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

Question

Quizplus · 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.

The Phase Windings of an Alternator Are Connected in Wye