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

3ϕ balanced system

In a balanced three-phase system, the sum of the three phase voltage phasors is zero. This is a fundamental property of balanced three-phase systems. 

Explanation:

• In a balanced three-phase system, the three phase voltages have equal magnitudes and are 120 degrees apart from each other. 
• When the three phasors are added vectorially, they will cancel each other out, resulting in a sum of zero. 

Concept

The capacitive reactance XC for a series RC circuit is given by:

\(X_C={1\over 2\pi fC}\)

From the above expression, we found that the frequency is inversely proportional to the capacitive reactance.

Calculation

Given, f₂ = 2f₁

\({X_{C2}\over X_{C1}}={f_1\over f_2}\)

\({X_{C2}\over X_{C1}}={f_1\over 2f_1}\)

\({X_{C2}}={X_{C1}\over 2}\)

If the frequency of an AC source connected to a pure capacitor is doubled, the capacitive reactance will decrease by half.

3ϕ star connection with neutral

The neutral current is given by:

\(I_N=I_R+I_Y+I_B\)

• If all phase loads are equal, the currents in each phase are balanced. The neutral point stays stable (zero potential shift), and the phase voltages remain symmetrical.
• However, if the loads on the three phases are unequal, this causes the neutral point to "shift" from its ideal position — this is called neutral shift.
• So, by distributing the total load equally across the three phases, you eliminate the root cause of neutral shift.

Explanation:

Three-Phase Balanced Star-Connected Load

Definition: In a three-phase balanced star-connected load, each phase carries the same current, and the phase voltages are equal in magnitude but are phase-shifted by 120 degrees from each other. The star connection means that one end of each of the three loads is connected to a common point called the neutral point.

Working Principle: In a star connection, the line voltage (the voltage between any two lines) is √3 times the phase voltage (the voltage between any line and the neutral point). For a balanced load, the currents in each phase are equal in magnitude and phase angle, and the vector sum of the currents at the neutral point is zero. This means that the neutral point is at the same potential as the ground in a balanced system.

Advantages:

• Provides a stable neutral point.
• Allows for the use of a lower voltage between any phase and neutral, which can be safer for certain applications.
• Facilitates the distribution of both single-phase and three-phase power.

Disadvantages:

• Requires more conductors compared to a delta connection for the same power level.
• In case of an unbalanced load, the neutral point can shift, causing voltage imbalances.

Applications: Three-phase star-connected systems are widely used in power transmission and distribution, as well as in various industrial applications where balanced loads are common, such as in motors and other heavy machinery.

Correct Option Analysis:

The correct option is:

Option 1: 0 V

This option correctly describes the potential of the neutral point of the load with respect to the ground in a balanced three-phase star-connected system. Since the system is balanced, the vector sum of the currents at the neutral point is zero, making the neutral point at the same potential as the ground.

Additional Information

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

Option 2: 400 V

This option is incorrect because 400 V is the line voltage in the three-phase system, not the potential of the neutral point. The line voltage is the voltage between any two lines, and in a balanced system, the neutral point is at 0 V with respect to the ground.

Option 3: 230 V

This option is incorrect as well. 230 V is approximately the phase voltage of the system (400 V / √3), but it is not the potential of the neutral point. The neutral point in a balanced system remains at 0 V with respect to the ground.

Option 4: 680 V

This option is incorrect because 680 V is not a standard voltage value in a three-phase system and does not relate to the potential of the neutral point. The neutral point potential is 0 V in a balanced star-connected system.

Conclusion:

Understanding the behavior of a three-phase balanced star-connected load is essential for correctly identifying the potential of the neutral point. In such a system, the neutral point is at 0 V with respect to the ground due to the balanced nature of the load, where the vector sum of the currents at the neutral point is zero. This makes the correct option 1, which states that the potential of the neutral point is 0 V.

Explanation:

In a three-phase system, the delta connection is one of the two standard methods of connecting the windings of a three-phase system, the other being the star (or wye) connection. The delta connection is characterized by the end of each winding being connected to the start of the next, forming a closed loop or triangle. This configuration has several implications for the relationship between the phase currents and the line currents in the system.

Delta Connection:

In a delta connection, the three windings of the electrical system are connected end-to-end to form a closed loop. Each corner of the delta is connected to a line conductor. The voltage across each winding is the same as the line voltage, but the currents in the windings (phase currents) and the currents in the line conductors (line currents) have a specific relationship that is different from the star connection.

Phasor Relation between Line Current and Phase Current:

In a delta connection, the line current (IL) is related to the phase current (Iph) by both magnitude and phase angle. The key points to understand are:

• The line current is the vector sum of the currents in the two windings that meet at the corresponding line connection point.
• The magnitude of the line current is √3 times the magnitude of the phase current.
• The line current lags the phase current by 30 degrees.

To illustrate, consider the following:

• If Iph is the current in one phase winding, the line current IL can be found by summing the currents from the two windings that share a common point at the line connection. This involves vector addition, taking into account the phase angles.
• Mathematically, this can be expressed as IL = √3 * Iph, with a phase difference where IL lags Iph by 30 degrees.

Correct Option Analysis:

The correct option is:

Option 2: IL lags Iph by 30 degrees.

This option correctly describes the phasor relationship between the line current and the phase current in a delta connection. The line current is not only √3 times the magnitude of the phase current but also lags the phase current by 30 degrees.

Additional Information

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

Option 1: IL leads Iph by 30 degrees.

This option is incorrect because, in a delta connection, the line current actually lags the phase current by 30 degrees, not leads.

Option 3: IL is in phase with Iph.

This option is incorrect as well because the line current is not in phase with the phase current. There is a 30-degree phase shift where the line current lags the phase current.

Option 4: IL leads Iph by 90 degrees.

This option is incorrect because a 90-degree phase shift is not characteristic of the relationship between line current and phase current in a delta connection.

Conclusion:

Understanding the phasor relationships in a three-phase delta connection is crucial for analyzing and designing electrical systems. The correct relationship is that the line current lags the phase current by 30 degrees and has a magnitude of √3 times the phase current. This fundamental knowledge helps in ensuring proper system operation and troubleshooting in practical 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.