JEE Main 2024 — Alternating Current Question with Solution

To solve this problem, we need to use the concept of resonance in an LCR (inductor-capacitor-resistor) circuit. At resonance, the inductive reactance and capacitive reactance cancel each other out. The condition for resonance in an LCR circuit is given by:

$$\omega L=\frac{1}{\omega C}$$

Where:

• $$\omega$$ is the angular frequency
• $$L$$ is the inductance
• $$C$$ is the capacitance

Rewriting for angular frequency:

$${\omega }^2=\frac{1}{LC}$$

The angular frequency $$\omega$$ is related to the frequency $f$ by:

$$\omega =2\pi f$$

Substituting this into the equation for angular frequency gives:

$$(2\pi f{)}^2=\frac{1}{LC}$$

Therefore, the frequency $f$ can be found by:

$$f=\frac{1}{2\pi \sqrt{LC}}$$

Given values:

• Inductance, $$L=100\mathrm{mH}=100\times 10^{-3}\mathrm{H}$$
• Capacitance, $$C=2.5\mathrm{nF}=2.5\times 10^{-9}\mathrm{F}$$

Plug these values into the frequency equation:

$$f=\frac{1}{2\pi \sqrt{(100\times 10^{-3})(2.5\times 10^{-9})}}$$

First, calculate the product of $L$ and $C$:

$$L\cdot C=100\times 10^{-3}\cdot 2.5\times 10^{-9}=2.5\times 10^{-10}$$

Now, take the square root of the product:

$$\sqrt{2.5\times 10^{-10}}=\sqrt{2.5}\times 10^{-5}$$

Given that $${\mathrm{a}}^2=10$$, we have:

$$\mathrm{a}=\sqrt{10}$$

Since $\sqrt{2.5}=\frac{\sqrt{10}}{2}$:

$$\sqrt{2.5}\times 10^{-5}=\frac{\sqrt{10}}{2}\times 10^{-5}$$

Now substitute back into the frequency formula:

$$f=\frac{1}{2\pi \left(\frac{\sqrt{10}}{2}\times 10^{-5}\right)}=\frac{1}{\pi \sqrt{10}\times 10^{-5}}$$

Simplify the equation:

$$f=\frac{10^5}{\pi \sqrt{10}}$$

Given that $$\pi \approx 3.14$$, we get:

$$f\approx \frac{10^5}{3.14\times 3.162}$$

Simplify further:

$$f\approx \frac{10^5}{9.93}\approx 10\times 10^3\mathrm{Hz}$$

Thus, the frequency of the AC source is approximately:

$$10\times 10^3\mathrm{Hz}$$