Wow, what a huge amount of manual bean counting to get to the answer. Two conversions to orthogonal coordinates for phasor addition, to get each of three lines. Its nice to know how this is done, but thank goodness for computers....
For unbalanced circuits voltage will not be apportioned equally as demonstrated by the superposition theorem. AC power uses the relative phase shift between voltage and current. Check out the AC power lecture at: m.ua-cam.com/video/nse2vSlOoME/v-deo.html
Yes, you'd think the two identical impedances would carry the same current but they don't because imbalance current continues to circulate unevenly throughout the circuit. Another interesting observation you can make about unbalanced 3 wire Ys is that the "Y" connections (ie: the far right node and far left node) actually exhibit a voltage difference between them. (it's largely this reason why unbalanced 4 wire Ys aren't as difficult to analyze, the 4th neutral wire ensures the two Ys are at the same potential). To determine the voltage differential between the Ys use Kirchhoff's Voltage Law. Let's call the node on the left A and the node on right B. Start at A, rise L2, fall V2, fall VAB. VAB would equal L2 -V2 = 20.8/-138.7. Other loops should yield identical results.
Ordinarily I don’t play “math detective” however this one is too good. The issue you’re having most likely stems from the last inverse tangent. Recall in the rectangular to polar conversion lecture: m.ua-cam.com/video/spnKjJtV5y4/v-deo.html that the inverse tangent will give you the “nearest” angle and not the “only” angle. This, being the second phase, should have a phase shift of close to -120 in the third quadrant and your answer will require active intervention. That’s why I’d recommend getting a scientific calculator or using an online resource that will do polar math without having to resort to manual rectangular operations and subsequent conversion. While not impossible, this manner of calculation is like traveling cross country on a donkey. You’ll get there ... but it’ll take a long time.
Wow, what a huge amount of manual bean counting to get to the answer. Two conversions to orthogonal coordinates for phasor addition, to get each of three lines. Its nice to know how this is done, but thank goodness for computers....
@11:03 Why is V2 and V3 not equal to the given voltage? Also why is I2 and I3 used in solving the powers not matching to the solved values
For unbalanced circuits voltage will not be apportioned equally as demonstrated by the superposition theorem. AC power uses the relative phase shift between voltage and current. Check out the AC power lecture at: m.ua-cam.com/video/nse2vSlOoME/v-deo.html
At around 25:25 why isn’t I2 and I3 the same? I understand the maths and got the same results as you this surprised my instinctual expectation
Yes, you'd think the two identical impedances would carry the same current but they don't because imbalance current continues to circulate unevenly throughout the circuit.
Another interesting observation you can make about unbalanced 3 wire Ys is that the "Y" connections (ie: the far right node and far left node) actually exhibit a voltage difference between them. (it's largely this reason why unbalanced 4 wire Ys aren't as difficult to analyze, the 4th neutral wire ensures the two Ys are at the same potential).
To determine the voltage differential between the Ys use Kirchhoff's Voltage Law. Let's call the node on the left A and the node on right B. Start at A, rise L2, fall V2, fall VAB.
VAB would equal L2 -V2 = 20.8/-138.7. Other loops should yield identical results.
At 23:15. Where does the 100.5 V (-116.2 degrees) come from? 74.9 (-123.5 deg) - 31 (-17.2 deg) - 45.4 (125.8 deg) yields a different answer.
Check your calculations. When left to right is the assumed polarity the 3 entries should be PLUS MINUS MINUS to yield 100V at -116 degrees.
@@bigbadtech This is what I did. imgur.com/a/DEgyy7L
Ordinarily I don’t play “math detective” however this one is too good. The issue you’re having most likely stems from the last inverse tangent. Recall in the rectangular to polar conversion lecture: m.ua-cam.com/video/spnKjJtV5y4/v-deo.html that the inverse tangent will give you the “nearest” angle and not the “only” angle. This, being the second phase, should have a phase shift of close to -120 in the third quadrant and your answer will require active intervention. That’s why I’d recommend getting a scientific calculator or using an online resource that will do polar math without having to resort to manual rectangular operations and subsequent conversion. While not impossible, this manner of calculation is like traveling cross country on a donkey. You’ll get there ... but it’ll take a long time.
@@bigbadtech You're correct. My calculations are faulty as the calculator confirms Thanks for clarifying.
16:48 “bullet proof vest and night vision goggles”. Nah man you just need night vision goggles and the stealth of Sam Fisher to survive