Chapter 21, Problem 89GP

To determine

Part (a)To Show:

The Q factor is given by $Q=\frac{1}{R}\times \sqrt{\frac{L}{C}}$

Answer to Problem 89GP

Solution:

The Q factor is given by $Q=\frac{1}{R}\times \sqrt{\frac{L}{C}}$

Explanation of Solution

Given:

The Q-factor can be defined as the ratio of the voltage across the capacitor (or inductor)to the voltage across the resistance, at resonance.

Formula Used:

Let $f_o$ be the resonant frequency

$\begin{array}{l}{f}_o=\frac{1}{2\pi \sqrt{LC}}\\ \text{also, }Q=\frac{V_L}{V_R}=\frac{I_{RMS}{X}_L}{I_{RMS}R}\end{array}$

Where inductive reactance $X_L=2\pi f_{o}L$

Calculation:

$X_L=2\pi \times \frac{1}{2\pi \sqrt{LC}}\times L$

$X_L=\sqrt{\frac{L}{C}}$

$Q=\frac{X_L}{R}=\frac{1}{R}\sqrt{\frac{L}{C}}$

To determine

Part (b) To Determine:

The values of L and R to determine the Q value of 650.

Answer to Problem 89GP

Solution:

$L=2.5\times {10}^{-6}\text{H}$ and $R=0.029\Omega$

Explanation of Solution

Given:

$f_o=1\mu H_z,Q=550,C=0.01\mu F$

Formula Used:

$L=\frac{1}{{\left(2\pi f_o\right)}^{2}C}$

$Q=\frac{1}{R}\sqrt{\frac{L}{C}}$

Where,

Symbols have their usual meanings.

Calculation:

$\begin{array}{l}L=\frac{1}{{\left(2\pi f_o\right)}^{2}C}\\ L=\frac{1}{{(2\pi \times 1\times {10}^6)}^2(0.01\times {10}^{-6})}\text{H}\\ L=2.5\times {10}^{-6}\text{H}\end{array}$

$Q=\frac{1}{R}\sqrt{\frac{L}{C}}$

$\begin{array}{l}R=\frac{1}{Q}\sqrt{\frac{L}{C}}\\ R=\frac{1}{550}\sqrt{\frac{2.5\times {10}^{-6}}{0.01\times {10}^{-6}}}\Omega \\ R=0.029\Omega \end{array}$