Hi sir, may i clarify something regarding Q3 and V3 equations at bus 3. Why Y31 or Y13 and V1 were been considered since bus 3 is not connected with bus 1? Is it not only the mutual admittances, Y32 or Y23, as well as the self-admittance, Y33, should be used in bus 3?
In the previous lecture of gsiedal where s= 0.6 + 0.3i , why didn't u take - (0.6 + 0.3i) while calculating as in that question this was the output power and not injected power...? Please reply
Sir. In order to find V3 we need Q3. However in the Q3 equation there is delta3 ( angle for V3). How does this possible because we do not know delta3 but it is in the Q3 equation. Does it assumes to be 0 degree in Q3 equation?
since delta3 is angle of V3, we can use the older value of V3 and obtain its angle delta3 and use it for calculation of Q3. Then we use that Q3 for calculation of new value of V3. (which includes the new value of delta3).
Hye Sir. In L12 lecture at minute 1:06:43 when you were calculating Q3, delta3 was taken to be 0 degree. Referring to your previous reply, In this case the older value of V3 is 0.98 with the delta3 unknown? A bit confuse because delta3 is the unknown and only the mag of V3 defined eaerlier.
Great work sir...best video for this topic!
Thanks a lot....u saved my time and mind 😀😀😀
Thanku sir you provide a good lecture
Why the current direction is always towards bus instead of load.
Thank you very much sir.
Hi sir, may i clarify something regarding Q3 and V3 equations at bus 3. Why Y31 or Y13 and V1 were been considered since bus 3 is not connected with bus 1? Is it not only the mutual admittances, Y32 or Y23, as well as the self-admittance, Y33, should be used in bus 3?
Also while calculating Q3 why u didn't take y31 its conjugate value why u only take its magintude
while solving manually what should be written in place of delta 2 and delta 3 while calculating q3..since v2 and v3 are unknown
you can start the iterations with delta2 = 0 and delta3 = 0
In the previous lecture of gsiedal where s= 0.6 + 0.3i , why didn't u take - (0.6 + 0.3i) while calculating as in that question this was the output power and not injected power...?
Please reply
i love u man thank u
super explanation sir
why get a complex conjugate of V1 and V2 ie apostrophe used in
code line:
disp([V2 V3]')
transpose
matrix
Simple thing , complex explanation.
Sir. In order to find V3 we need Q3. However in the Q3 equation there is delta3 ( angle for V3). How does this possible because we do not know delta3 but it is in the Q3 equation. Does it assumes to be 0 degree in Q3 equation?
since delta3 is angle of V3, we can use the older value of V3 and obtain its angle delta3 and use it for calculation of Q3. Then we use that Q3 for calculation of new value of V3. (which includes the new value of delta3).
Hye Sir.
In L12 lecture at minute 1:06:43 when you were calculating Q3, delta3 was taken to be 0 degree.
Referring to your previous reply, In this case the older value of V3 is 0.98 with the delta3 unknown?
A bit confuse because delta3 is the unknown and only the mag of V3 defined eaerlier.
you can get the value from the initial guess which specify the values of magnitude and angles of all the bus voltages
SIR ,final value of Q3 ????
Final Value of Q3 is -0.11991 pu. That means bus 3 actually consuming reactive power.
v2 = 1.0058 at -1.9943 degrees
v3 = 0.9800 at 0.91348 degrees
thanks for reply sir
it means generator at bus 3 is consuming power ... am i right sir?
it means generator at bus 3 is consuming "reactive" power and at the same time supplying "active" power. S3 = (0.25 - j 0.11991) pu
k sir , i understood thankssssss .....
nice
15:53 i lost my mind
Tadel terefe
Sir dont use of matlab explain the gauss seidal method
Honestly you should try actually solving them. Matlab or other software's aren't allowed in quizes.
Ts
Wah sir simple chiz ko complex karna koi apse sikhe
why get complex conjugate transpose of V2 and V3
ie the apostrophe in code line
"disp([V2 V3]')"
Ts