For an R-L-C circuit driven with voltage of amplitude v_m and frequency w_0 = 1/sqrt(LC), the curren

Q: Why is it called 'Quality' factor?

It measures the 'quality' of a resonant circuit. A higher $Q$ means a sharper resonance peak, less energy loss per cycle, and better selectivity for tuning into specific frequencies (like in a radio).

Q: Can the Q-factor be less than 1?

Yes, in heavily damped circuits where the resistance $R$ is very high, the $Q$-factor can drop below 1. This means the circuit is not very selective, and the resonance peak is very broad and flat.

Q: How does resistance affect the Q-factor?

Resistance is inversely proportional to $Q$. Higher resistance increases energy dissipation (damping), which widens the bandwidth and consequently lowers the $Q$-factor.

Q: Does the Q-factor have any units?

No, it is a dimensionless quantity. It is defined as the ratio of two frequencies ($\omega_0$ and $2\Delta\omega$), so the units cancel out.

Q: What is the physical meaning of $\omega_0 L$ in the formula?

The term $\omega_0 L$ is the inductive reactance ($X_L$) at the resonant frequency. Therefore, the $Q$-factor can also be thought of as the ratio of the inductive reactance to the resistance at resonance ($Q = \frac{X_L}{R}$).

Question

For an R-L-C circuit driven with voltage of amplitude $v_m$ and frequency $\omega_0 = \frac{1}{\sqrt{LC}}$, the current exhibits resonance. The quality factor, $Q$ is given by

Options:

(A) $\frac{\omega_0 L}{R}$

(B) $\frac{\omega_0 R}{L}$

(C) $\frac{R}{\omega_0 C}$

(D) $\frac{CR}{\omega_0}$

Sigma Percentile · Accepted Answer

### The Magic of Resonance

Imagine you are pushing a child on a swing. If you push at random times, the swing doesn't go very high. But if you time your pushes perfectly to match the swing's natural rhythm, the child goes higher and higher with very little effort. This is the essence of **resonance**.

In an electrical R-L-C series circuit, the "pushes" come from the AC voltage source, and the "swing" is the oscillating current. When the frequency of the AC source exactly matches the natural frequency of the circuit, $\omega_0 = \frac{1}{\sqrt{LC}}$, the inductive and capacitive reactances cancel each other out. The circuit behaves as if it's purely resistive, and the current reaches its absolute maximum peak, $I_{max} = \frac{v_m}{R}$.

### The Sharpness of the Peak: Quality Factor

But not all resonance peaks are created equal. Some are broad and gentle, while others are incredibly sharp and narrow. How do we measure this sharpness? Enter the **Quality Factor**, or **$Q$-factor**.

The $Q$-factor is a dimensionless number that describes how under-damped an oscillator is. In simpler terms, it tells us how "good" or "selective" the circuit is at picking out its resonant frequency while rejecting others.

Mathematically, it is defined as the ratio of the resonant frequency to the bandwidth:

$$Q = \frac{\omega_0}{2\Delta\omega}$$

Here, the bandwidth $2\Delta\omega$ is the range of frequencies over which the power dissipated in the circuit is at least half of its maximum value at resonance. This corresponds to the frequencies where the current is $\frac{I_{max}}{\sqrt{2}}$.

### Deriving the Formula

For a series R-L-C circuit, the bandwidth is directly related to the resistance and inductance. A higher resistance means more energy is lost as heat per cycle, which "damps" the oscillation and widens the peak. The exact relationship is:

$$2\Delta\omega = \frac{R}{L}$$

Now, let's substitute this bandwidth back into our definition of the $Q$-factor:

$$Q = \frac{\omega_0}{\left(\frac{R}{L}\right)}$$

Flipping the denominator, we arrive at the beautiful and elegant formula for the Quality Factor:

$$Q = \frac{\omega_0 L}{R}$$

This tells us that to get a highly selective circuit (a high $Q$), we need a large inductance and a very small resistance.

*(Bonus insight: Since $\omega_0 = \frac{1}{\sqrt{LC}}$, you can substitute $\omega_0$ to find an alternative form: $Q = \frac{1}{\omega_0 CR}$. Both are perfectly valid, but the first one matches our options!)*

Animated Solution for Physics - Alternating Current: For an R-L-C circuit driven with voltage of amplitude $v_{m}$ and frequency $ω_{0} = \frac{1}{L C}$ , the current exhibits resonance. The quality factor, $Q$ is given by

Select Answer:

Visualized Solution

$Resonance in RLC Circuit$

$Definition of Quality Factor (Q)$

$Bandwidth of RLC Circuit$

$Calculating Q$

$Alternative Form$

The Sigma Insight: AC Circuits and Power in AC Circuits

The Magic of Resonance

The Sharpness of the Peak

Deriving the Formula

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Resonance in RLC Circuit

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Definition of Quality Factor (Q)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Bandwidth of RLC Circuit

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Q

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Alternative Form