What is the Q of a parallel RLC circuit, if it is resonant at 4.468 MHz, L is 47 microhenrys and R is 180 ohms?

The Quality factor (Q) for a parallel RLC circuit is determined by the ratio of its parallel resistance to its reactance: Q = R / X_L. First, compute the inductive reactance using X_L = 2 π f L. Substituting the given values gives 2 × π × 4.468 × 10⁶ Hz × 47 × 10⁻⁶ H. As the micro and mega prefixes cancel, this becomes 2 × π × 4.468 × 47, which equals approximately 1319.4 ohms. Next, apply the parallel Q formula. Divide the given resistance of 180 ohms by the calculated reactance of 1319.4 ohms. This results in 180 / 1319.4 ≈ 0.136. Because the parallel resistance is lower than the reactance, the circuit is heavily damped and has a Q of less than 1.