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

R = 7.286 Ω

Step-by-step explanation:

given,

C = 4.80 m F

L = 0.52 H

V = 56 V

`\omega = (1)/(√(LC))`

`\omega = \frac{1}{\sqrt{0.52* 4.80 * 10^(-3)}}`

`\omega =20\ rad/s`

now,

at resonance V_c=IωL

`I = (V_c)/(\omega L)`

`I = (80)/(20 * 0.52)`

I = 7.686 A

Resistance of the resistor

V = I R

`R = (V)/(I)`

`R = (56)/(7.686)`

R = 7.286 Ω