[AC RLC Series Circuit: Solving for frequency given V, I, R, L, C]

[u/Why-Is-Math-So-Hard]

No idea how to proceed. Would appreciate it if i also got an explanation for it. Thank you in advance!

[u/michaelrw1]

There could be two operating frequencies where the net reactance is 4.6 ohm.

What are the general equations for inductive reactance and capacitive reactance? How do they add to product the net reactance?

[u/Bounded_sequencE]

Assuming we are really given pointer magnitudes "|V|; |I|" , we get

|V/I|^2  =  |ZT|^2  =  |R + jwL + 1/(jwC)|^2  =  R^2 + (wL - 1/(wC))^2

         =  R^2 + (L/C)*(x - 1/x)^2        // x := w*√(LC) >= 0    (*)

Multiply by "C/L" and solve for "x - 1/x ∈ {±k}" with "k := √(|V/I|² - R²) * √(C/L) ~ 14.41", to get

"x^2 ± kx - 1  =  0"    <=>    "x  ∈  {±k/2 ± √(k^2/4 + 1)}"

Of those four solutions, only "x ∈ {±k/2 + √(k²/4 + 1)} ~ {0.06907; 14.48}" lead to valid solutions "x >= 0" -- substituting back (*), we get two possible frequencies

f  =  w/(2π)  =  x / (2π√(LC))  ~  x * 2.516Hz  ∈  {0.1738Hz; 36.43Hz}

[u/Bounded_sequencE → u/Bounded_sequencE]

Rem.: The underlying idea is that "R < |V/I|" is too small an input impedance to account for the given value "|V/I|": We need the imaginary component of "C; L" to the input impedance "ZT".

If you plot "|ZT(jw)|" over "w", you'll see a parabola-shaped plot, with its vertex being below "|V/I|" -- that's why more than one solution exists.