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

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To calculate R, L, and tan θ in a circuit with sinusoidal currents, phasors are recommended for simplifying the analysis. The voltage source e can be derived from the given current i1(t) using known values for R and C. The discussion emphasizes the importance of considering phase angles in calculations, as neglecting them can lead to incorrect results. After determining the voltage E and the currents I1 and I2, the next step involves expressing these in polar form to find the argument for tan θ. The final calculations yield R = 2 and L = 3/2, with tan θ calculated as 1/5.
  • #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.
 
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  • #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.
 
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  • #33
OK, so here's my final and complete work! I hope it is correct. Is it?
 

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  • #34
Yup. Good work!
 

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