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 ______. (rounded off to nearest integer)

The correct answer is 48.

To find the overall Q-factor of the series circuit consisting of a lossy inductor and a lossy capacitor, we can model the losses as equivalent series resistances.

Let:
$Q_L$ be the Q-factor of the inductor = 60
$Q_C$ be the Q-factor of the capacitor = 240

The quality factor (Q-factor) of a component is defined by the ratio of its reactance to its equivalent series resistance. At resonance, the inductive reactance and capacitive reactance are equal in magnitude:

$X_L=X_C=X$

Thus, we can express the equivalent series resistances of the inductor ($R_L$) and the capacitor ($R_C$) as:

$R_L=\frac{X}{Q_L}$

$R_C=\frac{X}{Q_C}$

When the inductor and capacitor are connected in series, the total equivalent series resistance ($R_s$) of the circuit is the sum of the individual resistances:

$R_s=R_L+R_C$

The overall Q-factor of the series circuit ($Q_s$) is given by:

$Q_s=\frac{X}{R_s}=\frac{X}{R_L+R_C}$

Substituting the expressions for $R_L$ and $R_C$ into the equation:

$Q_s=\frac{X}{\frac{X}{Q_L}+\frac{X}{Q_C}}$

This gives the reciprocal relation for the overall Q-factor:

$\frac{1}{Q_s}=\frac{1}{Q_L}+\frac{1}{Q_C}$

Rearranging this formula to solve for $Q_s$ yields:

$Q_s=\frac{Q_L·Q_C}{Q_L+Q_C}$

Now, substitute the given values ($Q_L=60$ and $Q_C=240$):

$Q_s=\frac{60·240}{60+240}=\frac{14400}{300}=48$

Thus, the overall Q-factor of the circuit is 48.