Question

Problema 10, subpunctul c, va rog.

Answer 1

$tg\varphi=\frac{|X_L-X_C|}{R} =>X_L-X_C=\pm Rtg\varphi=>\omega L-\frac{1}{\omega C}\pm Rtg\varphi=0=>\omega^2LC\pm RC\omegatgtg\varphi-1=0=>\omega=\frac{\pm RCtg\varphi\pm\sqrt{R^2+C^2+tg^2\varphi+4LC}}{2LC}=>$

$\omega=\pm\frac{Rtg\varphi}{2L}\pm\frac{1}{2L} \sqrt{R^2tg^2\varphi+4\frac{L}{C} }\\\nu=\frac{\omega}{2\pi}$