Question

For a given series LCR circuit, it is found that maximum current is drawn when value of variable capacitance is $2.5 \, \text{nF}$. If resistance of $2000 \, \Omega$ and $100 \, \text{mH}$ inductor is being used in the given circuit. The frequency of AC source is ____ $\times \, 10^3 \, \text{Hz}$. (Given $\pi^2 = 10$)

Answer 1

Correct Answer: 10

The problem describes a series LCR circuit where the current is maximum, which occurs at resonance. We are given the values of capacitance, resistance, and inductance, and we need to find the frequency of the AC source.

Concept Used:

In a series LCR circuit, the current is maximum when the circuit is in resonance. This occurs when the inductive reactance (\(X_L\)) equals the capacitive reactance (\(X_C\)).

\[ X_L = X_C \]

The formulas for the reactances are:

\[ X_L = \omega L \quad \text{and} \quad X_C = \frac{1}{\omega C} \]

where \(\omega\) is the angular frequency of the AC source, \(L\) is the inductance, and \(C\) is the capacitance. At resonance, the angular frequency (\(\omega_0\)) is given by:

\[ \omega_0 L = \frac{1}{\omega_0 C} \implies \omega_0 = \frac{1}{\sqrt{LC}} \]

The relationship between angular frequency (\(\omega\)) and frequency (\(f\)) is \(\omega = 2\pi f\). Therefore, the resonant frequency (\(f_0\)) is:

\[ f_0 = \frac{\omega_0}{2\pi} = \frac{1}{2\pi\sqrt{LC}} \]

At this frequency, the impedance of the circuit is minimum (\(Z = R\)), and thus the current is maximum.

Step-by-Step Solution:

Step 1: List the given values and convert them to SI units.

• Capacitance, \(C = 2.5 \, \text{nF} = 2.5 \times 10^{-9} \, \text{F}\)
• Inductance, \(L = 100 \, \text{mH} = 100 \times 10^{-3} \, \text{H} = 0.1 \, \text{H}\)
• Resistance, \(R = 2000 \, \Omega\) (This value is not needed to calculate the resonant frequency).
• Given approximation, \(\pi^2 = 10\).

Step 2: Use the formula for the resonant frequency.

The frequency of the AC source must be the resonant frequency \(f_0\) for the current to be maximum.

\[ f_0 = \frac{1}{2\pi\sqrt{LC}} \]

To simplify the calculation, let's square both sides of the equation:

\[ f_0^2 = \frac{1}{(2\pi)^2 LC} = \frac{1}{4\pi^2 LC} \]

Step 3: Substitute the given values into the squared formula.

Using \(L = 0.1 \, \text{H}\), \(C = 2.5 \times 10^{-9} \, \text{F}\), and \(\pi^2 = 10\):

\[ f_0^2 = \frac{1}{4(10) (0.1) (2.5 \times 10^{-9})} \]

Now, let's simplify the denominator:

\[ \text{Denominator} = 40 \times 0.1 \times 2.5 \times 10^{-9} = 4 \times 2.5 \times 10^{-9} = 10 \times 10^{-9} = 10^{-8} \]

So, the equation for \(f_0^2\) becomes:

\[ f_0^2 = \frac{1}{10^{-8}} = 10^8 \]

Step 4: Calculate the resonant frequency \(f_0\).

Taking the square root of both sides:

\[ f_0 = \sqrt{10^8} = 10^4 \, \text{Hz} \]

So, the frequency of the AC source is \(10000 \, \text{Hz}\).

Step 5: Express the result in the required format.

The problem asks for the frequency in the form _____ \(\times \, 10^3 \, \text{Hz}\).

\[ 10^4 \, \text{Hz} = 10 \times 10^3 \, \text{Hz} \]

The value to be filled in the blank is 10.

Answer 2

For maximum current, the circuit must be in resonance.
The resonant frequency \( f_0 \) is given by:
\[f_0 = \frac{1}{2\pi\sqrt{L \cdot C}}\]
Substitute \( L = 100 \times 10^{-3} \, \text{H} \) and \( C = 2.5 \times 10^{-9} \, \text{F} \):
\[f_0 = \frac{1}{2\pi\sqrt{100 \times 10^{-3} \times 2.5 \times 10^{-9}}}\]
Simplify under the square root:
\[f_0 = \frac{1}{2\pi\sqrt{25 \times 10^{-11}}}\]
\[f_0 = \frac{1}{2\pi \times 5 \times 10^{-6}}\]
Using \( \pi^2 = 10 \):
\[f_0 = \frac{10^5 \sqrt{10}}{10}\]
\[f_0 = 10 \times 10^3 \, \text{Hz}\]