Question

An inductor having a Q-factor of 60 is connected in series with a capacitor having a Q-factor of 240. The overall Q-factor of the circuit is ________. (round off to nearest integer)

Hint:

To find the total Q-factor of a series LC circuit, use the reciprocal formula: \[ \frac{1}{Q_{\text{total}}} = \frac{1}{Q_L} + \frac{1}{Q_C}. \] This formula works for any series combination of inductors and capacitors in resonance circuits.

Answer 1

Correct Answer: 48

In this question, we are asked to find the overall Q-factor of a circuit that consists of an inductor and a capacitor connected in series. The Q-factor (Quality Factor) is a measure of the "sharpness" or "selectivity" of a resonant system, with higher Q-values indicating less energy loss per cycle. The formula for the total Q-factor for a series combination of an inductor and a capacitor is: \[ \frac{1}{Q_{\text{total}}} = \frac{1}{Q_L} + \frac{1}{Q_C} \] where:
- \( Q_L \) is the Q-factor of the inductor,
- \( Q_C \) is the Q-factor of the capacitor.
Given:
- \( Q_L = 60 \),
- \( Q_C = 240 \).
Substitute these values into the formula: \[ \frac{1}{Q_{\text{total}}} = \frac{1}{60} + \frac{1}{240}. \] To add these fractions, find the least common denominator (LCD), which in this case is 240: \[ \frac{1}{Q_{\text{total}}} = \frac{4}{240} + \frac{1}{240} = \frac{5}{240}. \] Now, take the reciprocal to find \( Q_{\text{total}} \): \[ Q_{\text{total}} = \frac{240}{5} = 48. \] Therefore, the overall Q-factor of the series combination of the inductor and capacitor is \( \boxed{48} \).