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:

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