Delta 3 phase power

[huntfishride247]

I am new to all of this delta and wye stuff. I have a basic knowledge of Ohms Law etc. One of my homework questions is as follows: (I dont expect you to do my homework for me) Three 10 ohm resistors are connected in delta on a 208V, 3 phase line. What is the power supplied to the line? I am assuming that it is a balanced load. I know in a delta S= E x I x square root of 3 so 208V/ 30 ohms = 6.9333333 A. Using the formula I get 2494.89 Watts. If the question means there are 10 ohms in each line 208V/ 10ohm = 20.8A Using the formula I get 7484.67 Watts. Their answer (in book) is 13kW. What am I doing wrong? Any help would be greatly appreciated.

 

[Besoeker]

I am new to all of this delta and wye stuff. I have a basic knowledge of Ohms Law etc. One of my homework questions is as follows: (I dont expect you to do my homework for me) Three 10 ohm resistors are connected in delta on a 208V, 3 phase line. What is the power supplied to the line? I am assuming that it is a balanced load. I know in a delta S= E x I x square root of 3 so 208V/ 30 ohms = 6.9333333 A. Using the formula I get 2494.89 Watts. If the question means there are 10 ohms in each line 208V/ 10ohm = 20.8A Using the formula I get 7484.67 Watts. Their answer (in book) is 13kW. What am I doing wrong? Any help would be greatly appreciated.

Each resistor has 208V across it.
Go from there.

 

[charlie b]

Three 10 ohm resistors are connected in delta on a 208V, 3 phase line. . . . I am assuming that it is a balanced load.

Since they told you that you had three equal resistors in this connection pattern, you don’t need to assume the load is balanced. It is, because they told you it is.

I know in a delta S= E x I x square root of 3 so 208V/ 30 ohms = 6.9333333 A.

Why did you add the three resistors, 10 ohms each, to get a total of 30 ohms? That only works if the resistors are in series. These are not. For things to be in series, current that passes through one must have no other option than to go through the next one, and then through the next one. They are connected in a delta. That means that current passing through one can go through the second or through the third or through the line leading back to the source. If current has a choice of paths to take, then you don’t have a series circuit.

Their answer (in book) is 13kW. What am I doing wrong?

The voltage across each of the resistors is 208 volts. Power equals (voltage squared) divided by resistance. P = (208) * (208) / 10 = 4,326 watts. You have three of them, and power adds arithmetically (no vectors or phase angles to play with). Three times 4,326 equals 12,979 watts, or about 13kW.

Welcome to the forum.

 

[huntfishride247]

I appreciate the help. I guess I just wasn't thinking. With all the formulas that they are teaching me it gets all confusing for me. I was trying to find the current then use the formula S=E*I*1.73. I thought the question was saying there was three 10 ohm resistors on one line. I have been told that I make things harder then they really are. Hopefully I will be able to do the rest of my homework on my own. I really have no one to ask for help on electrical problems. So I didn't know where to turn. Again thanks for the input.