written 8.3 years ago by
teamques10
★ 68k
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modified 8.3 years ago
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Given:
$$R = 10 Ω \hspace{4cm} \text{Series inductive coil} \\
L= 0.1H \hspace{3.8cm} \text{Series inductive coil} \\
\hspace{1.5cm}C = 150µF = 150 \times 10-6 \hspace{1cm} \text{in parallel with inductive coil} \\
\hspace{1.5cm}V = 200 V \hspace{3.5cm} \text{in parallel with inductive coil}$$
Resonant frequency,
$f_r=\dfrac{1}{2 \pi}\sqrt{\dfrac{1}{LC}-\dfrac{R^2}{L^2}}=\dfrac{1}{2 \pi}\sqrt{\dfrac{1}{0.1 \times150\times10^{-6}}}-\dfrac{10^2}{(0.1)^2} \\
f_r=37.9 Hz$
Dynamic impedance,
$Z_c=\dfrac{L}{CR}=\dfrac{0.1}{150 \times10^{-6}\times10} \\
Z_c=66.67 \Omega$
Circuit current at resonance
$I_r=\dfrac{V}{Z_r}=\dfrac{200}{66.67}\\
I_r=3Amp$
For phasor diagram,
$I_L=\dfrac{V}{Z_L}=\dfrac{200}{\sqrt{10^2+2\times\pi37.9\times0.1}} \\
I_L=17.91 Amp \\
I_c=\dfrac{V}{X_c}=\dfrac{200}{\dfrac{1}{2 \pi \times37.9\times150\times10^{-6}}}$
Phase angle of the coil