An L-R-C series circuit has L = 0.450 H, C=2.50×10^−5F, and resistance R.

Required:
a. What is the angular frequency of the circuit when R = 0?
b. What value must R have to give a decrease in angular frequency of 10.0 % compared to the value calculated in Part a.

Respuesta :

Answer:

298rad/s and 116.96 ohms

Explanation:

Given an L-R-C series circuit where

L = 0.450 H,

C=2.50×10^−5F, and resistance R= 0

In this situation we have a simple LC circuit with angular frequency

Wo = 1√LC

= 1/√(0.450)(2.50×10^-5)

= 1/√0.00001125

= 1/0.003354

= 298rad/s

B) Now we need to find the value of R such that it gives a 10% decrease in angular frequency.

Wi/W° = (100-10)/100

Wi/W° = 90/100

Wi/W° = 0.90 ............... 1

Angular frequency of oscillation

The complete aspect of the solution is attached, please check.

Ver imagen isaacoranseola
Lanuel

a. The angular frequency of the circuit when R = 0 Ohms is 294.12 rad/s.

b. The value R must have to give a decrease in angular frequency of 10.0 % compared to the initial value is equal to 116.96 Ohms.

Given the following data:

  • Inductance, L = 0.450 Henry
  • Capacitance, C = [tex]2.50\times 10^{-5}[/tex] Farads

a. To determine the angular frequency of the circuit when R = 0 Ohms:

Mathematically, the angular frequency of a LC circuit is given by the formula:

[tex]\omega = \frac{1}{\sqrt{LC} } \\\\\omega =\frac{1}{\sqrt{0.450 \times 2.50\times 10^{-5}}} \\\\\omega =\frac{1}{\sqrt{1.125 \times 10^{-5}}} \\\\\omega = \frac{1}{0.0034} \\\\\omega = 294.12\;rad/s[/tex]

b. To find the value R must have to give a decrease in angular frequency of 10.0 % compared to the value calculated above:

The mathematical expression is given as follows:

[tex]\frac{\omega_f}{\omega_i} = \frac{100-10}{100} \\\\\frac{\omega_f}{\omega_i} =\frac{90}{100} \\\\\frac{\omega_f}{\omega_i} =0.9[/tex]

[tex](\frac{\omega_f}{\omega_i})^2 = 1 - \frac{R^2C}{4L} \\\\0.90^2=1 - \frac{R^2C}{4L}\\\\R=\sqrt{\frac{4L(1-0.81)}{C}} \\\\R=\sqrt{\frac{4\times 0.450 \times (0.19)}{2.50\times 10^{-5}}}\\\\R = \sqrt{\frac{0.342}{2.50\times 10^{-5}} }\\\\R =\sqrt{13680}[/tex]

R = 116.96 Ohms.

Read more: https://brainly.com/question/23754122