re: formula wheel
I teach an electrical apprenticeship class.
I was teaching basic electrical math. We were on the section of the PIE and the EIR formulas. One of my students said that they had a customer ask them if the voltage was low to their business and the amperage went up would their electric bill be any higher? Of course he said yes!!!
Since the question came up I proceeded to show them that if the amperage went up and the voltage went down that there would be no change in KW on the service. Since the power company bills based on KW this would not effect the amount of their bill.
Since I like to do things a little different in class I gave them a question on the test to see if they were listening or just reading the lesson out of the book . The question goes as follows
An industrial service has a voltage of 465V and an amperage of 1350A. If the voltage is raised to 487V what is the new amperage on the service.
465V x 1350A = 627,750W
627,750W/ 487V = 1289.01A
One of my students gave the answer as follows
465V/ 1350A = 0.34 ohms
487/ 0.34 ohms = 1432.35A
Since he figured it with a formula I gave him, how do I explain that this method is wrong. Everyone else in the class got it right.
My next question was as follows;
If the voltage on a circuit to a piece of equipment is raised; what does this do to the amperage?
Everyone answered "lowers" including the one that missed the other question.
All opinions and reasons are greatly appreciated!
I teach an electrical apprenticeship class.
I was teaching basic electrical math. We were on the section of the PIE and the EIR formulas. One of my students said that they had a customer ask them if the voltage was low to their business and the amperage went up would their electric bill be any higher? Of course he said yes!!!
Since the question came up I proceeded to show them that if the amperage went up and the voltage went down that there would be no change in KW on the service. Since the power company bills based on KW this would not effect the amount of their bill.
Since I like to do things a little different in class I gave them a question on the test to see if they were listening or just reading the lesson out of the book . The question goes as follows
An industrial service has a voltage of 465V and an amperage of 1350A. If the voltage is raised to 487V what is the new amperage on the service.
465V x 1350A = 627,750W
627,750W/ 487V = 1289.01A
One of my students gave the answer as follows
465V/ 1350A = 0.34 ohms
487/ 0.34 ohms = 1432.35A
Since he figured it with a formula I gave him, how do I explain that this method is wrong. Everyone else in the class got it right.
My next question was as follows;
If the voltage on a circuit to a piece of equipment is raised; what does this do to the amperage?
Everyone answered "lowers" including the one that missed the other question.
All opinions and reasons are greatly appreciated!