An alternating EMF of frequency `(1)/(2pi sqrt(LC))` is applied to a series LCR circuit. For this frequency of the applied EMF,

Options

A. The circuit is at resonance and its impedance is made up only of a reactive part
B. the current in the circuit is in phase with the applied EMF and the voltage across R not equal this applied EMF potential differences
C. the sum of the potential difference across the inductance and capacitance equals the applied EMF which is `180^(@)` ahead of phase of the current in the circuit
D. the quality factor of the circuit is `omegaL//R` or `1//omegaCR` and this is a measure of the voltage magnification (produced by the circuit at resonance) as well as sharpness of resonance of the circuit

Explanation

To solve the problem, we need to analyze the behavior of a series LCR circuit when an alternating EMF of frequency \( \frac{1}{2\pi \sqrt{LC}} \) is applied. This frequency is significant because it corresponds to the resonant frequency of the circuit. Here’s a step-by-step solution:

### Step 1: Understand the Resonant Frequency
The resonant frequency \( f_0 \) of a series LCR circuit is given by the formula:
\[ f_0 = \frac{1}{2\pi \sqrt{LC}} \]
This frequency is where the circuit will resonate, meaning the inductive reactance \( X_L \) and capacitive reactance \( X_C \) will be equal.

### Step 2: Identify the Condition for Resonance
At resonance, the inductive reactance \( X_L \) and capacitive reactance \( X_C \) are equal:
\[ X_L = X_C \]
This means that:
\[ \omega L = \frac{1}{\omega C} \]
where \( \omega = 2\pi f \) is the angular frequency.

### Step 3: Substitute the Given Frequency
Given that the frequency of the applied EMF is:
\[ f = \frac{1}{2\pi \sqrt{LC}} \]
We can convert this to angular frequency:
\[ \omega = 2\pi f = 2\pi \left(\frac{1}{2\pi \sqrt{LC}}\right) = \frac{1}{\sqrt{LC}} \]

### Step 4: Confirm Resonance Condition
Now, substituting \( \omega \) into the resonance condition:
\[ X_L = \omega L = \frac{1}{\sqrt{LC}} L = \frac{L}{\sqrt{LC}} = \sqrt{\frac{L}{C}} \]
\[ X_C = \frac{1}{\omega C} = \frac{1}{\frac{1}{\sqrt{LC}} C} = \sqrt{\frac{L}{C}} \]
Since \( X_L = X_C \), the circuit is indeed at resonance.

### Step 5: Analyze the Impedance
At resonance, the impedance \( Z \) of the circuit is purely resistive, as the reactive components cancel each other out:
\[ Z = R \]
where \( R \) is the resistance in the circuit.

### Conclusion
Thus, the circuit is at resonance when the alternating EMF of frequency \( \frac{1}{2\pi \sqrt{LC}} \) is applied, and the impedance is purely resistive.

### Final Answer
The correct option is that the circuit is at resonance.

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