Hello Sir...thank you for your interest in my RU-vid site. It was put together a while ago with an interest in reaching out to those interested in power systems. I have since expanded these as part of 14 courses related to electrical power and they are located on EEP - (Electrical Engineering Portal). My courses may be found by going to: academy.electrical-engineering-portal.com/courses...and selecting Graham Van Brunt P. Eng., B.Sc. after hovering over the "Authors" heading. The costs look expensive, but they are always heavily discounted. You will not be disappointed in any of them.
Hello...thank you for your interest in my RU-vid site. It was put together a while ago with an interest in reaching out to those interested in power systems. I have since expanded these as part of 14 courses related to electrical power and they are located on EEP - (Electrical Engineering Portal). My courses may be found by going to: academy.electrical-engineering-portal.com/courses...and selecting Graham Van Brunt P. Eng., B.Sc. after hovering over the "Authors" heading. The costs look expensive, but they are always heavily discounted. You will not be disappointed in any of them.
The equations are not messed up...each component has an A, B & C phase. The Positive sequence A, B & C phases are separated by 120 degrees as the Negative sequence A, B & C phases are separated by 120 degrees. The 0 sequence A, B & C phases are in phase with each other. The A phases of each component will vary relative to each other as do the B & C phases of each component.
If you plug in the phasors into the expressions, you cannot possibly arrive at the results. Your results are correct, but you have displayed the expressions incorrectly
Hello Sir...thank you for your interest in my RU-vid site. It was put together a while ago with an interest in reaching out to those interested in power systems. I have since expanded these as part of 14 courses related to electrical power and they are located on EEP - (Electrical Engineering Portal). My courses may be found by going to: academy.electrical-engineering-portal.com/courses...and selecting Graham Van Brunt P. Eng., B.Sc. after hovering over the "Authors" heading. The costs look expensive, but they are always heavily discounted. You will not be disappointed in any of them.
Very intuitive explanations and very profesional. Can you just explain further on the start of 39:08, on how a to power negative2 become simply a, thanks.
From 16:40 u can see that a^(-1) = a^2, from this we understand that a^2=1/a hence while finding a^(-2) we substitute 1/a instead of a^2 in the equation a^(-2)=1/a^2 therefore we get a^(-2)=1/(1/a) which is nothing but 'a'
Hello...thank you for your interest in my RU-vid site. It was put together a while ago with an interest in reaching out to those interested in power systems. I have since expanded these as part of 14 courses related to electrical power and they are located on EEP - (Electrical Engineering Portal). My courses may be found by going to: academy.electrical-engineering-portal.com/courses...and selecting Graham Van Brunt P. Eng., B.Sc. after hovering over the "Authors" heading. The costs look expensive, but they are always heavily discounted. You will not be disappointed in any of them.
