For a given series LCR circuit it is found that maximum current is drawn when value of variable capacitance is $2.5 \mathrm{~nF}$. If resistance of $200 \Omega$ and $100 \mathrm{~mH}$ inductor is being used in the given circuit. The frequency of ac source is _________ $\times 10^3 \mathrm{~Hz}$ (given $\mathrm{a}^2=10$)

<p>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:</p> <p>$\omega L = \frac{1}{\omega C}$</p> <p>Where: <ul> <li>$\omega$ is the angular frequency</li> <li>$L$ is the inductance</li> <li>$C$ is the capacitance</li> </ul> </p> <p>Rewriting for angular frequency:</p> <p>$\omega^2 = \frac{1}{LC}$</p> <p>The angular frequency $\omega$ is related to the frequency $ f $ by:</p> <p>$\omega = 2 \pi f$</p> <p>Substituting this into the equation for angular frequency gives:</p> <p>$(2 \pi f)^2 = \frac{1}{LC}$</p> <p>Therefore, the frequency $ f $ can be found by:</p> <p>$f = \frac{1}{2 \pi \sqrt{LC}}$</p> <p>Given values: <ul> <li>Inductance, $L = 100 \mathrm{~mH} = 100 \times 10^{-3} \mathrm{~H}$</li> <li>Capacitance, $C = 2.5 \mathrm{~nF} = 2.5 \times 10^{-9} \mathrm{~F}$</li> </ul> </p> <p>Plug these values into the frequency equation:</p> <p>$f = \frac{1}{2 \pi \sqrt{(100 \times 10^{-3}) (2.5 \times 10^{-9})}}$</p> <p>First, calculate the product of $ L $ and $ C $:</p> <p>$L \cdot C = 100 \times 10^{-3} \cdot 2.5 \times 10^{-9} = 2.5 \times 10^{-10}$</p> <p>Now, take the square root of the product:</p> <p>$\sqrt{2.5 \times 10^{-10}} = \sqrt{2.5} \times 10^{-5}$</p> <p>Given that $\mathrm{a}^2 = 10$, we have:</p> <p>$\mathrm{a} = \sqrt{10}$</p> <p>Since $ \sqrt{2.5} = \frac{\sqrt{10}}{2} $:</p> <p>$\sqrt{2.5} \times 10^{-5} = \frac{\sqrt{10}}{2} \times 10^{-5}$</p> <p>Now substitute back into the frequency formula:</p> <p>$$ f = \frac{1}{2 \pi \left( \frac{\sqrt{10}}{2} \times 10^{-5} \right) } = \frac{1}{\pi \sqrt{10} \times 10^{-5}} $$</p> <p>Simplify the equation:</p> <p>$f = \frac{10^5}{\pi \sqrt{10}}$</p> <p>Given that $\pi \approx 3.14$, we get:</p> <p>$f \approx \frac{10^5}{3.14 \times 3.162}$</p> <p>Simplify further:</p> <p>$f \approx \frac{10^5}{9.93} \approx 10 \times 10^3 \mathrm{~Hz}$</p> <p>Thus, the frequency of the AC source is approximately:</p> <p>$10 \times 10^3 \mathrm{~Hz}$</p>