In an L-R-C series circuit, 305 , 0.401 , and 5.98×10-8 . When the ac source operates at the resonance frequency of the circuit, the current amplitude is 0.491A
What is the voltage amplitude of the source?
What is the amplitude of the voltage across the resistor?
What is the amplitude of the voltage across the inductor?
What is the amplitude of the voltage across the capacitor?
What is the average power supplied by the source?

Explanation:

Given;

inductance, L = 305 H

Resistance , R = 0.401 Ω

Capacitance, C = 5.98 x 10⁻⁸ F

Amplitude current, I₀ = 0.491A

resonance frequency (F₀) = ?

Part (a) What is the voltage amplitude of the source?

V₀ = I₀ Z

Z is the impedance, since the circuit is in resonance Z = R

V₀ = I₀ Z = I₀ R = 0.491 x 0.401 = 0.197 V

Part (b) What is the amplitude of the voltage across the resistor?

V₀ =  I₀ R = 0.491 x 0.401 = 0.197 V

Part (c) What is the amplitude of the voltage across the inductor?

Where XL is the inductive reactance

Part (d) What is the amplitude of the voltage across the capacitor?

Where XC is the capacitive reactance

Part (e) What is the average power supplied by the source?

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2

higher potential

explanation:

because it is in a high speed so it must move into a of high potential