The current in series `LCR` circuit will be the maximum when `omega` is

Options

A. As large as possible
B. Equal `o` natural frequency of `LCR` system
C. `sqrt(LC)`
D. `sqrt(1//LC)`

Explanation

To find the value of `omega` at which the current in a series LCR circuit is maximum, we can follow these steps:

### Step 1: Understand the relationship between current, voltage, and impedance
In a series LCR circuit, the current (I) is given by the formula:
\[ I = \frac{V}{Z} \]
where \( V \) is the voltage and \( Z \) is the impedance of the circuit.

### Step 2: Identify when the current is maximum
The current will be maximum when the impedance \( Z \) is minimum. Therefore, we need to find the condition under which the impedance is minimized.

### Step 3: Impedance in a series LCR circuit
The impedance \( Z \) in a series LCR circuit is given by:
\[ Z = R + j(X_L - X_C) \]
where:
- \( R \) is the resistance,
- \( X_L = \omega L \) is the inductive reactance,
- \( X_C = \frac{1}{\omega C} \) is the capacitive reactance.

### Step 4: Condition for resonance
At resonance, the inductive reactance equals the capacitive reactance:
\[ X_L = X_C \]
This implies:
\[ \omega L = \frac{1}{\omega C} \]

### Step 5: Solve for omega
Rearranging the equation gives:
\[ \omega^2 = \frac{1}{LC} \]
Taking the square root of both sides, we find:
\[ \omega = \frac{1}{\sqrt{LC}} \]

### Step 6: Conclusion
Thus, the current in the series LCR circuit will be maximum when:
\[ \omega = \frac{1}{\sqrt{LC}} \]