Connect with us

Determining the source voltage

Discussion in 'Electronics Homework Help' started by electronicsLearner77, Oct 10, 2020.

Scroll to continue with content
  1. electronicsLearner77

    electronicsLearner77

    94
    1
    Jul 2, 2015
    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?
     
  2. Harald Kapp

    Harald Kapp Moderator Moderator

    10,344
    2,247
    Nov 17, 2011
    What is E2? Why would you use an RMS value + angle here? Makes no sense. Use the vectors as given in the task description.

    Not directly. You can paste an image of a set of equations generated by LaTeX.
    Or see this list of BB Codes that allow a bit more text formatting than the menu items above this edit window.
     
  3. electronicsLearner77

    electronicsLearner77

    94
    1
    Jul 2, 2015
    upload_2020-10-10_16-23-20.png

    I am not sure if this is what you are asking. I have represented in phasor notation for E2
    Generally all AC circuit analysis is done using phasor notation. Am I correct?
     
  4. Harald Kapp

    Harald Kapp Moderator Moderator

    10,344
    2,247
    Nov 17, 2011
    No. What you show here is e2, but for E2 you use the RMS. Why? That's not correct
    Up to this part everything is correct.
     
  5. bertus

    bertus Moderator

    1,000
    371
    Nov 8, 2019
  6. electronicsLearner77

    electronicsLearner77

    94
    1
    Jul 2, 2015
    But book says for phasors we have to use the RMS values, that is the reason i converted.
    upload_2020-10-10_23-34-0.png
    And from the examples i followed the same methodology.
     
  7. Ratch

    Ratch

    1,079
    330
    Mar 10, 2013
    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
     
  8. electronicsLearner77

    electronicsLearner77

    94
    1
    Jul 2, 2015
    Could you please let me know how to handle below equation
    upload_2020-10-15_14-56-1.png
    I think it is always easy to go with the method i mentioned, that i hope is the standard followed by all.
     
  9. Ratch

    Ratch

    1,079
    330
    Mar 10, 2013
    To each his own way.
    Ratch
    electronicsLearne.JPG
     
  10. electronicsLearner77

    electronicsLearner77

    94
    1
    Jul 2, 2015
    I don't want to drag it further, but my point is here we are dealing signals and our topic is electronics and not trigonometry. So solving like a trigonometry problem you miss the signal analysis. This is only my point of view.
     
  11. Harald Kapp

    Harald Kapp Moderator Moderator

    10,344
    2,247
    Nov 17, 2011
    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
     

    Attached Files:

    electronicsLearner77 likes this.
  12. Harald Kapp

    Harald Kapp Moderator Moderator

    10,344
    2,247
    Nov 17, 2011
    What is your new result? Do you know the expected result?
     
  13. electronicsLearner77

    electronicsLearner77

    94
    1
    Jul 2, 2015
    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.
     
  14. Harald Kapp

    Harald Kapp Moderator Moderator

    10,344
    2,247
    Nov 17, 2011
    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).

    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.
     
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day

-