Question

In a series $LCR$ circuit connected to an AC source of variable frequency, the values are given as $L = 2\,\text{H}$, $C = 32\,\mu\text{F}$, and $R = 10\,\Omega$. What is the resonant angular frequency $\omega_r$ of this circuit system?}

• A. \( 125\,\text{rad/s} \)
• B. \( 250\,\text{rad/s} \)
• C. \( 500\,\text{rad/s} \)
• D. \( 1000\,\text{rad/s} \)

Hint:

Always double-check whether a question asks for the angular frequency (\(\omega_r\) in rad/s) or the linear cyclic frequency (\(f_r\) in Hz). If it asks for linear frequency, you must divide your angular result by \(2\pi\) (\(f_r = \frac{\omega_r}{2\pi}\)).

Answer 1

The Correct Option is A

Concept: In a series electrical circuit containing an inductor (\(L\)), a capacitor (\(C\)), and a resistor (\(R\)), the total opposition to alternating current is called impedance (\(Z\)). This impedance is calculated using the formula: \[ Z = \sqrt{R^2 + (X_L - X_C)^2} \] where \(X_L = \omega L\) represents the inductive reactance, and \(X_C = \frac{1}{\omega C}\) represents the capacitive reactance. As you change the driving frequency of the connected AC source, the values of both reactances shift in opposite directions. Electrical resonance happens at a specific frequency where the inductive reactance matches the capacitive reactance exactly: \[ X_L = X_C \] At this point, the two reactive components cancel each other out completely. This drops the total circuit impedance to its absolute minimum value, which is just equal to the pure thermal resistance (\(Z = R\)). Consequently, the current amplitude peaks. Setting these reactances equal lets us isolate the resonant angular frequency (\(\omega_r\)): \[ \omega_r L = \frac{1}{\omega_r C} \quad \Rightarrow \quad \omega_r^2 = \frac{1}{LC} \quad \Rightarrow \quad \omega_r = \frac{1}{\sqrt{LC}} \]

Step 1: Convert all given metrics into standard SI units. Before plugging numbers into the formula, ensure all components are in standard units:
• Inductance, \(L = 2\,\text{H}\) (already in Henries)
• Capacitance, \(C = 32\,\mu\text{F} = 32 \times 10^{-6}\,\text{F}\) (converted from microfarads to Farads)
• Resistance, \(R = 10\,\Omega\) (already in Ohms)

Step 2: Calculate the product of inductance and capacitance. Let's find the value of the \(LC\) product under the square root: \[ LC = 2 \times (32 \times 10^{-6}) \] \[ LC = 64 \times 10^{-6}\,\text{s}^2 \]

Step 3: Calculate the square root of the product. Take the square root of our result: \[ \sqrt{LC} = \sqrt{64 \times 10^{-6}} = \sqrt{64} \times \sqrt{10^{-6}} \] \[ \sqrt{LC} = 8 \times 10^{-3}\,\text{s} \]

Step 4: Invert the result to find the resonant angular frequency (\(\omega_r\)). Now, complete the calculation by taking the reciprocal: \[ \omega_r = \frac{1}{8 \times 10^{-3}} = \frac{1000}{8} \] \[ \omega_r = 125\,\text{rad/s} \] The resonant angular frequency of this system is exactly \(125\,\text{rad/s}\). Note that the resistor's value (\(10\,\Omega\)) has no impact on this frequency point; it only dictates the height and sharpness of the current peak during resonance.