Animated Solution for Physics - Electromagnetic Induction: In a series L-C-R resonant circuit, the quality factor is measured as 100. If the inductance is increased by two fold and resistance is decreased by two fold, then the quality factor after this change will be ……… .

The Quality Factor (or Q-factor) is one of the most fascinating concepts in alternating current circuits. It tells us how "sharp" or "selective" a resonant circuit is. Imagine tuning your radio to a specific station; a high Q-factor means you hear that station perfectly without interference from neighboring frequencies!

In our problem, we are given a series L-C-R resonant circuit with an initial Quality Factor of $100$.

The formula for the Quality Factor in terms of inductive reactance is:

Here, ${\omega }_0$ is the resonant angular frequency, $L$ is the inductance, and $R$ is the resistance.

A Lesson in Exam Intent

Before we jump into the calculation, we need to address a critical subtlety. Strictly speaking, the resonant frequency is given by ${\omega }_0=\frac{1}{\sqrt{LC}}$. If we change the inductance $L$, the resonant frequency ${\omega }_0$ should also change!

If we used the absolute rigorous formula $Q=\frac{1}{R}\sqrt{\frac{L}{C}}$, doubling $L$ and halving $R$ would give us a new Q-factor of $2\sqrt{2}\times 100\approx 282.8$.

However, in the context of this specific JEE question, the capacitance $C$ is not mentioned, and the answer is expected to be an integer. This is a classic scenario where we must understand the examiner's intent: we are expected to treat ${\omega }_0$ as a constant.

Let's proceed with the assumption that ${\omega }_0$ is constant. We are given two modifications: 1. The inductance is increased by two fold: $L_f=2L$ 2. The resistance is decreased by two fold: $R_f=\frac{R}{2}$

Let's substitute these new values into our Quality Factor formula:

Now, it's just simple algebra. The $2$ in the denominator of the resistance flips up to the numerator:

Notice that the term in the parentheses is exactly our initial Quality Factor, $Q_i$!

Since we know $Q_i=100$, we simply multiply:

The final Quality Factor after the changes is 400. This massive increase in the Q-factor means our circuit has become incredibly sharp and highly selective!