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An ac series circuit contains a resistor of 25 ohms, an inductor of 10.5 mH and a variable capacitor. If the frequency of the applied voltage is 190 Hz, to what setting should the capacitor be set if resonance is achieved? A) 61.75μF B) 66.83 μF C) 29 μF D) 84μF E) 52.725 μF

Respuesta :

Answer:

B) 66.83 μF

Explanation:

R = resistance of the resistor = 25 ohm

L = inductance of the inductor = 10.5 mH = 0.0105 H

f = resonance frequency achieved = 190 Hz

C = Capacitance of the capacitor

For resonance to be possible

[tex]2\pi fL = \frac{1}{2\pi fC}[/tex]

[tex] 2\pi fL = \frac{1}{2\pifC}[/tex]

[tex] 2 (3.14)(190)(0.0105) = \frac{1}{2 (3.14)(190)C}[/tex]

C = 66.83 x 10⁻⁶ F

C = 66.83 μF

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