Question 1

The reason for the difference between line current and phase current in a delta connection is that current flows through different windings at different times in a three-phase circuit.

Accepted Answer

In a delta connection, the line current is the sum of two phase currents, but the phase currents are not equal in magnitude or direction. This is because the current flows through different windings at different times, resulting in a phase shift of 120 degrees between each winding.

Question 2

A three-phase motor is connected to a 480V, 60-Hz line. A clamp-on ammeter indicates a running current of 68A at full load, and a three-phase wattmeter indicates a true power of 40,277W. To solve this problem, first find the amount of apparent power in the circuit using the formula ____.
A) VA = 1.732 / (EL x IL)
B) VA = (EL x IL) / 1.732
C) VA = 1.732(EL + IL)
D) VA = 1.732 x EL x IL

Accepted Answer

The correct formula for calculating the apparent power (VA) in a three-phase circuit is VA = 1.732 x EL x IL, where EL is the line voltage and IL is the line current.

Question 3

A formula used to compute the voltage in a wye-connected system is ____.
A) EPhase = ELine / 1.732
B) EPhase = ELine x 1.732
C) EPhase = 1.732 / ELine
D) ELine = EPhase / 1.732

Accepted Answer

The formula to find voltage in a wye-connected system involves dividing the line voltage by 1.732 (the square root of 3), which gives the phase voltage. Therefore, the correct formula is EPhase = ELine / 1.732.

Question 4

The line current of a delta connection is higher than the phase current by a factor of the square root of 3 (1.732).

Accepted Answer

This is a well-known fact in electrical engineering that the line current of a delta connection is sqrt(3) times the phase current.

Question 5

The wye, or ____, connection is made by connecting one end of each of the three-phase windings together.
A) triangle
B) square
C) star
D) rectangle

Accepted Answer

The wye, or star, connection is the standard method for connecting three-phase windings, where one end of each winding is connected to form a common point. This results in a balanced load and allows for the neutral point to be grounded for safety. The other options (triangle, square, rectangle) are not commonly used for three-phase connections.

Question 6

Correcting the power factor of a three-phase circuit is similar to the procedure used to correct the power factor of a single-phase circuit.

Accepted Answer

Correcting the power factor of a three-phase circuit involves using capacitors or other devices to reduce reactive power and improve the overall power factor, similar to the process used in a single-phase circuit. However, the actual implementation may be different due to the different configurations and requirements of three-phase systems.

Question 7

The amount of phase current can now be determined using the Ohm's law equation, IP(Load) = Z / EP(Load).

Accepted Answer

The correct Ohm's law equation for determining the amount of phase current is IP(Load) = EP(Load) / Z, where IP(Load) is the phase current, EP(Load) is the phase voltage, and Z is the impedance.

Question 8

In a wye-connected system, the line voltage is lower than the phase voltage by a factor of the square root of 3 (1.732).

Accepted Answer

In a wye-connected system, the line voltage is higher than the phase voltage by a factor of the square root of 3 (1.732).

Question 9

The phase windings of an alternator are connected in wye. The alternator produces a line voltage of 440V and supplies power to two resistive loads. One load contains resistors with a value of 4 ohm each, connected in wye. The second load contains resistors with a value of 6 ohm each, connected in delta. Both loads are connected directly to the output of the alternator. To determine the amount of line current needed for load 2, use the following formula: IL(Load 2) = IP(Load 2) x ____.
A) 1.732
B) 3
C) 73.33
D) 440

Accepted Answer

Since Load 1 is connected in wye, its current will directly contribute to the line current. The current flowing through each resistor in Load 1 will be Vline/4 = 440/4 = 110 A (where Vline is the line voltage). Since there are three such resistors in parallel, the total current contributing to the line current will be 3x110 = 330 A.

For Load 2, we need to determine the phase current (IP) and then use the appropriate formula to find the line current (IL). Since Load 2 is connected in delta, the phase current will be equal to the line current (assuming a balanced load). To determine the phase current, we first need to find the equivalent resistance of the delta load. The equivalent resistance of a delta-connected load is simply equal to the resistance of each resistor in the load. Therefore, the equivalent resistance of Load 2 is 6 ohms (since each resistor has a value of 6 ohms).

The phase voltage (VP) of the delta load will be Vline/sqrt(3) = 440/sqrt(3) = 254 V (where sqrt(3) is the square root of 3). Using Ohm's Law, the phase current flowing through the delta load will be IP = VP/Req = 254/6 = 42.33 A.

To find the line current for Load 2, we can use the formula IL = sqrt(3) x IP. Plugging in the value of IP, we get:

IL(Load 2) = sqrt(3) x 42.33 = 73.33 A

Therefore, the answer is A) 1.732.

Question 10

The horsepower rating of three-phase motors and the kilovolt-amp rating of three-phase transformers are about 75% greater than for single-phase motors or transformers with a similar frame size.

Accepted Answer

The horsepower rating of three-phase motors and the kilovolt-amp rating of three-phase transformers are generally about 1.5 times (or 50% greater) than their single-phase counterparts for a similar frame size, not 75% greater.

Question 11

A delta-connected alternator is connected to a wye-connected resistive load. The alternator produces a line voltage of 240V, and the resistors have a value of 6 ohm each. If EL(Load) = 240V, then EP(Load) = ____V.
A) 80
B) 138.57
C) 160
D) 415.68

Accepted Answer

In a wye-connected system, the line voltage (EL) is related to the phase voltage (EP) by the equation EL = √3 * EP. Given that EL(Load) = 240V, solving for EP gives EP = EL / √3 = 240V / √3 ≈ 138.57V.

Question 12

If the phase values of voltage and current in a three-phase circuit are known, the apparent power can be computed using the formula VA = ____.
A) 3 x EPhase x IPhase
B) (EPhase x IPhase) / 3
C) 3 / (EPhase x IPhase)
D) EPhase x IPhase

Accepted Answer

The apparent power (VA) in a three-phase circuit is calculated using the formula VA = 3 x EPhase x IPhase, where EPhase is the phase voltage and IPhase is the phase current. This formula accounts for the three phases in the system, hence the multiplication by 3.

Question 13

A wye-connected three-phase alternator supplies power to a delta-connected resistive load. The alternator has a line voltage of 480V. Each resistor of the delta load has 8 ohm of resistance. The load is connected directly to the alternator. Therefore, the line voltage supplied by the alternator is 160V.

Accepted Answer

The line voltage supplied by the alternator would be the same as the phase voltage since the load is delta-connected. Therefore, the line voltage would be 480V, not 160V.

Question 14

One reason why three-phase power is superior to single-phase power is that in a balanced three-phase system, the conductors need be only about ____% of the size of conductors for a single-phase two-wire system of the same kVA (kilovolt-amp) rating.
A) 33.3
B) 60
C) 75
D) 85

Accepted Answer

The answer of One reason why three-phase power is superior...

Question 15

The phase windings of an alternator are connected in wye. The alternator produces a line voltage of 440V and supplies power to two resistive loads. One load contains resistors with a value of 4 ohm each, connected in wye. The second load contains resistors with a value of 6 ohm each, connected in delta. Both loads are connected directly to the output of the alternator. The line voltage for both loads 1 and 2 will be the same, ____V.
A) 73.33
B) 219.09
C) 220
D) 440

Accepted Answer

When connected in wye, the line voltage of the alternator is equal to the phase voltage. Therefore, the phase voltage of the alternator is 440V / √3 = 254.03V.

For load 1 (connected in wye), the line voltage is equal to the phase voltage. Therefore, the line voltage for load 1 is also 254.03V.

For load 2 (connected in delta), the line voltage is √3 times the phase voltage. Therefore, the line voltage for load 2 is 254.03V x √3 = 440V.

Since both loads are connected directly to the output of the alternator, they share the same line voltage. Therefore, the line voltage for both loads is 254.03V, or approximately 220V. The closest answer choice is C, 220V.

Question 16

Motor power factor can be computed using this formula: PF = VA / P.

Accepted Answer

The correct formula for power factor (PF) is PF = P / VA, where P is the real power in watts and VA is the apparent power in volt-amperes.

Question 17

The inductive VARs in a three-phase circuit can be computed using the formula VARsL equals the square root of VA2 - P2.

Accepted Answer

The formula VARsL = √(VA² - P²) is used to calculate the reactive power (VARs) in a circuit, where VA is the apparent power and P is the real power.

Question 18

A delta-connected alternator is connected to a wye-connected resistive load. The alternator produces a line voltage of 240V, and the resistors have a value of 6 ohm each. Using Ohm's law, we determine the amount of phase current, IP(Load) = ____A.
A) 13.3
B) 23.1
C) 26.6
D) 69.28

Accepted Answer

In a wye-connected load, the phase voltage is the line voltage divided by √3. Thus, the phase voltage is 240V/√3 = 138.56V. Using Ohm's law, the phase current is 138.56V / 6Ω = 23.1A.

Question 19

In a(n) ____ connection, line voltage and phase voltage are the same.
A) alpha
B) beta
C) delta
D) gamma

Accepted Answer

In a delta connection, line voltage and phase voltage are the same. This means that the voltage across each load is equal to the voltage between any two line wires. In contrast, in an alpha, beta, or gamma connection, line voltage is higher than phase voltage.

Question 20

In a wye-connected system, phase current and line current are the same.

Accepted Answer

In a three-phase wye-connected system, the phase current is the current flowing through each individual phase of the load, while line current is the current flowing through the power supply lines. In a balanced wye-connected system, where the load is evenly distributed among the three phases, the phase current and the line current are equal. Therefore, in this case, the statement is true.

Quiz 7: Three-Phase Circuits

The reason for the difference between line current and phase current in a delta connection is that current flows through different windings at different times in a three-phase circuit.

A three-phase motor is connected to a 480V, 60-Hz line. A clamp-on ammeter indicates a running current of 68A at full load, and a three-phase wattmeter indicates a true power of 40,277W. To solve this problem, first find the amount of apparent power in the circuit using the formula ____.

A formula used to compute the voltage in a wye-connected system is ____.

The line current of a delta connection is higher than the phase current by a factor of the square root of 3 (1.732).

The wye, or ____, connection is made by connecting one end of each of the three-phase windings together.

Correcting the power factor of a three-phase circuit is similar to the procedure used to correct the power factor of a single-phase circuit.

The amount of phase current can now be determined using the Ohm's law equation, IP(Load) = Z / EP(Load).

In a wye-connected system, the line voltage is lower than the phase voltage by a factor of the square root of 3 (1.732).

The horsepower rating of three-phase motors and the kilovolt-amp rating of three-phase transformers are about 75% greater than for single-phase motors or transformers with a similar frame size.

A delta-connected alternator is connected to a wye-connected resistive load. The alternator produces a line voltage of 240V, and the resistors have a value of 6 ohm each. If EL(Load) = 240V, then EP(Load) = ____V.

If the phase values of voltage and current in a three-phase circuit are known, the apparent power can be computed using the formula VA = ____.

One reason why three-phase power is superior to single-phase power is that in a balanced three-phase system, the conductors need be only about ____% of the size of conductors for a single-phase two-wire system of the same kVA (kilovolt-amp) rating.

Motor power factor can be computed using this formula: PF = VA / P.

The inductive VARs in a three-phase circuit can be computed using the formula VARsL equals the square root of VA2 - P2.

A delta-connected alternator is connected to a wye-connected resistive load. The alternator produces a line voltage of 240V, and the resistors have a value of 6 ohm each. Using Ohm's law, we determine the amount of phase current, IP(Load) = ____A.

In a(n) ____ connection, line voltage and phase voltage are the same.

In a wye-connected system, phase current and line current are the same.