Question 80: AC fundamentals

Question

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 _________ $$	imes 10^3 \mathrm{~Hz}$$ (given $$\mathrm{a}^2=10$$)

Accepted Answer

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, $$L = 100 \mathrm{~mH} = 100 	imes 10^{-3} \mathrm{~H}$$
Capacitance, $$C = 2.5 \mathrm{~nF} = 2.5 	imes 10^{-9} \mathrm{~F}$$
Plug these values into the frequency equation:
$$ f = \frac{1}{2 \pi \sqrt{(100 	imes 10^{-3}) (2.5 	imes 10^{-9})}} $$
First, calculate the product of $ L $ and $ C $:
$$ L \cdot C = 100 	imes 10^{-3} \cdot 2.5 	imes 10^{-9} = 2.5 	imes 10^{-10} $$
Now, take the square root of the product:
$$ \sqrt{2.5 	imes 10^{-10}} = \sqrt{2.5} 	imes 10^{-5} $$
Given that $$ \mathrm{a}^2 = 10 $$, we have:
$$ \mathrm{a} = \sqrt{10} $$
Since $ \sqrt{2.5} = \frac{\sqrt{10}}{2} $:
$$ \sqrt{2.5} 	imes 10^{-5} = \frac{\sqrt{10}}{2} 	imes 10^{-5} $$
Now substitute back into the frequency formula:
$$ f = \frac{1}{2 \pi \left( \frac{\sqrt{10}}{2} 	imes 10^{-5} ight) } = \frac{1}{\pi \sqrt{10} 	imes 10^{-5}} $$
Simplify the equation:
$$ f = \frac{10^5}{\pi \sqrt{10}} $$
Given that $$ \pi \approx 3.14 $$, we get:
$$ f \approx \frac{10^5}{3.14 	imes 3.162} $$
Simplify further:
$$ f \approx \frac{10^5}{9.93} \approx 10 	imes 10^3 \mathrm{~Hz} $$
Thus, the frequency of the AC source is approximately:
$$ 10 	imes 10^3 \mathrm{~Hz} $$

AC fundamentals

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