Respuesta :
Given:
Inductance, L = 150 mH
Capacitance, C = 5.00 mF
[tex]\Delta V_{max}[/tex] = 240 V
frequency, f = 50Hz
[tex]I_{max}[/tex] = 100 mA
Solution:
To calculate the parameters of the given circuit series RLC circuit:
angular frequency, [tex]\omega[/tex] = [tex]2\pi f = 2\pi \times50 = 100\pi [/tex]
a). Inductive reactance, [tex]X_{L}[/tex] is given by:
[tex]\X_{L} = \omega L = 100\pi \times 150\times 10^{-3} = 47.12\Omega [/tex]
[tex]X_{L} = 47.12\Omega[/tex]
b). The capacitive reactance, [tex]X_{C}[/tex] is given by:
[tex]\X_{C} = \frac{1}{\omega C} = \frac{1}{2\pi fC} = \frac{1}{2\pi \times 50\times 5.00\times 10^{-3}} = 0.636\Omega [/tex]
[tex]X_{C} = 0.636\Omega[/tex]
c). Impedance, Z = [tex]\frac{\Delta V_{max}}{I_{max}} = \frac{240}{100\times 10^{-3}} = 2400\Omega [/tex]
[tex]Z = 2400\Omega[/tex]
d). Resistance, R is given by:
[tex]Z = \sqrt {R^{2} + (X_{L} - X_{C})}[/tex]
[tex]2400^{2} = R^{2} + (47.12 - 0.636)^{2}[/tex]
[tex]R = \sqrt {5757839.238}[/tex]
[tex]R = 2399.5\Omega[/tex]
e). Phase angle between current and the generator voltage is given by:
[tex]tan\phi = \frac{X_{L} - X_{C}}{R}[/tex]
[tex]\phi =tan^{-1}( \frac{X_{L} - X_{C}}{R})[/tex]
[tex]\phi =tan^{-1}( \frac{47.12 - 0.636}{2399.5}) = tan^{-1}{0.0.01937}[/tex]
[tex]\phi = 1.11^{\circ}[/tex]
