How Do You Calculate R, L, and Tan θ in a Circuit with Sinusoidal Currents?

  • Context: Engineering 
  • Thread starter Thread starter MissP.25_5
  • Start date Start date
  • Tags Tags
    Circuit
Click For Summary

Discussion Overview

The discussion revolves around calculating resistance (R), inductance (L), and the tangent of the phase angle (tan θ) in a circuit with sinusoidal currents. Participants explore different methods for solving the problem, including the use of phasors and time-domain functions, while addressing the complexities involved in the calculations.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants suggest using phasors to find the voltage source e from the given current functions i1(t) and i2(t).
  • Others argue that the phases of the currents are critical and cannot be ignored, emphasizing the need to work with complex numbers in calculations.
  • A participant expresses uncertainty about the correct expression for e and whether they have calculated it correctly.
  • There is a discussion about converting phasors to rectangular form and the implications of doing so on the calculations.
  • Some participants question the validity of negative resistance values obtained during calculations, indicating potential errors in earlier steps.
  • There is a request for clarification on how to calculate the sum of complex exponentials in the context of finding tan θ.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best method for solving the problem, with multiple competing views on the use of phasors versus time-domain functions and the importance of phase angles in calculations. Unresolved questions about the correctness of certain calculations persist.

Contextual Notes

Participants express uncertainty regarding the assumptions made in their calculations, particularly about the treatment of phase angles and the conversion between different forms of complex numbers. There are indications of missing steps or potential errors in calculations that lead to unexpected results.

Who May Find This Useful

This discussion may be useful for students and practitioners in electrical engineering or physics who are grappling with circuit analysis involving sinusoidal currents and phasor representations.

  • #31
MissP.25_5 said:
But calculator is not used in my class and it's not allowed in exams. We don't need exact answers, we only need to calculate it to its simplest form as it can get.

I understand what you're saying. I've seen your rectangular form for E. I've seen your rectangular form for both I1 and I2, so I know that you can form I = I1 + I2.

Write out E/I and start simplifying. You might be surprised by how it (eventually) boils down to something tractable.
 
Physics news on Phys.org
  • #32
rude man said:
You need to get comfortable moving between polar and rectangular forms of complex numbers.

OK, sermon over.

If E = a + jb, then E = sqrt(a^2 + b^2)exp(jθ) where θ = tan-1b/a).

With what you're doing the coefficient sqrt(a^2 + b^2) is not important. The important quantity is exp(jθ).

Word of warning: keep track of the signs of a and b separately.
tan-1(b/-a) is not the same phase angle as tan-1(-b/a), for example.

Yes, I know how to write E in its polar complex form, what i meant was for you to write it out for me just to see if it matches with mine. But I guess the easiest way is finding arg(E/I) instead of subtracting it separately because then you won't get tan theta. The only problem of solving E/I is separating its Re and I am part. Anyway, I did manage to find E/I and finally got the answer as tan theta= 1/5. Do you get the same answer as I do?

EDIT: i will attach my work later because I am using iphone now.
 
Last edited:
  • #33
OK, so here's my final and complete work! I hope it is correct. Is it?
 

Attachments

  • IMG_4407.jpg
    IMG_4407.jpg
    42.8 KB · Views: 441
  • IMG_4408.jpg
    IMG_4408.jpg
    49.2 KB · Views: 424
  • #34
Yup. Good work!
 

Similar threads

Replies
19
Views
3K
  • · Replies 26 ·
Replies
26
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
7
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K