For an RLC series circuit, R = 100Ω, L = 150mH, and C = 0.25μF. (a) If an ac source of variable frequency is connected to the circuit, at what frequency is maximum power dissipated in the resistor? (b) What is the quality factor of the circuit?

Respuesta :

Answer:

[tex]\omega_O = 0.16 rad /sec[/tex]

Q = 0.24

Explanation:

given data:

resonant  angular frequency is given as  \omega_O = \frac{1}{\sqrt{LC}}

where L is inductor = 150 mH

C is capacitor = 0.25\mu F

[tex]\omega = \frac{1}{\sqrt{150*10^{6}*0.25*10^{-6}}}}[/tex]

[tex]\omega_O = 0.16 rad /sec[/tex]

QUALITY FACTOR is given as

[tex]Q = \frac{1}{R}{\sqrt\frac{L}{C}}[/tex]

Putting all value to get quality factor value

Q =[tex] \frac{1}{1000}{\sqrt\frac{150*10^{6}}{0.25*10^{-6}}}[/tex]

Q = 0.24