:
- s + 1 = 0 => s₁ = -1
- s + 2 = 0 => s₂ = -2
- s - 3 = 0 => s₃ = +3
3. **Stability Criterion**: In control systems, a linear time-invariant (LTI) system is stable if and only if all the poles of its transfer function (roots of the characteristic equation) lie in the left-half of the s-plane (LHP). This means the real part of every pole must be negative.
4. **Analyze the Poles**:
- s₁ = -1 (Real part is negative, stable)
- s₂ = -2 (Real part is negative, stable)
- s₃ = +3 (Real part is positive, unstable)
5. **Conclusion**: Since there is at least one pole (s = 3) located in the right-half of the s-plane (RHP), the system is unstable. Even a single pole in the RHP causes the system's impulse response to grow exponentially over time.
Comparison with other options:
- (a) Stable: Incorrect, as not all poles are in the LHP.
- (b) Marginally stable: Incorrect, as this requires poles on the imaginary axis (jω axis) with no poles in the RHP.
- (c) Not necessarily stable: Incorrect, as the stability is definitively determined to be unstable based on the given equation.
The best method for determining the stability and transient response is :
A)
Bode plot
B)
Nyquist plot
C)
Root locus
D)
None of these
To determine both stability and transient response, we need information about the location of the closed-loop poles in the s-plane.
1. **Root Locus**: This method plots the trajectories of the closed-loop poles as a system parameter (usually the gain K) varies. Since the positions of the poles directly determine both the stability (whether poles are in the left-half plane) and the transient response (damping ratio ζ and natural frequency ωn, which dictate overshoot, settling time, and rise time), the Root Locus is the best tool for analyzing both simultaneously.
2. **Bode Plot**: This is a frequency response method. While it is excellent for determining relative stability (Gain Margin and Phase Margin), it does not provide a direct visualization of the transient response (time-domain behavior) as intuitively as the Root Locus does.
3. **Nyquist Plot**: Like the Bode plot, this is a frequency domain technique. It is primarily used to determine the absolute and relative stability of a closed-loop system by analyzing the open-loop transfer function, but it does not directly provide transient response characteristics.
Therefore, because the Root Locus maps the movement of closed-loop poles, it is the most comprehensive method for evaluating both stability and transient performance.
The gain at the breakaway point of the root locus of a unity feedback system with open loop transfer function G(s) = Ks / ((s-1)(s-4)) is :
A)
1
B)
2
C)
5
D)
9
Step-by-step solution:
1. **Identify the Open Loop Transfer Function**:
G(s) = Ks / ((s-1)(s-4))
Since it is a unity feedback system, H(s) = 1.
2. **Characteristic Equation**:
The characteristic equation is 1 + G(s)H(s) = 0
1 + Ks / ((s-1)(s-4)) = 0
(s-1)(s-4) + Ks = 0
s^2 - 5s + 4 + Ks = 0
s^2 + (K-5)s + 4 = 0
3. **Express K in terms of s**:
Ks = -(s^2 - 5s + 4)
K = -(s^2 - 5s + 4) / s
K = -s + 5 - 4/s
4. **Find the Breakaway Point**:
The breakaway point occurs where dK/ds = 0.
dK/ds = -1 + 0 + 4/s^2 = 0
4/s^2 = 1
s^2 = 4
s = ±2
5. **Determine the valid Breakaway Point**:
The poles of G(s) are at s = 1 and s = 4. The root locus exists on the real axis to the left of an odd number of poles/zeros. Here, the segment between s=1 and s=4 contains the root locus.
Therefore, s = +2 is the valid breakaway point (since s = -2 is not between the poles).
6. **Calculate the Gain K at the Breakaway Point**:
Substitute s = 2 into the expression for K:
K = -(2^2 - 5(2) + 4) / 2
K = -(4 - 10 + 4) / 2
K = -(-2) / 2
Wait, let's re-evaluate the characteristic equation:
(s-1)(s-4) + Ks = 0
(2-1)(2-4) + K(2) = 0
(1)(-2) + 2K = 0
-2 + 2K = 0
2K = 2
K = 1
Correction: Let's re-check the calculation.
Characteristic Equation: s^2 - 5s + 4 + Ks = 0.
At s=2: 2^2 - 5(2) + 4 + 2K = 0
4 - 10 + 4 + 2K = 0
-2 + 2K = 0
2K = 2
K = 1.
Wait, let me re-read the transfer function: G(s) = Ks / ((s-1)(s-4)).
If the options are 1, 2, 5, 9, and my calculation gives 1, then option (a) is correct.
Let me double check the derivative:
K = -s + 5 - 4/s
dK/ds = -1 + 4/s^2 = 0 => s=2.
K = -2 + 5 - 4/2 = 3 - 2 = 1.
However, if the question intended G(s) = K / ((s-1)(s-4)), the result would be different. But based on the provided G(s) = Ks / ((s-1)(s-4)), the gain K at s=2 is 1.
Re-evaluating the options provided in the prompt: a. 1, b. 2, c. 5, d. 9.
My calculation yields 1.
Wait, let me check if there is a sign error in the characteristic equation.
1 + G(s) = 0
1 + Ks/((s-1)(s-4)) = 0
(s^2 - 5s + 4) + Ks = 0
K = -(s^2 - 5s + 4)/s = -s + 5 - 4/s.
Correct.
Let me re-verify the root locus segment. Poles at 1 and 4. Zero at 0.
Real axis segments: (-inf, 0] and [1, 4].
Breakaway point must be between 1 and 4. s=2 is correct.
K = 1 is correct.
Wait, looking at common textbook problems of this type, if the function was G(s) = K / (s(s-1)(s-4)), the answer would be different. But for the given function, K=1.
Let me re-calculate one more time.
(s-1)(s-4) = s^2 - 5s + 4.
At s=2: (2-1)(2-4) = 1 * (-2) = -2.
Characteristic equation: -2 + K(2) = 0 => 2K = 2 => K = 1.
Final check: Option (a) is 1. My result is 1.
No load current in a transformer :
A)
lags the applied voltage by 90°
B)
lags the applied voltage by somewhat less than 90°
C)
leads the applied voltage by 90°
D)
leads the applied voltgaes by somewhat less than 90°
The no-load current (I₀) of a transformer consists of two components: the magnetizing current (Iₘ) and the core-loss current (I꜀).
1. **Magnetizing Current (Iₘ):** This component is responsible for creating the magnetic flux in the core. Since the core is primarily inductive, Iₘ lags the applied voltage by exactly 90°.
2. **Core-loss Current (I꜀):** This component accounts for hysteresis and eddy current losses. It is essentially a resistive component and is in phase with the applied voltage (0° lag).
The total no-load current I₀ is the phasor sum of these two components: I₀ = √(Iₘ² + I꜀²). Because I₀ is the resultant of a component at 90° (Iₘ) and a component at 0° (I꜀), the resulting phasor I₀ will lag the voltage by an angle (φ₀) that is less than 90°. Typically, for a power transformer, this angle is very large (around 75° to 80°), but it is always 'somewhat less than 90°'.
- Option (a) is incorrect because it ignores the core-loss current.
- Option (c) and (d) are incorrect because the transformer is an inductive load, and current in an inductor always lags the voltage.