Hi Jim. I have an observation in the unbalanced, 3-phase , 3-wire section of this video. More specifically, at the 20 minute mark in the video where your using the voltage divider rule and transfer function between the the Z" (double prime) impedance and the Z2 impedance. The Z" is now in series with the Z2 impedance as you correctly pointed out. But should not the Z" be in the numerator of the transfer function instead of Z2? In other words, you have source in series with Z2 followed by Z" then the return to the source. I think Z" should be in the numerator of the transfer function not Z2. Can you advise?
Good question. The AC Voltage Divider Rule necessitates the impedance of interest be in the numerator (ie: top part of fraction). If you're solving for V2 it'd be Z2. If you're solving for V' it'd be Z'. It can be used in both fashions. Importantly, you should check your work with Kirchhoff's Voltage Law (or Ohm's Law) to verify your answer. Check out the AC VDR lecture at: ua-cam.com/video/msrHsLlSF_E/v-deo.html
H Jim. At the 16 min mark, S2 should have angle -80 instead of 40, and S3 should have angle 160, instead of 40. Is that right? S2 = I V * = (282 mA, angle -160) (120 v , angle -120) = 33.9 VA angle -80, not 40. Same with S3, or else I'm missing something. Thank you!
the angle of S is the same as the angle of Z, which is relative. The combining in the calculator of the voltage angle, which is absolute, with the current angle which is relative, spits out a different angle than the Z angle to begin, with unless the voltage angle is 0.
Not a calculator problem. Check the “AC Power” lecture. AC power calculations do not use absolute phase shift but rather the “relative” phase shift. For these loads current lags voltage by a relative 40 degrees.
Hi Jim. I have an observation in the unbalanced, 3-phase , 3-wire section of this video. More specifically, at the 20 minute mark in the video where your using the voltage divider rule and transfer function between the the Z" (double prime) impedance and the Z2 impedance. The Z" is now in series with the Z2 impedance as you correctly pointed out. But should not the Z" be in the numerator of the transfer function instead of Z2? In other words, you have source in series with Z2 followed by Z" then the return to the source. I think Z" should be in the numerator of the transfer function not Z2. Can you advise?
Good question. The AC Voltage Divider Rule necessitates the impedance of interest be in the numerator (ie: top part of fraction). If you're solving for V2 it'd be Z2. If you're solving for V' it'd be Z'. It can be used in both fashions. Importantly, you should check your work with Kirchhoff's Voltage Law (or Ohm's Law) to verify your answer. Check out the AC VDR lecture at: ua-cam.com/video/msrHsLlSF_E/v-deo.html
Hi Jim, @ minute 23:41, shouldn't the total for V2 add up to 36V
No. One must take into account magnitude, phase shift, AND direction. Assuming L2 to be the dominant phase left to right it's L2-L1-L3.
H Jim. At the 16 min mark, S2 should have angle -80 instead of 40, and S3 should have angle 160, instead of 40. Is that right?
S2 = I V * = (282 mA, angle -160) (120 v , angle -120) = 33.9 VA angle -80, not 40. Same with S3, or else I'm missing something. Thank you!
I am doing the conjugate, btw
I can see that it should be the same like you have it, but my calculator wont give it to me.
It's a calculator problem
the angle of S is the same as the angle of Z, which is relative. The combining in the calculator of the voltage angle, which is absolute, with the current angle which is relative, spits out a different angle than the Z angle to begin, with unless the voltage angle is 0.
Not a calculator problem. Check the “AC Power” lecture. AC power calculations do not use absolute phase shift but rather the “relative” phase shift. For these loads current lags voltage by a relative 40 degrees.
beautiful