Respuesta :
Answer:
(a) So range of frequency [tex]f > 43.27[/tex] Hz
(b) the reactance is 89.75 Ω
Explanation:
Given:
(a)
Capacitance of a capacitor [tex]C= 23 \times 10^{-6}[/tex] F
Reactance of capacitive circuit [tex]X_{C} =[/tex] 160 Ω
From the formula of reactance,
[tex]X_{C} = \frac{1}{\omega C}[/tex]
[tex]X_{C} = \frac{1}{2\pi fC}[/tex]
[tex]f = \frac{1}{2\pi X_{C} C }[/tex]
[tex]f = \frac{1}{6.28 \times 160 \times 23 \times 10^{-6} }[/tex]
[tex]f = 43.27[/tex] Hz
So range of frequency [tex]f > 43.27[/tex] Hz
(b)
Capacitance [tex]C = 41 \times 10^{-6}[/tex] F
Frequency [tex]f = 43.27[/tex] Hz
From the formula of reactance,
[tex]X_{C} = \frac{1}{2\pi fC}[/tex]
[tex]X_{C} = \frac{1}{6.28 \times 43.27 \times 41 \times 10^{-6} }[/tex]
[tex]X_{C} =[/tex] 89.75 Ω
Therefore, the reactance is 89.75 Ω
