Question

A resistance if 40 ohm is connected in a series with an inductor of self- inductance 5H and a capacitor of capacitance 80uF . This combo us then connected to an AC source of rms voltage 220V . Frequency of AC source can be changed continuously .
1. What should be the frequency of source which drives circuit to resonance ?
2. What is the impedance of circuit in the state of resonance ? ​

Answer 1

Explanation:

Given A resistance if 40 ohm is connected in a series with an inductor of self- inductance 5H and a capacitor of capacitance 80uF . This combo us then connected to an AC source of rms voltage 220V . Frequency of AC source can be changed continuously .

• So a resistance of 40 ohm is connected
• Inductor of self inductance 5 H
• Capacitor of capacitance 80 micro F = 80 x 10^-6 F
• Rms voltage is 220 V
• Now there is a resistance R, inductance and capacitance are connected.
• So the magnitude of net impedance will be the vector sum  
• So Z = √(Xl – Xc)^2 + R^2
• Now in a resonance condition we have
•                 Xl = Xc
•                So Z = R
• Now resonant frequency
•                              f = 1 / 2π √LC
•                                 = 1 / 2 x 3.14 √5 x 80 x 10^-6
•                                 = 1 / 6.28 √400 x 10^-6
•                                 = 1 / 6.28 x 20 x 10^-3
•                                = 50 / 6.28
•                               f = 7.96 hertz
• So impedance will be 40 ohms since Z = R

Reference link will be

https://brainly.in/question/3436228