Question

Problema 2.4 va rog frumos! Am aflat I=24 A și apoi m-am blocat...

Answer 1

$L=6,37 mH=6,37*10^{-3}H\\C=159\mu F=159*10^{-6} F\\X_L=2\pi \nu L=6,28*200*6,37*10^{-3}=8\Omega\\X_C=\frac{1}{2\pi \nu C}=\frac{1}{6,28*200*159*10^{-6}}=\frac{1}{199704*10^{-6}}=\frac{10^6}{199704}=5\Omega\\ Z^2=R^2+(X_C-X_L)^2=4^2+(5-8)^2=16+(-3)^2=16+9=25=>Z=\sqrt{25}=5\Omega\\U=ZI=>I=\frac{U}{Z}=\frac{120}{5}= 24A=>I=24A$

$U_R=RI=4*24=96 V\\U_{R_{max}}=\sqrt{2}U_R=\sqrt{2}*96=135,7V=>U_{R_{max}}=135,7V\\U_L=X_LI=8*24=192 V\\U_{L_{max}}=\sqrt{2}U_L=\sqrt{2}*192=271,5V=>U_{L_{max}}=271,5V\\U_C=X_CI=5*24=120 V\\U_{C_{max}}=\sqrt{2}U_C=\sqrt{2}*120=169,7V=>U_{C_{max}}=169,7V$

$tg\varphi=\frac{X_L-X_C}{R}=\frac{8-5}{4} =\frac{3}{4}= 0,75=>tg\varphi=0,75\\X_C=X_L<=>\frac{1}{2\pi \nu_o C}= 2\pi \nu_o L<=>4\pi^2\nu_o^2LC=1<=>\nu_o=\frac{1}{2\pi \sqrt{LC}}=\frac{1}{6,28\sqrt{6,37*10^{-3}*159*10^{-6}}}= 0,159*\frac{1}{\sqrt{1012,83*10^{-9}}}=\frac{0,159}{0,001}\approx158Hz=>\nu_o=158Hz$