What is the Q (Quality factor) of a series circuit that resonates at 10 kHz, has equal reactance of 5 kilo-ohms each, and a resistor value of 50 ohms?

Concept:

The quality factor is defined as the ratio of the maximum energy stored to maximum energy dissipated in a cycle

\(Q = 2\pi \frac{{Maximum\;energy\;stored}}{{Total\;energy\;lost\;per\;period}}\)

\(Q = \frac{Resonant ~frequency}{Bandwidth}\)

In a series RLC, 

Bandwidth = \(\frac{R}{L}\)

Resonant frequency = \(\frac{1}{\sqrt {LC}}\)

The quality factor of the series RLC circuit is given as:

\(Q=\frac{L}{R~\sqrt{LC}}\)

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

Calculation:

\(X_L=\omega L\)

\(X_C = \frac{1}{\omega ~C}\)

X_L = 5 kΩ

X_C = 5 kΩ 

\(L=\frac{X_L}{\omega}\)

= \(\frac{5}{2\pi\times 10}\)

\(C=\frac{1}{\omega ~X_C}\)

= \(\frac{1}{2\pi \times 10 \times 5}\)

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

\(Q = \frac{1}{50}\sqrt{\frac{5\times 2\pi \times 10 \times 5 \times 10^6}{2\pi \times 10}}\)

= 100

Note:

We can also use the formula, 

\(Q=\frac{X_L}{R} = \frac{5 \times 10^3}{50}\)

= 100