Question

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?

Answer 1

The maximum power dissipation in the resistor occurs at a frequency of approximately 1175.5 Hz in an RLC series circuit with the given values of R, L, and C. The quality factor of the circuit is approximately 57.74.

Step-by-step explanation:

In an RLC series circuit, the maximum power dissipation in the resistor is achieved at the resonant frequency, which is given by the formula:

fr = 1 / (2π √LC)

Substituting the given values:

R = 100Ω, L = 150mH (or 0.15H), and C = 0.25μF (or 0.00000025F), we can calculate the resonant frequency:

fr = 1 / (2π √(0.15 x 0.00000025))

fr ≈ 1175.5 Hz

Therefore, the maximum power dissipation in the resistor occurs at a frequency of approximately 1175.5 Hz.

The quality factor (Q) of the circuit is a measure of its damping ability. It is given by the formula:

Q = R √C / L

Substituting the given values:

R = 100Ω, L = 150mH (or 0.15H), and C = 0.25μF (or 0.00000025F), we can calculate the quality factor:

Q = 100 √(0.00000025) / 0.15

Q ≈ 57.74

Therefore, the quality factor of the circuit is approximately 57.74.

Answer 2

`\omega_O = 0.16 rad /sec`

Q = 0.24

Step-by-step 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

`\omega = \frac{1}{\sqrt{150*10^(6)*0.25*10^(-6)}}}`

`\omega_O = 0.16 rad /sec`

QUALITY FACTOR is given as

`Q = (1)/(R){\sqrt(L)/(C)}`

Putting all value to get quality factor value

Q =
`(1)/(1000){\sqrt(150*10^(6))/(0.25*10^(-6))}`

Q = 0.24