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

This problem concerns a series LCR circuit experiencing maximum current, a condition synonymous with resonance. Given capacitance, resistance, and inductance values, the objective is to determine the AC source frequency.

Underlying Principle:

In a series LCR circuit, maximum current flow is achieved at resonance. This occurs when inductive reactance (\(X_L\)) is precisely equal to capacitive reactance (\(X_C\)).

\[ X_L = X_C \]

The mathematical expressions for these reactances are:

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

Here, \(\omega\) represents the AC source's angular frequency, \(L\) denotes inductance, and \(C\) signifies capacitance. At resonance, the specific angular frequency (\(\omega_0\)) is defined by:

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

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

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

At this resonant frequency, the circuit's impedance (\(Z\)) reaches its minimum value, equalling \(R\), thereby maximizing the current.

Solution Procedure:

Step 1: Consolidate given parameters and convert 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, \(R = 2000 \, \Omega\) (Note: Resistance is irrelevant for resonant frequency calculation).
• Provided approximation: \(\pi^2 = 10\).

Step 2: Apply the resonant frequency formula.

Maximum current occurs when the AC source frequency matches the resonant frequency \(f_0\).

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

To facilitate calculation, we square both sides of the equation:

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

Step 3: Substitute known values into the squared formula.

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

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

Simplify the denominator:

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

The equation for \(f_0^2\) is thus:

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

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

Taking the square root of both sides yields:

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

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

Step 5: Format the result as requested.

The problem requires the frequency to be presented in the format _____ \(\times \, 10^3 \, \text{Hz}\).

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

The blank should be filled with the value 10.