Three Phase Circuits MCQ Quiz - Objective Question with Answer for Three Phase Circuits - Download Free PDF

Explanation:

Assertion (A): The neutral wire in a star-connected balanced three-phase system carries no current.

Reason (R): In a balanced system, the sum of the three-phase current phasors is always zero.

Correct Option Analysis:

The correct option is:

Option 4: Both A and R are true, and R is the correct explanation of A.

Explanation:

In a star-connected balanced three-phase system, the load on all three phases is equal in magnitude and phase angle. This means the currents in the three phases are equal in magnitude but are separated by a phase difference of 120°. The phasor sum of the three-phase currents in such a system is always zero. Mathematically, if the three currents are represented as:

• I_a = I∠0°
• I_b = I∠-120°
• I_c = I∠120°

Then, the phasor sum of the currents is:

I_a + I_b + I_c = I∠0° + I∠-120° + I∠120° = 0

This means that the current flowing through the neutral wire, which is meant to carry the unbalanced current in the system, is zero in a balanced system. Therefore, the neutral wire does not carry any current in such cases.

The reason (R) directly explains the assertion (A). Since the phasor sum of the three-phase currents is zero in a balanced system, there is no resultant current to flow through the neutral wire. Hence, both the assertion and the reason are true, and the reason is the correct explanation of the assertion.

Additional Information:

To further understand the analysis, let’s evaluate the other options:

Option 1: A is true, but R is false.

This option is incorrect because the reason (R) is also true. The phasor sum of the three-phase currents in a balanced system is indeed zero, and this is the fundamental explanation for the neutral wire carrying no current in such systems.

Option 2: A is false, but R is true.

This option is incorrect because the assertion (A) is true. In a balanced system, the neutral wire does not carry any current, as explained earlier.

Option 3: Both A and R are true, but R is not the correct explanation of A.

This option is incorrect because the reason (R) is directly related to and explains the assertion (A). The fact that the phasor sum of the three-phase currents is zero is the reason why the neutral wire carries no current in a balanced system.

Conclusion:

In a star-connected balanced three-phase system, the neutral wire carries no current because the phasor sum of the three-phase currents is always zero. This is a fundamental property of balanced three-phase systems and highlights the importance of symmetry in electrical systems. The correct option is Option 4, as both the assertion and the reason are true, and the reason correctly explains the assertion.

Explanation:

Phasor Representation in a Pure Inductive Circuit

Definition: In electrical engineering, a pure inductive circuit is one where the circuit contains only an inductor, and the resistance and capacitance are negligible or absent. The phasor representation of current and voltage in such a circuit is essential for analyzing the behavior of alternating current (AC) signals.

In AC circuits, the voltage and current are sinusoidal in nature, and their magnitudes and phase relationships can be represented using phasors. A phasor is a complex number that represents the amplitude and phase angle of sinusoidal signals.

Correct Option: Voltage leads current by 90°

In a pure inductive circuit, the voltage across the inductor leads the current passing through it by 90°. This phase difference occurs due to the fundamental nature of inductors and how they interact with changing currents.

Explanation:

When an alternating current passes through an inductor, the inductor opposes the change in current by inducing a voltage across itself, as described by Faraday’s Law of Electromagnetic Induction. The induced voltage is proportional to the rate of change of current:

V = L × (dI/dt)

Where:

• V = Voltage across the inductor
• L = Inductance of the inductor
• dI/dt = Rate of change of current

In a sinusoidal AC signal, the current waveform is represented as:

I = I_m × sin(ωt)

Where:

• I_m = Maximum current amplitude
• ω = Angular frequency
• t = Time

Taking the derivative of current with respect to time gives:

dI/dt = I_m × ω × cos(ωt)

From the above equation, the voltage across the inductor becomes:

V = L × I_m × ω × cos(ωt)

Since cos(ωt) can be expressed as sin(ωt + 90°), the voltage waveform leads the current waveform by 90°. This phase relationship is a defining characteristic of pure inductive circuits.

Phasor Representation:

Using phasors, the voltage and current can be represented as vectors in the complex plane:

• Current (I): A vector pointing along the reference axis (typically the positive real axis).
• Voltage (V): A vector leading the current by 90°, pointing along the positive imaginary axis.

Advantages of Phasor Representation:

• Simplifies the analysis of AC circuits by focusing on amplitude and phase relationships.
• Allows easy visualization of phase differences between voltage and current.

Applications:

• Used in the design and analysis of electrical circuits involving inductors.
• Helps in understanding the behavior of transformers, motors, and other inductive devices.

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 2: Current and voltage are in phase

This option is incorrect because, in a pure inductive circuit, the voltage and current are never in phase. The voltage always leads the current by 90° due to the inductive property of the circuit.

Option 3: Voltage leads current by 180°

This option is also incorrect. A phase difference of 180° would mean the voltage and current are completely out of phase, which occurs in certain resistive-capacitive or resistive-inductive circuits under specific conditions but not in a pure inductive circuit.

Option 4: Current leads voltage by 90°

This option is incorrect because, in a pure inductive circuit, the voltage leads the current by 90°, not the other way around. Current leading voltage by 90° is a characteristic of pure capacitive circuits.

Conclusion:

Understanding the phase relationship between voltage and current in various types of AC circuits is fundamental in electrical engineering. In a pure inductive circuit, the voltage leads the current by 90°, which is a direct consequence of the inductive property of the circuit. The phasor representation provides a valuable visualization tool for analyzing and designing such circuits effectively.

Explanation:

Three-Phase Balanced System Voltage Analysis

Definition: In a three-phase balanced system, the voltages of the three phases are equal in magnitude and symmetrically displaced by 120° in phase. This configuration ensures uniform power delivery and efficient operation, which is why three-phase systems are commonly used in electrical power distribution and industrial applications.

Given: Phase voltage for phase A is represented as VA = 230 ∠0°. We need to determine the voltage for phase B.

Analysis:

In a balanced three-phase system, the phase voltages are separated by 120° in phase. This means:

• The voltage for phase A is given as VA = 230 ∠0°.
• The voltage for phase B will lag phase A by 120° (or equivalently lead by -120°).
• The voltage for phase C will lag phase B by 120° (or equivalently lead phase A by -240°, which is equivalent to +120° in the phasor cycle).

Using the phasor representation:

• VA = 230 ∠0°
• VB = 230 ∠-120°
• VC = 230 ∠120°

Therefore, the voltage for phase B is VB = 230 ∠-120°.

This corresponds to Option 3, which is the correct answer.

Important Note: The negative sign in the angle for VB indicates that phase B lags phase A by 120°. This phase relationship is essential for maintaining the symmetry of the three-phase system.

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 1: VB = 230 ∠90°

This option incorrectly states the phase angle of VB. In a balanced three-phase system, the phase angles of the voltages are separated by 120°, not 90°. A phase angle of 90° would correspond to a scenario different from the given balanced three-phase system.

Option 2: VB = 230 ∠120°

This option suggests that phase B leads phase A by 120°. However, in a balanced three-phase system, phase B lags phase A by 120°, not leads. This makes Option 2 incorrect.

Option 4: VB = 230 ∠100°

This option provides an arbitrary phase angle of 100°, which does not align with the symmetrical 120° phase separation in a balanced three-phase system. Hence, this option is incorrect.

Conclusion:

Understanding the phase relationships in a three-phase balanced system is crucial for correctly identifying the voltages of different phases. In this case, the voltage for phase B is determined to be VB = 230 ∠-120°, which corresponds to Option 3. The symmetrical phase separation of 120° between the voltages ensures efficient power delivery in three-phase systems.

Explanation:

Apparent Power Calculation in a Three-Phase System:

Definition: Apparent power is the total power flowing in an electrical circuit and is a combination of both active power (real power) and reactive power. It is measured in volt-amperes (VA) or kilovolt-amperes (kVA). Apparent power represents the vector sum of active power and reactive power in a circuit.

Formula: The relationship between apparent power (S), active power (P), and reactive power (Q) in an electrical circuit is given by the equation:

S = √(P² + Q²)

Where:

• S = Apparent Power (in kVA)
• P = Active Power (in kW)
• Q = Reactive Power (in kVAR)

Given Data:

• Reactive Power (Q) = 12 kVAR
• Active Power (P) = 16 kW

Step-by-Step Solution:

1. Substitute the values of active power (P) and reactive power (Q) into the formula for apparent power:

S = √(P² + Q²)

2. Calculate the squares of active power and reactive power:

P² = (16 kW)² = 256 kW²

Q² = (12 kVAR)² = 144 kVAR²

3. Add the squared values:

P² + Q² = 256 + 144 = 400

4. Find the square root of the sum to calculate the apparent power:

S = √400 = 20 kVA

Correct Answer:

The apparent power of the three-phase system is 20 kVA. Hence, the correct option is:

Option 1: 20 kVA

Additional Information

Analysis of Other Options:

To understand why the other options are incorrect, let’s analyze them:

Option 2: 28 kVA

This value is incorrect because it does not match the calculation based on the formula for apparent power. If we substitute the given values of active and reactive power into the formula, the result is clearly 20 kVA, not 28 kVA. This option might arise from a misunderstanding of the formula or incorrect calculations.

Option 3: 4 kVA

This value is far below the calculated apparent power and does not correspond to the given active and reactive power values. It could be derived from an erroneous assumption or misinterpretation of the relationship between active power, reactive power, and apparent power.

Option 4: 10 kVA

While this value is closer to the actual apparent power, it is still incorrect. The apparent power must be calculated using the formula √(P² + Q²), which yields 20 kVA for the given values of active and reactive power. This option might result from halving or incorrectly approximating the calculated apparent power.

Conclusion:

The correct calculation of apparent power, based on the formula and the given values of active and reactive power, leads to a result of 20 kVA. Understanding the relationship between active power, reactive power, and apparent power is essential for accurate calculations in electrical systems. This ensures that the correct option is chosen, and incorrect options are identified as arising from calculation errors or misunderstandings of the formula.

Explanation:

Active Power in a Three-Phase System

Definition: Active power (P), also known as real power, is the actual power consumed by electrical devices to perform useful work, such as lighting, heating, or rotating a motor. In a three-phase electrical system, active power is calculated based on the voltage, current, and the power factor of the system.

Correct Formula: The active power (P) in a balanced three-phase system is given by:

P = √3 × VL × IL × cosΦ

Where:

• √3: A constant derived from the relationship between line-to-line and line-to-neutral voltages in a three-phase system.
• VL: Line-to-line voltage (in volts).
• IL: Line current (in amperes).
• cosΦ: Power factor, which represents the phase angle between voltage and current.

Explanation:

The formula P = √3 × VL × IL × cosΦ is derived based on the principles of three-phase power systems. In a balanced three-phase system, the power delivered by each phase is equal. The total active power is the sum of the active power contributions from all three phases. Using trigonometric relationships and electrical theory, the factor √3 is introduced to account for the phase difference between the line voltages and currents.

Why Option 1 is Correct:

Option 1 correctly states the formula for calculating active power in a three-phase system. The inclusion of the constant √3 accounts for the relationship between the line-to-line and line-to-neutral voltages in a three-phase system, ensuring accurate calculation of the total active power. The power factor (cosΦ) is a critical component, as it reflects the efficiency of power usage in the system. This formula is widely used in electrical engineering for power calculations in industrial and commercial applications.

Applications:

• Used for calculating power consumption in industrial machinery and equipment.
• Applicable in power system analysis for load flow studies.
• Essential for designing energy-efficient systems by optimizing the power factor.

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 2: VL × IL

This formula is incomplete and incorrect for a three-phase system. While it may represent apparent power in a single-phase system, it does not account for the phase relationships and power factor in a three-phase system. Therefore, this option cannot be used to calculate active power.

Option 3: 3 × VL × IL × cosΦ

This formula is incorrect for a three-phase system. The factor "3" implies direct addition of the power contributions from each phase without considering the √3 constant, which is necessary to relate line-to-line voltage and line-to-neutral voltage. Thus, this formula leads to incorrect results for active power calculation.

Option 4: VL × IL × cosΦ

This formula represents the active power calculation for a single-phase system, not a three-phase system. It does not incorporate the √3 constant required for three-phase systems, nor does it account for the contributions from all three phases. Hence, it is unsuitable for three-phase power calculations.

Conclusion:

The correct formula for calculating active power in a balanced three-phase system is P = √3 × VL × IL × cosΦ, as stated in Option 1. This formula accurately accounts for the contributions of all three phases, the relationship between line-to-line and line-to-neutral voltages, and the power factor. Understanding the correct formula is essential for efficient power system analysis and energy management in industrial and commercial applications.

• In a three-phase system, the order in which the voltages attain their maximum positive value is called Phase Sequence.
• There are three voltages or EMFs in the three-phase system with the same magnitude, but the frequency is displaced by an angle of 120° electrically.
• Consider the R, Y, and B the three phases of the supply system.
• Taking an example, if the phases of any coil are named as R, Y, B then the Positive phase sequence will be RYB, YBR, BRY also called clockwise sequence and similarly the Negative phase sequence will be RBY, BYR, YRB respectively and known as an anti-clockwise sequence.

• For a three-phase system, there are only two possible phase sequences RYB and RBY corresponding to the two possible directions of alternator rotation.
• Phase rotation has no impact on resistive loads, but it will have an impact on unbalanced reactive loads, as shown in the operation of a phase rotation detector circuit.
• Phase rotation can be reversed by swapping any two of the three terminals supplying three-phase power to a three-phase load.

Calculation:

Load impedance = 6 + j9

Line impedance = 1 + j2

Total impedance = 7 + j11

Magnitude of total impedance = 13.038 Ω

Supply voltage (V_S) = 400 V

Concept

The active power in a 3ϕ star-connected load is given by:

\(P=3V_pI_pcosϕ=\sqrt{3}V_LI_Lcosϕ\)

where, \(cosϕ={R\over \sqrt{R^2+X^2}}\)

Calculation

Given, Z = 4 + j3 Ω

\(cosϕ={4\over \sqrt{4^2+3^2}}=0.8\)

\(I_p={V_p\over Z}={230\over 5}\)

\(P=3\times 230\times {230\over 5}\times 0.8\)

P = 25.4 kW

Concept: 

Let the three-phase supply voltages be:

\(V_1=V_m\space sin(\omega t)\)

\(V_2=V_m\space sin(\omega t-120)\)

\(V_3=V_m\space sin(\omega t+120)\)

Now, the three-phase supply currents are:

\(i_1=I_m\space sin(\omega t-\phi)\)

\(i_2=I_m\space sin(\omega t-120-\phi)\)

\(i_3=I_m\space sin(\omega t+120-\phi)\)

The instantaneous power is:

\(P=V_mI_m[sin(\omega t) sin(\omega t-\phi)+sin(\omega t-120) sin(\omega t-120-\phi)+sin(\omega t+120) sin(\omega t+120-\phi)]\)

\(P=3V_mI_m\space cos\phi\) = constant

Additional InformationThe instantaneous power in a single-phase system has a sinusoidal variation with double the supply frequency.

The order in which the three-phase voltages attain their positive peak values is known as the phase sequence. Conventionally the three phases are designated as red-R, yellow-Y, and blue-B phases.

The phase sequence is said to be RYB if R attains its peak or maximum value first with respect to the reference as shown in the counter-clockwise direction followed by Y phase 120° later and B phase 240° later than the R phase.

Important Points:

• The phase sequence of the voltages applied to a load is determined by the order in which the 3 phase lines are connected
• The phase sequence can be reversed by interchanging any one pair of lines without causing any change in the supply sequence
• Reversal of sequence results in reversal of the direction of rotation in case of induction motor

Calculation:

In the given figure, a γ - connected balanced source is connected to γ - connected unbalanced load.

In a balanced source, voltages are equal in magnitude and phase displaced by 120°.

E_A = 100 ∠0°

E_B = 100 ∠-120°

E_C = 100 ∠120° = 100 ∠-240°

In γ - connection, phase voltage = \(\rm \frac{Line\ voltage}{\sqrt3}\)

Given, Z_A = 10 Ω = 10 ∠0°, ZB = 20 ∠60°

∴ \(\rm I_A=\frac{E_A}{Z_A}=\frac{100 ∠ 0^{\circ}}{10∠ 0^{\circ}}=10∠ 0^\circ\ \)

∴ \(\rm I_B=\frac{E_B}{Z_B}=\frac{100 ∠ -120^{\circ}}{20∠ 60^{\circ}}=5∠ -180^\circ\ \)

Since the potential difference between n and n' is zero

therefore,

\(\vec{I_A}+\vec{I_B}+\vec{I_C}=0\)

⇒ \(\vec{I_C}=-(\vec{I_A}+\vec{I_B})\ \)

\(=-(10∠ 0^{\circ}+5∠ -180^{\circ})\)

= 5 ∠180°

∴ \(\rm Z_C=\frac{E_C}{I_C}=\frac{100∠ -240^{\circ}}{5∠ 180^{\circ}}\)

= 20 ∠-60° Ω

Therefore, Correct option is (b)

Concept:

In a star-connected three-phase system,

V_L = √3 × V_ph

And I_L = I_ph

\({I_{ph}} = \frac{{{V_L}}}{{\surd 3Z}}\)

In a delta connected three-phase system,

V_L = V_ph

I_L = √3 × I_ph

\({I_{ph}} = \frac{{{V_L}}}{Z}\)

Where,

V_L is line voltage

V_ph is phase voltage

I_L is line current

I_ph is the phase current

Calculation:

The given load is star connected load.

V_L = 400 V

Inductive reactance X = 3 Ω

R = 4 Ω

Impedance Z_ph = 4 + j3 = 5∠36.86°

Phase voltage \({V_{ph}} = \frac{{400}}{{\surd 3}} = 230.94\;V\)

Concept:

In star connection,

\({V_{ph}} = \frac{{{V_L}}}{{\sqrt 3 }}\)

IL = Iph

In delta connection,

\({I_{ph}} = \frac{{{I_L}}}{{\sqrt 3 }}\)

VL = Vph

Calculation:

V_RN = 100 V

V_phY = 100 V

V_LY = 100√3 V

V_LΔ = 100√3 V

V_phΔ = 100√3 V

Load impedance, Z_L = (8 + j6) Ω/phase

\({I_{ph{\rm{\Delta }}}} = \frac{{{V_{ph{\rm{\Delta }}}}}}{Z} = \frac{{100\sqrt 3 }}{{\sqrt {{8^2} + {6^2}} }} = 10\sqrt 3 \;A\)

Instantaneous power :

• The product of the instantaneous voltage and the instantaneous current for a circuit.
• The power at any instant of time.
• In Ac circuit the instantaneous electric power is given by P = V I
• For a balanced three-phase system the instantaneous power at any instant of voltage or current will be same or the total power Pa + Pb + Pc = P = 3V_phI_ph cos ϕ will be same.

Instantaneous Power of three-phase balanced load

When phase A is at its peak value

P  = 2000 W 

So power 30° later of phase A 

P = 2000 W = 2 kW

Concept:

• The delta in a three-phase system is formed by connecting one end of the winding to the starting end of other winding and the connections are continued to form a closed loop.
• Line voltage in delta connection is equal to phase voltage. Therefore first statement is true.

​In a delta-connected three-phase system:

\(V_{line} = V_{phase}\)

\(I_{line} =\sqrt 3 I_{phase}\)

Total Power \(= √3 × V_L × I_L×cosϕ\)

• The figure given below shows the phasor diagram of Line and Phase voltages in ABC phase sequence:

• For abc phase sequence, Line voltages of each phase will always lead the phase voltages by 30^o. Therefore the second statement is false.