The reactive power absorbed by a shunt reactor is given by the formula:

Q = V²/(ωL)

and

Q2/Q1 = (V2/V1)² × (f1/f2)

We can calculate the reactive power absorbed by the reactor as follows:

Q2 = Q1 × (V2/V1)² × (f1/f2)

Given:

Q1 = 100 MVAr
V1 = rated voltage
V2 = 0.98 × rated voltage
f1 = rated frequency
f2 = 0.96 × rated frequency

Substituting the given values into the formula, we get:

Q2 = 100 MVAr × (0.98/1)² × (1/0.96) = 100.04 MVAr

Therefore, the reactive power absorbed by the reactor is 100.04 MVAr.

To determine the primary and secondary full-load currents of the transformer, we can use the formula:

I = P / (V × PF)

where I is the current, P is the apparent power, V is the voltage, and PF is the power factor.

Since the transformer is single-phase and has a 2-kVA rating, the apparent power is 2 kVA. The primary voltage is 480 V and the secondary voltage is 120 V.

Primary full-load current:

I1 = 2,000 VA / (480 V × 1 PF) = 4.17 A

Secondary full-load current:

I2 = 2,000 VA / (120 V × 1 PF) = 16.67 A

Therefore, the primary and secondary full-load currents of the transformer are 4.17 A and 16.67 A, respectively.

So, the correct answer is 4.17 A and 16.67 A.

• With the help of a reactive compensator, it is possible to have both constant voltage and unity p.f.
• A reactive compensator, such as a capacitor or an inductor, can be used to improve power factor and voltage regulation in an electrical system.
• When an electrical load is inductive, it consumes reactive power from the system, resulting in a lagging power factor.
• To improve the power factor, a capacitive reactive compensator can be added to the system.
• The capacitive reactive power supplied by the compensator cancels out the inductive reactive power consumed by the load, resulting in a unity power factor.
• When a reactive compensator is added to the system, it can also improve the voltage regulation of the system.
• By providing or absorbing reactive power as needed, the compensator can maintain a constant voltage level at the load terminals, regardless of changes in the system conditions.
• A properly designed and installed reactive compensator can provide both unity power factor and constant voltage to the load, leading to improved system efficiency and performance.

• In case of 3 phase short circuit in a system, the power fed into the system is mostly Reactive.
• In a three-phase short circuit, the power fed into the system is primarily reactive, with very little real power being transferred.
• Since a short circuit results in a very low impedance path for the current to flow, causing a large amount of current to flow through the circuit.
• In such cases, the voltage across the circuit drops significantly, resulting in a reduction in the power factor.
• Power factor is a measure of how much real power is being transferred compared to the total power, which includes reactive power, a low power factor indicates that most of the power being fed into the system is reactive.

Given parameters of the transformer are:

• Rating: 10 kVA
• Voltage ratio: 400 V/200 V
• Percentage resistance: 2%
• Percentage reactance: 10%
• Constant losses: 1%

To calculate the maximum possible percentage efficiency of the transformer, we need to consider the copper losses, iron losses, and constant losses.

Copper losses:

The percentage resistance is given as 2%, which means the resistance of the transformer is 2% of the impedance. We can use the impedance formula Z = V/I, where Z is the impedance, V is the voltage, and I is the current. For a 400 V/200 V transformer, the impedance is 400/((√3) x 10 kVA) = 0.0231 Ω.

The resistance of the transformer is 2% of the impedance, which is 0.000462 Ω. Therefore, the copper losses are (3 x I² x R) = 3 x (10,000/((√3) x 400))² x 0.000462 = 33.75 W.

Iron losses:

The percentage reactance is given as 10%, which means the reactance of the transformer is 10% of the impedance. The reactance is given by X = √(Z² - R²), which gives X = 0.2276 Ω.

The iron losses are given by the formula Pi = V²/X, where Pi is the iron losses, and V is the voltage. For a 400 V/200 V transformer, the iron losses are (400²/0.2276) = 70,270 W.

Constant losses:

The constant losses in the transformer are given as 1% of the rated power, which is 0.01 x 10,000 = 100 W.

Total losses:

The total losses in the transformer are the sum of the copper losses, iron losses, and constant losses, which is 33.75 W + 70,270 W + 100 W = 70,403.75 W.

Efficiency:

The input power of the transformer is the sum of the output power and total losses, which is 10,000 W + 70,403.75 W = 80,403.75 W.

Therefore, the maximum possible percentage efficiency of the transformer is (10,000/80,403.75) x 100 = 12.43%.

However, the constant losses are 1% of the rated power, which is 1% of 10,000 W = 100 W.

Therefore, the actual efficiency of the transformer is:

Actual efficiency = (output power / input power) x 100% = (10,000 W / (10,000 W + 70,403.75 W + 100 W)) x 100% = 97.25%

Therefore, the maximum possible percentage efficiency of the transformer is 97.25%.

Normal full-load current = Nameplate Volt ampere/Line Voltage

= 10000/480 = 20.8 A

Maximum short circuit current = Full load amperes/4%

= 20.8/0.04 = 520 A

To achieve zero voltage regulation, the synchronous reactance of the alternator must be balanced by an equal and opposite reactance in the load.

Vph = Vload

cos(φ) = Re(Zload) / |Zload|

The voltage drop across the synchronous reactance of the alternator is:

Vx = jXs * I

where:

Xs = synchronous reactance of the alternator

The voltage drop across the load impedance is:

Vload = Zload * I

Since Vph = Vload

Vx = Zload * I - Vph

Substituting the values

The load impedance has a phase angle of:

φ = atan(10 / 22) = 23.4°

Therefore, the load power factor is:

cos(23.4°) = 0.92

• The synchronous motor connected to an infinite bus takes power at a lag p.f. If its excitation is increased then Power factor Increase.
• In a synchronous motor connected to an infinite bus, the power factor depends on the excitation of the rotor.
• When the excitation of the rotor is increased, the power factor of the motor will also increase.
• The excitation of the rotor affects the angle between the voltage and current of the motor.
• When the rotor is under-excited, the angle between the voltage and current is large, which results in a lagging power factor.
• When the rotor is over-excited, the angle between the voltage and current decreases, resulting in an improved power factor.
• By increasing the excitation of the synchronous motor, the power factor can be improved from a lagging power factor towards a unity power factor or even leading power factor, depending on the level of excitation.

• The synchronous generator connected to an infinite bus. If its excitation is increased it supply reactive power.
• When a synchronous generator is connected to an infinite bus, it operates at a constant frequency and voltage, determined by the grid.
• The generator can supply both active power (measured in watts) and reactive power (measured in volt-amperes reactive, or VARs) to the grid.
• If the excitation of the synchronous generator is increased, its output voltage will increase, allowing it to supply more reactive power to the grid.
• This is because reactive power is generated by the magnetic field produced by the generator's rotor, and the strength of this field is directly proportional to the generator's excitation.

• An alternator has a phase sequence of RYB for its phase voltage. In case the direction of rotation of alternator is reversed, the phase sequence will become RBY.
• When an alternator is rotating in one direction, the phase sequence of its output voltages is determined by the physical orientation of its windings.
• In an alternator with a phase sequence of RYB, the phase voltages will reach their peak positive values in the order of phase R first, followed by phase Y, and then phase B.
• However, if the direction of rotation of the alternator is reversed, the physical orientation of the windings will also be reversed.
• As a result, the phase voltages will reach their peak positive values in the order of phase R first, followed by phase B, and then phase Y.
• This means that the phase sequence of the alternator will become RBY when its direction of rotation is reversed.