← MES Jr. Grade
Past Questions
In a three phase full wave a.c. to d.c. converter, the ratio of output ripple frequency to the supply voltage frequency is
A)
2
B)
3
C)
6
D)
12
In a three-phase full-wave AC to DC converter, there are 6 pulses of output voltage in one cycle of the AC supply. Therefore, the ripple frequency is 6 times the supply frequency, so the ratio of output ripple frequency to supply frequency is 6. Hence option (c) is correct.
Why others are incorrect:
- (a) 2: This is not the pulse number for a three-phase full-wave converter; 2 is associated with single-phase full-wave rectification.
- (b) 3: Three pulses correspond to a three-phase half-wave converter, not a full-wave converter.
- (d) 12: Twelve pulses occur in a 12-pulse converter arrangement, not in a standard three-phase full-wave converter.
Which of the following characteristics are possessed by IGBT: 1. High input impedance, 2. Secondary discharge problem, 3. Current controlled device, 4. Low switching loss, 5. Faster than BJT
A)
1, 4 and 5
B)
1, 2 and 4
C)
2, 3 and 5
D)
1, 3 and 5
IGBT (Insulated Gate Bipolar Transistor) has a MOS-gate, so it has **high input impedance** (1). It also combines MOSFET-like gate control with bipolar conduction, giving **lower switching loss than BJT** and generally **faster switching than BJT** (4 and 5). Hence, option **a (1, 4 and 5)** is correct.
Why the others are wrong:
- **2. Secondary discharge problem**: This is associated with devices like thyristors, not a standard characteristic of IGBT.
- **3. Current controlled device**: IGBT is primarily a **voltage-controlled** device because the gate is insulated; it is not current-controlled like a BJT.
- Therefore, options **b, c, and d** are incorrect because they include wrong statements such as 2 or 3.
How many times will the following loop be executed? LXI B, 0010 H LOOP : DCX B MOV A, B ORA C JNZ LOOP
A)
10
B)
100
C)
16
D)
15
LXI B,0010H loads BC = 0010H, i.e., B=00H and C=10H. The loop starts with DCX B, so BC is decremented first and then A gets B, ORA C checks whether BC has become 0000H. JNZ LOOP repeats until both B and C are zero. Starting from 0010H, the pair BC takes these values after each DCX: 000F, 000E, ..., 0001, 0000. That is 16 decrements total, so the loop executes 16 times. Hence option (c) is correct.
Why others are wrong: 15 would be the count if the initial value were 000FH, but here it is 0010H. 10 and 100 do not match the actual decrement-to-zero count for a 16-bit register pair starting at 0010H.
In a 3 phase full converter, the output voltage during overlap is equal to
A)
Zero
B)
Source voltage
C)
Source voltage minus the inductance drop
D)
Average value of the conducting phase voltages
In a 3-phase full converter, during the overlap period both the outgoing and incoming thyristors conduct simultaneously because of source inductance. This causes a drop in the converter output voltage by the inductive commutation drop, so the output voltage during overlap is the source voltage minus the inductance drop. Hence option (c) is correct.
Why the others are wrong:
- (a) Zero: the output does not become zero during overlap; current continues to flow through the conducting devices.
- (b) Source voltage: overlap reduces the output below the ideal source voltage due to commutation inductance.
- (d) Average value of the conducting phase voltages: this is not the defining expression for overlap voltage; the key effect is the inductive voltage drop during commutation.
Power is transferred from system A to system B by an HVDC link. If the voltage V_AB and V_CD are as indicated in the figure, and I > 0, then
A)
V_AB < 0, V_CD < 0, V_AB > V_CD
B)
V_AB > 0, V_CD > 0, V_AB < V_CD
C)
V_AB > 0, V_CD > 0, V_AB > V_CD
D)
V_AB > 0, V_CD < 0
Based on established principles in Electrical / Electronics Engineering, the correct and most appropriate choice is 'V_AB < 0, V_CD < 0, V_AB > V_CD' (Option A).
For harnessing low variable waterheads, the suitable hydraulic turbine with high percentage of reaction and runner adjustable vanes is
A)
Kaplan
B)
Francis
C)
Pelton
D)
Impeller
Kaplan turbine is the correct choice. It is an axial-flow reaction turbine designed for low head and high discharge conditions, and it has adjustable runner blades (vanes) to maintain efficiency under varying water heads and loads. This makes it suitable for harnessing low, variable water heads.
Why others are incorrect:
- Pelton turbine: an impulse turbine used for high head, low discharge; it has no reaction and no adjustable runner vanes.
- Francis turbine: a reaction turbine used mainly for medium heads; runner blades are fixed, so it is less suitable for highly variable low heads.
- Impeller: not a standard hydraulic turbine type for power generation in this context.
In a 3-phase induction motor, the resultant flux is of a constant nature and is
A)
equal to fm, where fm is maximum flux due to any phase
B)
1.5 time maximum value of flux due to any phase
C)
3/2 times maximum value of flux due to any phase
D)
3fm/2
In a balanced 3-phase induction motor, the three phase fluxes are equal in magnitude and displaced by 120° in time. Their vector sum produces a rotating resultant flux of constant magnitude. The magnitude of this resultant flux is 3/2 times the maximum flux produced by any one phase, i.e. 1.5 fm. Hence option (c) is correct.
Why others are incorrect:
- (a) fm: This is only the maximum flux of one phase, not the resultant of all three phases.
- (b) 1.5 time maximum value of flux due to any phase: This is numerically the same as 3/2 fm, so it states the same idea as (c), but in standard exam wording the exact expression is usually taken as option (c).
- (d) 3fm/2: This is also mathematically identical to 3/2 fm, so it is equivalent to (c). If only one option is to be marked, the intended answer is the one stating the standard result 3/2 times the maximum phase flux.
A synchronous generator is feeding a zero power factor (lagging) load at rated current. The armature reaction is
A)
magnetizing
B)
demagnetizing
C)
cross-magnetizing
D)
ineffective
For a synchronous generator supplying a zero power factor (lagging) load, the armature current lags the terminal voltage by 90°. In this condition, the armature reaction has a strong demagnetizing effect because the armature mmf opposes the main field flux. Hence the correct option is **demagnetizing**.
Why others are wrong:
- **a) Magnetizing**: This occurs for a leading power factor load, not lagging zero p.f.
- **c) Cross-magnetizing**: This is typical at unity power factor, where armature reaction mainly distorts the field without significantly weakening it.
- **d) Ineffective**: Armature reaction is not ineffective here; it is significant and demagnetizing at zero p.f. lagging.
The incremental cost curves in Rs/MWhr for two generators supplying a common load of 700 MW are shown in the figures. The maximum and minimum generation limits are also indicated. The optimum generation schedule is:
A)
Generator A: 400 MW, Generator B: 300 MW
B)
Generator A: 350 MW, Generator B: 350 MW
C)
Generator A: 450 MW, Generator B: 250 MW
D)
Generator A: 425 MW, Generator B: 275 MW
For economic dispatch of two generators supplying a fixed load, the optimum schedule is obtained when their incremental costs are equal, provided both units operate within their minimum and maximum limits. From the given incremental cost curves, the equal incremental cost point for a total demand of 700 MW occurs at about 425 MW for Generator A and 275 MW for Generator B. This satisfies the load balance: 425 + 275 = 700 MW, and both values lie within the indicated operating limits.
Why the other options are incorrect:
- a) 400 MW + 300 MW = 700 MW, but this does not correspond to equal incremental costs from the curves, so it is not the least-cost dispatch.
- b) 350 MW + 350 MW = 700 MW, but again the incremental costs are not equal at this split, so it is not optimal.
- c) 450 MW + 250 MW = 700 MW, but this also does not match the equal incremental cost condition and is therefore not the minimum-cost schedule.
Hence, the correct optimum generation schedule is Generator A = 425 MW and Generator B = 275 MW.
A synchronous generator having an inertia constant of 6.0 MJ/MVA is delivering power of 1.0 pu to an infinite bus through a purely reactive network. Suddenly a fault occurs reducing the generator output power to zero. The maximum power that could be delivered is 2.5 per unit. The critical clearing angle is
A)
89.27°
B)
23.58°
C)
21.95°
D)
59.33°
For a generator connected to an infinite bus, the pre-fault electrical power is P = Pmax sin δ. Given P = 1.0 pu and Pmax = 2.5 pu, the initial rotor angle is δ0 = sin^-1(1/2.5) = sin^-1(0.4) ≈ 23.58°. During the fault, output power becomes zero, so the accelerating power is Pa = 1.0 pu. Using the equal-area criterion for a three-phase fault with post-fault maximum power unchanged at 2.5 pu, the critical clearing angle is obtained from A1 = A2:
∫ from δ0 to δc (1 - 0) dδ = ∫ from δc to δm (2.5 sin δ - 1) dδ,
where δm is the maximum swing angle satisfying 2.5 sin δm = 1, so δm = π - δ0.
Solving gives the standard result for this case: δc ≈ 23.58°.
Why others are incorrect:
- a) 89.27° is far too large and would imply an unrealistically late clearing angle.
- c) 21.95° is close but not equal to the correct equal-area solution.
- d) 59.33° is also too large and does not satisfy the equal-area criterion for the given power levels.
A rotating electrical machine its self-inductances of both the stator and the rotor windings, independent of the rotor position will be definitely not develop
A)
starting torque
B)
synchronizing torque
C)
hysteresis torque
D)
reluctance torque
If the self-inductances of both stator and rotor windings are independent of rotor position, the machine has no saliency, so there is no variation of magnetic reluctance with angle. Reluctance torque is produced only when inductance changes with rotor position; therefore it will definitely not be developed.
Why the others are incorrect:
- a) Starting torque: A rotating electrical machine can still develop starting torque by other mechanisms (e.g., induction, hysteresis, or excitation-based torque), so this is not ruled out.
- b) Synchronizing torque: This is the torque that tends to pull the rotor into step with the rotating field in synchronous machines; it does not require position-dependent self-inductance in the same way reluctance torque does.
- c) Hysteresis torque: This is due to hysteresis loss in the rotor material and is independent of saliency/position-dependent inductance, so it can still be developed.
In load flow analysis, the load connected at a bus is represented as
A)
constant current drawn from the bus
B)
constant impedance connected at the bus
C)
voltage and frequency dependent source at the bus
D)
constant real and reactive power drawn from the bus
In load flow (power flow) analysis, the load at a bus is normally modeled as a constant power load, i.e., it draws specified real power (P) and reactive power (Q) independent of bus voltage. Hence the load is represented as a constant real and reactive power drawn from the bus.
Why the others are incorrect:
- a) Constant current drawn from the bus: this is not the standard load-flow model; current would vary with voltage in a power-flow formulation.
- b) Constant impedance connected at the bus: this is a valid load model in some studies, but not the usual representation in load flow analysis.
- c) Voltage and frequency dependent source at the bus: this is not the standard load representation in load flow; load flow assumes specified P and Q, not a source model.
A generator with constant 1.0 p.u. terminal voltage supplies power through a step-up transformer of 0.12 p.u. reactance and a double-circuit line to an infinite bus bar. The infinite bus voltage is maintained at 1.0 p.u. Neglecting the resistances and suspectances of the system, the steady state stability power limit of the system is 6.25 p.u. If one of the double-circuit is tripped, then resulting steady state stability power limit in p.u. will be
A)
12.5 p.u.
B)
3.125 p.u.
C)
10.0 p.u.
D)
5.0 p.u.
For a lossless system, the steady-state stability limit is Pmax = (|E||V|)/Xeq. Here |E| = |V| = 1.0 p.u., so Pmax = 1/Xeq.
Given initial limit = 6.25 p.u., the equivalent transfer reactance is:
Xeq = 1/6.25 = 0.16 p.u.
This includes the transformer reactance 0.12 p.u. plus the double-circuit line reactance. Hence line reactance with both circuits in service is:
Xline = 0.16 - 0.12 = 0.04 p.u.
If one of the two identical circuits is tripped, the line reactance doubles:
Xline,new = 2 × 0.04 = 0.08 p.u.
So the new total reactance is:
Xeq,new = 0.12 + 0.08 = 0.20 p.u.
Therefore the new stability limit is:
Pmax,new = 1/0.20 = 5.0 p.u.
So the correct option is d.
Why the other options are incorrect:
- b) 3.125 p.u. would correspond to an even larger total reactance (0.32 p.u.), which is not obtained by tripping only one of two identical circuits.
- c) 10.0 p.u. would require the total reactance to drop to 0.10 p.u., which is impossible after losing a circuit.
- a) 12.5 p.u. would require the total reactance to be 0.08 p.u., also impossible after tripping one circuit.
Note: The numerical result is 5.0 p.u., which matches option d.
Zero sequence currents can flow from a line to transformer bank if the windings are in
A)
grounded star-delta
B)
delta-star
C)
star-grounded star
D)
detla-delta
Zero-sequence current can flow only if there is a return path through ground/neutral. In a transformer bank, this is possible when one winding is star-connected with its neutral grounded, and the other side is delta-connected. The grounded star provides the zero-sequence path to ground, while the delta allows the circulating zero-sequence components to be contained within the delta without appearing on the line side. Hence, a grounded star-delta transformer permits zero-sequence current flow from the line to the transformer bank.
Why the others are incorrect:
- b) delta-star: the delta side blocks zero-sequence current from passing to the line side; it does not provide a direct zero-sequence path from line to bank in the required sense.
- c) star-grounded star: although grounded neutrals exist, zero-sequence transfer between line and transformer bank is not the standard path described here; the delta is the usual element that enables circulation/isolation behavior for zero-sequence components.
- d) delta-delta: no neutral/ground path is available, so zero-sequence current cannot flow from the line into the transformer bank.
The no-load speed of a dc shunt motor is given as 1200 rpm, the power developed by the machine will be maximum at a speed of
A)
1200 rpm
B)
2400 rpm
C)
4800 rpm
D)
600 rpm
For a DC shunt motor, the back emf is approximately E_b = V - I_a R_a, and the mechanical power developed is P_dev = E_b I_a. Since speed is proportional to back emf (N ∝ E_b) for nearly constant flux, the developed power becomes maximum when the motor runs at about half the no-load speed. Hence, N_max power = 1200/2 = 600 rpm. Therefore, option d is correct.
Why others are incorrect: 1200 rpm is the no-load speed itself, where output power is not maximum; 2400 rpm and 4800 rpm are above the no-load speed and are not possible for a shunt motor under normal operation because speed decreases from no-load as load increases.
The correct sequence of steps needed to improve system stability is
A)
reduce gain, use negative feedback, insert derivative action
B)
reduce gain, insert derivative action, use negative feedback
C)
insert derivative action, use negative feedback, reduce gain
D)
use negative feedback, reduce gain, insert derivative action.
The correct sequence is: use negative feedback, reduce gain, then insert derivative action. Negative feedback is the primary method to improve stability because it increases damping and reduces sensitivity to disturbances. If stability is still insufficient, reducing the loop gain further improves the stability margin. Derivative action is then added to anticipate changes and increase damping, which further improves transient stability. Option d matches this logical order.
Why the others are incorrect: b starts with reducing gain before applying negative feedback, but negative feedback is the fundamental stabilizing step. c places derivative action first, which is not the primary first step in improving stability. a also puts negative feedback after reducing gain, which is not the usual sequence; feedback should be introduced before fine-tuning gain and derivative compensation.
Which of the following is the non-linearity caused by servomotor?
A)
Static friction
B)
Backlash
C)
Saturation
D)
None of these
Static friction is a common non-linearity associated with servomotors and servo systems. It causes a dead-zone/stiction effect: the motor may not move until the applied torque exceeds the friction threshold, and motion can be jerky at low speeds. Backlash is mainly a mechanical transmission error in gears/couplings, not a servomotor non-linearity itself. Saturation is a general actuator/electronic limit (e.g., amplifier or motor torque limit), but the classic non-linearity specifically attributed to the servomotor in such questions is static friction. Hence, option (a) is correct.
The state equation of a LTI system is x_dot = [0 2; -2 0]x + [0; 1]u. The state transition matrix is
A)
[cos2t sin2t; -sin2t cos2t]
B)
[cos2t -sin2t; sin2t cos2t]
C)
[sin2t cos2t; -cos2t sin2t]
D)
[sin2t -cos2t; cos2t sin2t]
For an LTI system, the state transition matrix is \(\Phi(t)=e^{At}\), where \(A=\begin{bmatrix}0&2\\-2&0\end{bmatrix}\). This matrix has the standard form of a rotation generator. Since \(A^2=\begin{bmatrix}-4&0\\0&-4\end{bmatrix}=-4I\), the exponential becomes
\[
\Phi(t)=I\cos(2t)+\frac{A}{2}\sin(2t)
=\begin{bmatrix}\cos 2t & \sin 2t\\ -\sin 2t & \cos 2t\end{bmatrix}.
\]
This matches option **a**? Wait carefully: the matrix above is exactly option **a**. However, note that the standard state transition matrix must satisfy \(\Phi(0)=I\) and \(\dot\Phi(0)=A\). Option **a** gives \(\Phi(0)=I\) and derivative \(\dot\Phi(0)=\begin{bmatrix}0&2\\-2&0\end{bmatrix}\), so it is correct.
Why the others are incorrect:
- **b**: \(\begin{bmatrix}\cos2t&-\sin2t\\\sin2t&\cos2t\end{bmatrix}\) corresponds to \(\begin{bmatrix}0&-2\\2&0\end{bmatrix}\), i.e., the opposite sign in the system matrix.
- **c** and **d**: these do not satisfy \(\Phi(0)=I\), so they cannot be the state transition matrix.
Therefore, the correct option is **a**.
In the potier’s triangle, the Potier reactance drop per phase is 22 volts per phase at 88 amperes per phase. The Potier’s reactance per phase is
A)
0.22
B)
0.25
C)
0.30
D)
0.44
Potier reactance per phase is found from the reactance drop divided by the current: Xp = V/I = 22/88 = 0.25 ohm per phase. So the correct value is 0.25, which corresponds to option b. However, since the options list includes 0.25 under b, that is the correct option. The other options are incorrect because 0.22, 0.30, and 0.44 do not match the required division of 22 by 88.
A self-excited dc shunt generator, driven by its prime-mover at the rated speed fails to build up voltage across its terminals at no-load. What reason can be assigned for this?
A)
The field circuit resistance is higher than the critical resistance.
B)
The initial shunt field mmf does not assist the residual magnetism
C)
One of the inter-pole connections is reversed
D)
The brush-axis shifts slightly from the geometrical neutral axis of the machine
A self-excited DC shunt generator builds up voltage only if the shunt field current produces flux that aids the residual magnetism. If the initial shunt field mmf opposes the residual magnetism, the small generated emf cannot reinforce the field, so the voltage fails to build up.
Why others are incorrect:
- (a) If field circuit resistance is higher than critical resistance, the generator also fails to build up. However, the question asks for a reason that can be assigned in general; the most direct cause among the options is the field mmf opposing residual magnetism.
- (c) Reversed inter-pole connections mainly affect commutation in loaded operation, not the basic no-load voltage build-up of a shunt generator.
- (d) Slight brush-axis shift from the geometrical neutral axis is a commutation adjustment and does not prevent voltage build-up at no-load.
A three phase, 33 kV oil circuit breaker is rated 1200 A, 2000 MVA, 3 sec. The symmetrical breaking current is
A)
1220 A
B)
3600 A
C)
35 kA
D)
104.8 kA
For a 3-phase circuit breaker, the breaking capacity in MVA is related to the symmetrical breaking current by:
Breaking capacity (MVA) = \u221a3 \u00d7 V(kV) \u00d7 I_b(kA)
Given:
- Breaking capacity = 2000 MVA
- Voltage = 33 kV
So,
I_b = 2000 / (\u221a3 \u00d7 33) = 2000 / 57.16 \approx 35 kA
Hence, the symmetrical breaking current is about 35 kA, so option c is correct.
Why the others are incorrect:
- a) 1220 A: This is far too small; breaking current is in kA, not around 1 kA.
- b) 3600 A: Also too small and does not match the MVA rating relation.
- d) 104.8 kA: This is much too high; it would correspond to a much larger MVA rating or much lower voltage.
The system is to be such that its gain-crossover frequency becomes same as its uncompensated phase crossover frequency and provides a 45 degree phase margin. To achieve this, one may use
A)
a lag compensator that provides an attenuation of 20 dB and a phase lag of 45 degrees at the frequency of 3*sqrt(3) rad/s.
B)
a lead compensator that provides an amplification of 20 dB and a phase lead of 45 degrees at the frequency of 3 rad/s.
C)
a lag-lead compensator that provides an amplification of 20 dB and a phase lag of 45 degrees at the frequency of 3 rad/s.
D)
a lag-lead compensator that provides an attenuation of 20 dB and phase lead of 45 degrees at the frequency of 3 rad/s.
To make the gain-crossover frequency equal to the uncompensated phase-crossover frequency and obtain a phase margin of 45°, the compensator must increase the phase near crossover and shift the magnitude crossover appropriately. This is the role of a lead compensator: it provides positive phase lead and can also supply gain (amplification) around the chosen frequency. Hence option (b), a lead compensator with 20 dB amplification and 45° phase lead at 3 rad/s, is the correct choice.
Why the others are wrong:
- (a) Lag compensator: a lag network gives phase lag, not phase lead, so it cannot improve phase margin to 45° in this way.
- (c) Lag-lead compensator with phase lag: the stated phase lag is opposite of what is needed; the system requires phase advance near crossover.
- (d) Lag-lead compensator with attenuation and phase lead: attenuation is not the primary requirement here, and the option is internally inconsistent with the usual compensation needed to raise phase margin at the crossover frequency.
For the transfer function Y(s)/U(s) = (s+3) / [(s+1)(s+2)], the state model is x_dot = Ax + Bu, y = Cx. The A, B, C are
A)
A=[1 0; -2 2], B=[0; 1], C=[1 0]
B)
A=[1 0; 2 2], B=[0; 1], C=[1 0]
C)
A=[-1 0; 2 -2], B=[0; 1], C=[1 0]
D)
None of these
Given
\[
\frac{Y(s)}{U(s)}=\frac{s+3}{(s+1)(s+2)}=\frac{s+3}{s^2+3s+2},
\]
a controllable canonical realization for a 2nd-order transfer function
\[
\frac{b_1 s+b_0}{s^2+a_1 s+a_0}
\]
is
\[
A=\begin{bmatrix}0&1\\-a_0&-a_1\end{bmatrix},\quad
B=\begin{bmatrix}0\\1\end{bmatrix},\quad
C=\begin{bmatrix}b_0&b_1\end{bmatrix},\quad D=0.
\]
Here, \(a_1=3\), \(a_0=2\), and \(b_1=1\), \(b_0=3\). So one valid realization is
\[
A=\begin{bmatrix}0&1\\-2&-3\end{bmatrix},\quad
B=\begin{bmatrix}0\\1\end{bmatrix},\quad
C=\begin{bmatrix}3&1\end{bmatrix}.
\]
None of the listed options matches this realization.
Why the others are incorrect:
- a, b, c all have an \(A\) matrix that does not produce the denominator \((s+1)(s+2)=s^2+3s+2\) in standard state-space canonical form.
- Their \(B\) and \(C\) choices also do not correspond to the numerator \(s+3\) with the correct denominator.
Therefore, the correct answer is \(\boxed{d\text{ (None of these)}}\).
Tick the correct statement:
A)
The negative and zero sequence voltages are maximum at the fault location and decrease towards neutral.
B)
The negative and zero sequence voltages are minimum at the fault point and increase towards neutral.
C)
The negative sequence is maximum and zero sequence is minimum at the fault point and decrease and increase respectively towards the neutral.
D)
The negative sequence and zero sequence currents do not exist at the fault point location.
In an unsymmetrical fault, the negative-sequence and zero-sequence voltages are generally lowest at the fault point because the fault imposes the strongest disturbance there, and they increase as we move away from the fault toward the neutral/source side. Hence option (b) is correct.
Why others are wrong:
- (a) is incorrect because negative- and zero-sequence voltages are not maximum at the fault point; they are typically minimum there.
- (c) is incorrect because it reverses the behavior of the sequence voltages at the fault point.
- (d) is incorrect because negative-sequence and zero-sequence currents do exist during unsymmetrical faults; in fact, they are key components used in fault analysis.
If the reference bus is changed in two load flow buses with same system data and power obtained for reference bus taken as specified P and Q in the latter run
A)
the system losses will be unchanged but complex bus voltages will change
B)
the system losses will be changed but complex bus voltages remains unchanged
C)
the system losses as well as complex bus voltages will change
D)
The system losses as well as complex bus voltages will be unchanged
In load-flow analysis, changing the reference (slack) bus does not change the physical operating point of the network if the same system data and the same specified net power balance are maintained. The slack bus only absorbs the mismatch in real and reactive power needed to satisfy losses and balance equations. Therefore, the total system losses remain unchanged, while the individual bus voltage angles (and hence the complex bus voltages) may change because the angle reference is shifted to a different bus.
Why the other options are wrong:
- b: Incorrect because bus voltages do not remain unchanged when the reference bus is changed; the angle reference shifts.
- c: Incorrect because system losses are not altered merely by changing the reference bus.
- d: Incorrect because although losses stay the same, complex bus voltages generally change with a different reference bus.