A resonance circuit having inductance and resistance 2 x 10^-4 H and 6.28 ohm respectively oscillate

Q: Why don't we need the capacitance value to find the Q-factor?

At resonance, the inductive and capacitive reactances cancel each other out. The Q-factor can be expressed purely in terms of $L$, $R$, and the resonant frequency $\omega_0$, so $C$ is not strictly required if $\omega_0$ is already known.

Q: What does a Q-factor of 2000 actually mean?

It means the circuit is highly selective! The energy stored in the circuit is 2000 times greater than the energy dissipated per radian of the cycle. It will have a very sharp resonance peak.

Q: I got a different power of 10. Where did I go wrong?

Make sure you converted 10 MHz to Hertz correctly. $1\text{ MHz} = 10^6\text{ Hz}$, so $10\text{ MHz} = 10 \times 10^6 = 10^7\text{ Hz}$.

Q: Can I use $Q = \frac{1}{R} \sqrt{\frac{L}{C}}$ instead?

Yes, but you would first need to calculate $C$ using the resonant frequency formula $f_0 = \frac{1}{2\pi\sqrt{LC}}$. Using $Q = \frac{\omega_0 L}{R}$ is much more direct here.

Q: Why did the problem specify taking $\pi = 3.14$?

This was a massive hint! The resistance is exactly $6.28\text{ }\Omega$, which is $2 \times 3.14$. This allows for a clean, calculator-free cancellation during the exam.

Question

A resonance circuit having inductance and resistance $2 	imes 10^{-4}	ext{ H}$ and $6.28	ext{ }\Omega$ respectively oscillates at $10	ext{ MHz}$ frequency. The value of quality factor of this resonator is ......... .
[Take, $\pi = 3.14$]

Sigma Percentile · Accepted Answer

### The Anatomy of a Resonant Circuit

Imagine a circuit where energy dances back and forth between an inductor and a capacitor, perfectly in sync with an external alternating voltage. This is the magic of a resonance circuit. In this problem, we are given a circuit oscillating at a resonant frequency of $f_0 = 10	ext{ MHz}$, which is $10^7	ext{ Hz}$.

The circuit features an inductor with an inductance of $L = 2 	imes 10^{-4}	ext{ H}$ and a resistor with a resistance of $R = 6.28	ext{ }\Omega$. Our goal is to find the **Quality Factor** ($Q$), a dimensionless parameter that describes how 'sharp' or 'selective' this resonance is.

### The Master Equation: Quality Factor

The Quality Factor of a series resonant circuit is defined as the ratio of the resonant angular frequency times the inductance to the resistance. Mathematically, it is expressed as:

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

Since we are given the linear frequency $f_0$ instead of the angular frequency $\omega_0$, we must remember the fundamental relationship $\omega_0 = 2\pi f_0$. Substituting this into our master equation gives us a highly practical formula:

$$Q = \frac{2\pi f_0 L}{R}$$

### The Elegance of Numbers

Now, let's carefully substitute our known values into this equation. The problem explicitly instructs us to take $\pi = 3.14$. This is a massive hint from the examiner!

$$Q = \frac{2 	imes 3.14 	imes 10^7 	imes 2 	imes 10^{-4}}{6.28}$$

Look closely at the numerator. If we multiply $2$ by $3.14$, we get exactly $6.28$. This perfectly matches the resistance value in the denominator!

$$Q = \frac{6.28 	imes 10^7 	imes 2 	imes 10^{-4}}{6.28}$$

This allows for a beautiful, clean cancellation. The $6.28$ in the numerator and the $6.28$ in the denominator simply vanish, leaving us with a straightforward exponent calculation.

### The Final Verdict

After the cancellation, we are left with:

$$Q = 10^7 	imes 2 	imes 10^{-4}$$

Using the basic laws of exponents, we add the powers of $10$: $7 + (-4) = 3$.

$$Q = 2 	imes 10^3 = 2000$$

Our final Quality Factor is **2000**. A $Q$-factor this high indicates that the circuit is highly selective, meaning it has a very sharp resonance peak and loses very little energy per cycle compared to the energy it stores. It's a beautifully efficient resonator!

Animated Solution for Physics - Electromagnetic Induction: A resonance circuit having inductance and resistance $2 \times 1 0^{- 4} H$ and $6.28 Ω$ respectively oscillates at $10 MHz$ frequency. The value of quality factor of this resonator is ......... . [Take, $π = 3.14$ ]

Enter Numerical Value:

Visualized Solution

$C i r c u i t P a r am e t er s$

$Q u a l i t y F a c t or F or m u l a$

$S u b s t i t u t in g V a l u es$

$S im pl i f i c a t i o n$

$F ina l C a l c u l a t i o n$

$P h y s i c a l S i g ni f i c an ce$

The Sigma Insight: Alternating Current (AC) and Voltage

The Anatomy of a Resonant Circuit

The Master Equation

The Elegance of Numbers

The Final Verdict

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