An a.c. circuit has a coil, with resistance 10 Ω and inductance 100 mH, in parallel with a capacitor and has a resonant frequency of 120 Hz. The Q-factor of the circuit is:
A) 0.01
B) 0.08
C) 0.13
D) 7.5

Question

Quizplus · Accepted Answer

The Q-factor of a circuit at resonance is given by $Q = \frac{1}{R} \sqrt{\frac{L}{C}}$, where $R$ is the resistance, $L$ is the inductance, and $C$ is the capacitance. However, in a parallel resonant circuit like this, the Q-factor can also be expressed as $Q = \omega_0 L / R$, where $\omega_0$ is the resonant angular frequency. Given $R = 10 \Omega$, $L = 100 mH = 0.1 H$, and the resonant frequency $f_0 = 120 Hz$, we can find $\omega_0 = 2\pi f_0 = 2\pi \times 120$. Plugging these values into the formula for Q gives $Q = (2\pi \times 120 \times 0.1) / 10 = 7.5$.

An AC Circuit Has a Coil, with Resistance 10 Ω and and Inductance