Question

. An L-R-C series circuit has C = 4.80 mF, L = 0.520 H, and source voltage amplitude V = 56.0 V. The source is operated at the resonance frequency of the circuit. If the voltage across the capacitor has amplitude 80.0 V, what is the value of R for the resistor in the circuit?

Answer 1

Answer:

R = 7.286 Ω

Explanation:

given,

C = 4.80 m F

L = 0.52 H

V = 56 V

$\omega = \dfrac{1}{\sqrt{LC}}$

$\omega = \dfrac{1}{\sqrt{0.52\times 4.80 \times 10^{-3}}}$

$\omega =20\ rad/s$

now,

at resonance V_c=IωL

    $I = \dfrac{V_c}{\omega L}$

    $I = \dfrac{80}{20 \times 0.52}$

           I = 7.686 A

Resistance of the resistor

V = I R

$R = \dfrac{V}{I}$

$R = \dfrac{56}{7.686}$

   R = 7.286 Ω