An `AC` source of variable frequency `f` is connected to an `LCR` series circuit. Which one of the graphs in figure represents the variation of current of current `I` in the circuit with frequecy `f`?

Options

A.
B.
C.
D.

Explanation

To solve the problem, we need to analyze how the current \( I \) in an LCR series circuit varies with frequency \( f \). Here’s the step-by-step solution:

### Step 1: Understanding the LCR Circuit
In an LCR series circuit, the total current \( I \) is determined by the voltage \( V \) applied and the total impedance \( Z \) of the circuit. The impedance \( Z \) is given by:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
where:
- \( R \) is the resistance,
- \( X_L = \omega L = 2\pi f L \) is the inductive reactance,
- \( X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C} \) is the capacitive reactance.

### Step 2: Expression for Current
The current \( I \) can be expressed as:
\[ I = \frac{V}{Z} \]
Substituting the expression for \( Z \):
\[ I = \frac{V}{\sqrt{R^2 + (X_L - X_C)^2}} \]

### Step 3: Analyzing Resonance
At resonance frequency \( f_0 \), the inductive reactance equals the capacitive reactance (\( X_L = X_C \)). This occurs when:
\[ 2\pi f_0 L = \frac{1}{2\pi f_0 C} \]
From this, we can find the resonance frequency:
\[ f_0 = \frac{1}{2\pi \sqrt{LC}} \]
At this frequency, the impedance \( Z \) is minimized to \( R \), leading to maximum current:
\[ I_{\text{max}} = \frac{V}{R} \]

### Step 4: Current Variation with Frequency
- For frequencies below \( f_0 \), \( X_C > X_L \) leading to higher impedance and lower current.
- For frequencies above \( f_0 \), \( X_L > X_C \) also leading to higher impedance and lower current.
- Therefore, the current \( I \) reaches its maximum value at the resonance frequency \( f_0 \) and decreases as we move away from this frequency in either direction.

### Step 5: Identifying the Correct Graph
From the analysis, we conclude that the graph representing the current \( I \) as a function of frequency \( f \) will show a peak at the resonance frequency \( f_0 \). This corresponds to option C, where the current is maximum at a specific frequency and decreases on either side of that frequency.

### Conclusion
The correct answer is option C, which shows that the current \( I \) in the circuit varies with frequency \( f \) such that it is maximum at the resonance frequency and decreases for frequencies away from resonance.

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