Question

An AC circuit has $R=100\Omega$, $C=2\mu F$ and $L=80mH$, connected in series. The quality factor of the circuit is

• A. $20$
• B. $2$
• C. $0.5$
• D. $400$

Answer 1

The correct option is B

$2$

Step 1: Given Data:

Resistance, $R=100\Omega$

Capacitance, $C=2\mu F$

$=2\times {10}^{-6}F$

Inductance, $L=80mH$

$=80\times {10}^{-3}H$

Step 2: Formula Used:

Quality factor can be calculated using the formula,

$Q=\frac{\omega L}{R}$

where, $L=$ Inductance

$R=$Resistance

$\omega =$angular frequency

$=\frac{1}{\sqrt{LC}}$

$C=$Capacitance

Step 3: Calculating the Quality factor:

Putting the values in the above formula,

$Q=\frac{{\displaystyle \raisebox{1ex}{$L$}\!\left/ \!\raisebox{-1ex}{$\sqrt{LC}$}\right.}}{R}$

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

$=\frac{1}{100}\times \sqrt{\frac{80\times {10}^{-3}}{2\times {10}^{-6}}}$

$=\frac{1}{100}\times \sqrt{40\times {10}^3}$

$=\frac{1}{100}\times 200$

$=2$

Hence, option B is the correct answer.