A resistance (R), inductance (L) and capacitance (C) are connected in series to an ac source of voltage V having variable frequency. Calculate the energy delivered by the source to the circuit during one period if the operating frequency is twice the resonance frequency.

Answer

To solve the problem of calculating the energy delivered by the source to the circuit during one period when the operating frequency is twice the resonance frequency, we will follow these steps:

### Step 1: Identify Resonant Frequency
The resonant frequency \( f_0 \) for an RLC circuit is given by:
\[ f_0 = \frac{1}{2\pi\sqrt{LC}} \]
The angular frequency at resonance is:
\[ \omega_0 = 2\pi f_0 = \frac{1}{\sqrt{LC}} \]

### Step 2: Determine Operating Frequency
Given that the operating frequency is twice the resonance frequency, we have:
\[ \omega = 2\omega_0 = \frac{2}{\sqrt{LC}} \]

### Step 3: Calculate Impedance \( Z \)
The impedance \( Z \) of the RLC series circuit is calculated using:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
where \( X_L \) is the inductive reactance and \( X_C \) is the capacitive reactance.

- Inductive reactance:
\[ X_L = \omega L = \frac{2L}{\sqrt{LC}} = 2\sqrt{\frac{L}{C}} \]
- Capacitive reactance:
\[ X_C = \frac{1}{\omega C} = \frac{\sqrt{LC}}{2C} = \frac{1}{2}\sqrt{\frac{L}{C}} \]

Now substituting \( X_L \) and \( X_C \) into the impedance formula:
\[ Z = \sqrt{R^2 + \left(2\sqrt{\frac{L}{C}} - \frac{1}{2}\sqrt{\frac{L}{C}}\right)^2} \]
\[ = \sqrt{R^2 + \left(\frac{3}{2}\sqrt{\frac{L}{C}}\right)^2} \]
\[ = \sqrt{R^2 + \frac{9}{4}\frac{L}{C}} \]

### Step 4: Calculate Average Power \( P \)
The average power \( P \) delivered to the circuit is given by:
\[ P = \frac{V_{rms}^2}{Z^2} \cdot R \]
Substituting \( Z \):
\[ P = \frac{V_{rms}^2}{R^2 + \frac{9}{4}\frac{L}{C}} \cdot R \]

### Step 5: Calculate Energy Delivered Over One Cycle
The energy \( E \) delivered over one cycle is given by:
\[ E = P \cdot T \]
where \( T \) is the time period, \( T = \frac{2\pi}{\omega} \). Substituting \( \omega = \frac{2}{\sqrt{LC}} \):
\[ T = \frac{2\pi}{\frac{2}{\sqrt{LC}}} = \pi \sqrt{LC} \]
Now substituting \( P \) and \( T \):
\[ E = P \cdot T = \left(\frac{V_{rms}^2}{R^2 + \frac{9}{4}\frac{L}{C}} \cdot R\right) \cdot \left(\pi \sqrt{LC}\right) \]
\[ = \frac{V_{rms}^2 R \pi \sqrt{LC}}{R^2 + \frac{9}{4}\frac{L}{C}} \]

### Final Answer
The energy delivered by the source to the circuit during one period is:
\[ E = \frac{V_{rms}^2 R \pi \sqrt{LC}}{R^2 + \frac{9}{4}\frac{L}{C}} \]

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