Alternators
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.
True
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
138.57
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.
75
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.
False
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).
False
Motor power factor can be computed using this formula: PF = VA / P.
False
The amount of phase current can be determined using the Ohm's law equation, IP(Load) = Z / EP(Load).
False
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.
False
The line current of a delta connection is higher than the phase current by a factor of the square root of 3 (1.732).
True
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.
True
The wye, or ____, connection is made by connecting one end of each of the three-phase windings together.
Star
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 ____.
1.732
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.
220
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
23.1
A formula used to compute the voltage in a wye-connected system is ____.
EPhase = ELine / 1.73
The inductive VARs in a three-phase circuit can be computed using the formula VARsL equals the square root of VA2 - P2.
True
In a wye-connected system, phase current and line current are the same.
True
