I’m currently taking a course titled, “Principles of Industrial Electric Power” and so far we’ve been learning about three phase systems and how to calculate line and phase values. On my most recent quiz, I noticed one of the questions sparked some controversy. Here’s the problem in question:
“A delta-connected three-phase alternator is driving a wye-connected resistive load. The alternator output voltage is 480 V. The load resistors are 12 Ω each. What is the phase current in the alternator?”
Here’s a discussion where three people convinced each other that they were correct even though their answers had been marked incorrect.
We know that the line voltage is 480V because that is what is being supplied by the alternator. Then it says, that the load resistors are each 12ohms. Let’s draw what we have so far.
Given this information, it makes the most sense to start finding the values for the resistive load. Let’s start with the phase voltage. In wye connections, the phase voltage is calculated by dividing the line voltage by the square root of three.
480V= Ep x √3
Ep = 480V / √3
So, we know that the phase voltage of the load is about 277.128V. Using ohms law, we can calculate the phase current.
277.128V = 12 ohms x I
I = 277.128V / 12 ohms
This tells us that the phase current is about 23.094A.
We know that three phase systems are 120 degrees out of phase with each other so we can use the law of cosines to calculate the line current. Refer to the image below.
If you had used exact values instead of estimations, you would get a clean 40A. This method for finding line values is tedious, you could also just multiply the 23.094A by the square root of 3 and get the same result. This proves that the load draws 40 amps of current or said another way, the line current is 40A.