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Determining the source voltage

This the circuit i am trying to solve, I need to find the voltage e1
upload_2020-10-10_11-27-17.png
My work is like
e1 + e2 = v -> Applying the Kirchoff's Voltage Law
e1 = v - e2;
upload_2020-10-10_12-37-55.png
But the answer does not seem to match. What is the mistake?
Note: small clarification, can I use latex here?
 
Why get snarled up with RMS and phasors? Just subtract the trig expressions and deal with the result later if you want to express it into different views.electronicsLearne.JPG Now you can transform the above expression into a sin or cos function and multiply/divide by sqrt 2 or anything else you desire. Ratch
 

Harald Kapp

Moderator
Moderator
Generally all AC circuit analysis is done using phasor notation. Am I correct?
O.K..
But you don't need to convert to RMS for making the calculation and at the end convert back to peak. The first involves division by sqrt(2), the latter multiplication by sqrt(2). Both operations cancel each other. Simply using peak values simplifies the calculation.
As to your error look here:
upload_2020-10-21_17-43-34.png
You use 120.1 /_-60 = 120.2 × (cos(-60) - j sin(-60)) which should imho be 120.1 /_-60 = 120.2 × (cos(-60) + j sin(-60))
You then go on with
upload_2020-10-21_17-47-24.png
In this equation obviously cos(-60) = -1/2 and sin(-60) = -sqrt(3)/2 but cos(-60) is + 1/2, not -1/2
 

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Yes, i corrected what you pointed, but one confusion with the final angle calculation. The result is
upload_2020-10-23_17-15-18.png
The magnitude matches with the angle but struggling for the phase angle, i know i have to do some manipulation since the real and imaginary parts are -Ve. What exactly i need to do?
The answer is
upload_2020-10-23_17-17-7.png
I know some 180 degrees to be subtracted, but not sure of the sequence.

Note: It would really help if you upgrade to Latex editor.
 

Harald Kapp

Moderator
Moderator
You can see from real and imaginary parts in -26.5 -j × 154 that the phasor points into the 3rd quadrant. Therefore add or subtract 180°. It doesn't matter which ot the two you chose because you'll end with either 160.23 ° or -99,77 ° which is the same (if you don't believe me, draw the diagram).
The result matches closely my calculation (apart from some minor differences due to rounding).

Note: It would really help if you upgrade to Latex editor.
For a very few savvy ones like you: yes. LaTex isn't that common. You can always post an equation created in the favorite tool of your choice (there are others than LaTex, more specialized in primarily typesetting math equations) as an image, as you obviously have done.
 
Everybody has the wrong answer. Since all the voltages are in sin(wt) format, we don't have to convert anything. All we have to do is subtract e2 from v to get the answer. Subtracting is the same as flipping the e2 phasor over 180 degrees and adding up all the x-components of phasors v and e2 . Doing so we get -37.456. The y-components add up to -217.924. In polar form, the result is easily found by rectangular coordinate arithmetic to be 221.12/_-99.75525degrees. Any questions?

Ratch
 
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