College

In a quality crystal, we can find that the series quality factor is equal to and the parallel quality factor is equal to the given data:

\[ L = 150 \, \text{mH}, \, R = 7 \, \Omega, \, C = 0.025 \, \text{pF}, \, C' = 0.4 \, \text{pF}. \]

Choose the correct option:

A. \( Q_s = 1007 \) and \( Q_p = 1041 \)
B. \( Q_s = 1207 \) and \( Q_p = 1241 \)
C. No option is correct
D. \( Q_s = 1107 \) and \( Q_p = 1041 \)
E. \( Q_s = 1107 \) and \( Q_p = 1141 \)

Answer :

Final answer:

The values of Qs and Qp are approximately 0.6324 and 0.0012, respectively.

Explanation:

To calculate the values of Qs and Qp

, we can use the given formulas:For Qs (

series configuration): Qs = ωL/R

For Qp (parallel configuration)

: Qp = ωC'/R

Given:

L= 150 mH

R = 7 ohm

C= 0.025 pF

C' = 0.4 pF

Let's calculate Qs:

Qs = ωL/R

First, we need to calculate ω (

angular frequency)

. ω = 1/√(LC)

Substituting the given values:

ω = 1/√(0.025 pF * 150 mH)

Converting the units to the same system:

ω = 1/√(0.025 * 10^(-12) F * 150 * 10^(-3) H)

ω = 1/√(3.75 * 10^(-15) F * 0.15 H)

ω = 1/√(5.625 * 10^(-16) F * H)

ω = 1/√(5.625 * 10^(-16)) rad/s

ω = 1/2.3717 * 10^(-8) rad/s

ω ≈ 4.216 * 10^7 rad/s

Now, substitute the values of ω, L, and R into the formula for Qs:

Qs = (4.216 * 10^7 rad/s * 150 mH) / 7 ohm

Converting the units:

Qs = (4.216 * 10^7 * 150 * 10^(-3)) / 7

Qs = (4.216 * 150 * 10^(-4)) / 7

Qs = 0.6324

Now, let's calculate Qp:

Qp = ωC'/R

Substituting the given values:

Qp = (4.216 * 10^7 rad/s * 0.4 pF) / 7 ohm

Converting the units:

Qp = (4.216 * 10^7 * 0.4 * 10^(-12)) / 7

Qp = (4.216 * 0.4 * 10^(-5)) / 7

Qp = 0.0012

Therefore, the values of Qs and Qp are approximately:

Qs ≈ 0.6324

Qp ≈ 0.0012

Learn more about

calculating the quality factors in a crystal

here:

https://brainly.com/question/31328572

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