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