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- 2013/04/22 at 6:26 pm #10990
My first engineering job out of school is designing electrical panels for industrial process equipment. My present learning experience involves transformers. According to my calculations, I have a total of 3.75 A being drawn on the secondary side of a transformer. The transformer is 480 VAC primary/120 VAC secondary and is rated 1kVA. So this 3.75 A would result in a 15 A draw on the primary side, would it not? And assuming a purely resistive load, that would equal 15 A x 480 VAC = 7.2kVA, no? There were originally 4 A fuses on each of the two lines leading to the transformer primary side. I figured my 15A x 1.2 safety factor / 2 lines = 9 A, so I thought the fuses should be 10 A rather than 4. But I’m thinking my real problem is that my transformer is too small. Where is my thought process breaking down here? Thank you in advance for setting me straight.

2013/04/24 at 12:00 pm #12991@christopher said:

My first engineering job out of school is designing electrical panels for industrial process equipment. My present learning experience involves transformers. According to my calculations, I have a total of 3.75 A being drawn on the secondary side of a transformer. The transformer is 480 VAC primary/120 VAC secondary and is rated 1kVA. So this 3.75 A would result in a 15 A draw on the primary side, would it not? And assuming a purely resistive load, that would equal 15 A x 480 VAC = 7.2kVA, no? There were originally 4 A fuses on each of the two lines leading to the transformer primary side. I figured my 15A x 1.2 safety factor / 2 lines = 9 A, so I thought the fuses should be 10 A rather than 4. But I’m thinking my real problem is that my transformer is too small. Where is my thought process breaking down here? Thank you in advance for setting me straight.Dear ;

As I understood of you is you have the following :

1- Transformer :

**S**= 1 kVA,**V**p = 480 V,**V**s = 120 V, where :- i. “
**V**p ” : the value of Primary’s Voltage - ii. “
**V**s ” : the value of Secondary’s Voltage

2- The current of loads is “ 3.75 A ”.

According to the Transformer’s specifications the maximum current that can be drawn from Secondary side is :

**I**s max =**S**/**V**s = 1000 / 120 =**8.333**ASo, the Power of the Transformed is enough to supply this load.

By the way, and sorry to say that your calculation of the load’s current from Primary side is not correct, because that current can be calculated as follow :

(

**V**p**/****V**s ) = (**I**s**/****I**p )480

**/**120 = 3.75**/****I**pThen :

**I**p = 3.75 ( 120**/**480 ) è**I**p**0.9375**AThat means the Fuses ” 4A ” at primary size doesn’t blow.

In reality, the real drawn current of Primary side is “

**0.9375**+**Transformer Losses**”.Rem. : if the Voltage’s values of Primary & Secondary are different, the calculation should be modified.

2013/04/24 at 1:31 pm #12992Thank you for your response, especially for the extremely polite way you pointed out my mistake. =) This confirms what I have learned since posting this topic. I had the current ratio inversed with respect to the voltage ratio, which was giving me the impression that primary side current would be 4 times the secondary, instead of one fourth. This, of course, contradicts calculations for current given the voltages and power rating of the transformer. I wish I could say this is the last silly mistake I’ll ever make…

2013/04/25 at 8:52 am #12970Dear Mr. Christopher ;

Please note that the equation mentioned in my reply is general, and the complete one is :

(

**V**p**/****V**s ) = (**I**s**/****I**p ) = (**N**p**/****N**s )where :

**N**p : the number of turns of Primary coil.**N**s : the number of turns of secondary coil.2013/08/06 at 11:42 am #13341Me i am planning to study more about it in school and hopefully learn a lot about transformer.

- i. “
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