Find the impedance of the series L-C branch and check what happens to \(V_{out}/V_{in}\) at very low frequency, very high frequency, and exactly at resonance.
Step 1: Identify the circuit topology.
The circuit is a voltage divider. The source \(V_{in}\) drives a series path made of a resistor \(R\), and the output \(V_{out}\) is picked off at the node between \(R\) and a branch where an inductor \(L\) and a capacitor \(C\) are connected in series to the ground return. So the series combination of \(L\) and \(C\) sits between the output node and ground, while \(R\) sits between the source and the output node.
Step 2: Write the impedance of the L-C branch.
For a sinusoidal input at angular frequency \(\omega\), the impedance of the inductor is \(j\omega L\) and the impedance of the capacitor is \(\frac{1}{j\omega C}\). Since \(L\) and \(C\) are in series,
\[ Z_{LC} = j\omega L + \frac{1}{j\omega C} = j\left(\omega L - \frac{1}{\omega C}\right) \]
Step 3: Set up the voltage divider.
\(R\) and \(Z_{LC}\) form a divider from \(V_{in}\) to ground, with \(V_{out}\) taken across \(Z_{LC}\):
\[ \frac{V_{out}}{V_{in}} = \frac{Z_{LC}}{R + Z_{LC}} \]
Step 4: Check the behaviour at the two extreme frequencies.
As \(\omega \to 0\) (DC), the capacitor term \(\frac{1}{\omega C}\) grows without bound, so \(|Z_{LC}| \to \infty\), which makes \(Z_{LC}\) dominate over \(R\) and \(V_{out}/V_{in} \to 1\). So low frequencies pass through almost unattenuated.
As \(\omega \to \infty\), the inductor term \(\omega L\) grows without bound, so \(|Z_{LC}| \to \infty\) again, and once more \(V_{out}/V_{in} \to 1\). So high frequencies also pass through almost unattenuated.
Step 5: Check the behaviour at resonance.
At the resonant frequency \(\omega_0\), where \(\omega_0 L = \frac{1}{\omega_0 C}\), the two reactive terms cancel and \(Z_{LC} = 0\). The series L-C branch behaves like a short circuit to ground at this one frequency, so \(V_{out} = 0\) exactly at \(\omega_0\), while frequencies away from \(\omega_0\) pass through.
Final Answer:
The output stays close to \(V_{in}\) at both low and high frequencies, but drops to zero at the resonant frequency in between. That is exactly the behaviour of a band stop (notch) filter, so option (D) is correct. Option (A) is wrong because a genuine low pass filter would keep attenuating as frequency rises, not recover to \(V_{out} \approx V_{in}\) at high frequency. Option (B) is wrong for the same reason in reverse, since a high pass filter would stay attenuated at low frequency instead of passing it. Option (C), band pass, is the opposite behaviour of what this circuit gives, since here the middle band is rejected, not passed.
\[ \boxed{\text{Band stop filter}} \]
Find dV/dt