Esercizio 6 -

soluzione
Essendo in questo caso:

$P = V_p\cdot I \cos\varphi \ \Rightarrow\ \ I = \frac{P}{V_p \cdot \cos\varphi} = \frac{300}{40\cdot 0{,}731} = 10{,}27\ \text{A}.$

$I_{\max}=I\sqrt{2}=10{,}27\sqrt{2}=14{,}52\ \text{A}.$

$\text{Ad un } \cos\varphi = 0{,}731 \text{ corrisponde } \varphi=\arccos(0{,}731)=43^\circ.$

Mantenendo $V_p$ sull’asse reale in termini sinusoidali:

$i(t)=14{,}52,\sin(\omega t-43^\circ)\ \text{A}.$

$\text{Poi, sapendo che } Q_C=X_C I_2^{,2} ;\Rightarrow; I_2=\sqrt{\frac{Q_C}{X_C}} =\sqrt{\frac{90}{10}} =3\ \text{A}.$

Ai capi del bipolo ohmico–capacitivo:

$\ V_p^{,2}=V_{R2}^{,2}+V_C^{,2}, \qquad V_C=X_C I_2=10\cdot 3=30\ \text{V}.$

$V_{R2}=\sqrt{V_p^{,2}-V_C^{,2}} =\sqrt{40^{2}-30^{2}} =26{,}45\ \text{V}. \qquad V_{R2}=R_2 I_2 \Rightarrow R_2=\frac{V_{R2}}{I_2}=\frac{26{,}45}{3}=8{,}8\ \Omega.$

$\tan\theta=\frac{V_C}{V_{R2}}=\frac{30}{26{,}45} ;\Rightarrow; \theta=\arctan!\left(\frac{30}{26{,}45}\right)=48{,}6^\circ.$

$\text{Dal punto di vista vettoriale (valori efficaci):}\quad \tilde I=\tilde I_1+\tilde I_2.$

$\tilde I =10{,}27\big(\cos 43^\circ - j\sin 43^\circ\big) =7{,}51 - j,7, \qquad \tilde I_2 =3\big(\cos 48{,}6^\circ + j\sin 48{,}6^\circ\big) =1{,}98 + j,2{,}25.$

$\tilde I_1=\tilde I-\tilde I_2 =(7{,}51-j,7)-(1{,}98+j,2{,}25) =5{,}53 - j,9{,}25.$

$I_1=\sqrt{5{,}53^{2}+9{,}25^{2}}=10{,}77\ \text{A}, \qquad \beta=\arctan!\left(\frac{9{,}25}{5{,}53}\right) = -59{,}1^\circ.$

$P_2=R_2 I_2^{,2}=8{,}8\cdot 3^{2}=79{,}2\ \text{W} ;\Rightarrow; P_1=P-P_2=300-79{,}2=220{,}8\ \text{W}.$

$P_1=R_1 I_1^{,2} ;\Rightarrow; R_1=\frac{P_1}{I_1^{,2}} =\frac{220{,}8}{10{,}77^{2}} =1{,}9\ \Omega.$

$\textbf{Per il ramo ohmico-induttivo:}\qquad \tan\beta=\frac{Q_L}{P_1} ;\Rightarrow; Q_L=P_1\tan\beta =220{,}8\cdot \tan(-59{,}1^\circ) =-370\ \text{VAR}.$

$Q_L=X_L I_1^{,2} ;\Rightarrow; X_L=\frac{Q_L}{I_1^{,2}} =\frac{370}{10{,}77^{2}} =3{,}2\ \Omega, \qquad L=\frac{X_L}{2\pi f} =\frac{3{,}2}{2\pi\cdot 50} =10{,}2\ \text{mH}.$

Esercizio 6

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