Question

In the circuit shown below, it is observed that the amplitude of the voltage across the resistor is the same as the amplitude of the source voltage. What is the angular frequency \(\omega_0\) (in rad/s)?

• A. \(10^{4}\)
• B. \(10^{3}\)
• C. \(10^{3}\pi\)
• D. \(10^{4}\pi\)

Hint:

In a series \(RLC\) circuit, if the resistor voltage equals the source voltage, the circuit is at series resonance, giving \(\omega_0=1/\sqrt{LC}\).

Answer 1

The Correct Option is A

Step 1: For a series \(R\!-\!L\!-\!C\) circuit driven by \(V\), the current amplitude is \(|I|=\dfrac{|V|}{|Z|}\) with \(|Z|=\sqrt{R^{2}+(\omega L-1/\omega C)^{2}}\). Step 2: The resistor voltage amplitude equals the source amplitude: \(|V_R|=|I|R=|V|\Rightarrow |Z|=R\). Thus \[ \sqrt{R^{2}+(\omega L-1/\omega C)^{2}}=R \Rightarrow \omega L- \frac{1}{\omega C}=0, \] which is the resonance condition \(\omega_0=\dfrac{1}{\sqrt{LC}}\). Step 3: With \(L=10\,\text{mH}=0.01\,\text{H}\) and \(C=1\,\mu\text{F}=10^{-6}\,\text{F}\), \[ \omega_0=\frac{1}{\sqrt{(0.01)(10^{-6})}} =\frac{1}{10^{-4}}=10^{4}\ \text{rad/s}. \]