When an AC voltage, of variable frequency is applied to series L-C-R circuit , the current in the circuit is the same at 4 kHz and 9 kHz. The current in the circuit is maximum at

Options

A. 5 kHz
B. 6.5 kHz
C. 4.2 kHz
D. 6 kHz

Explanation

To solve the problem step by step, we will analyze the given information about the series L-C-R circuit and the conditions under which the current is the same at two different frequencies.

### Step-by-Step Solution:

1. **Understanding the Given Information:**
We know that the current in the circuit is the same at two different frequencies: \( f_1 = 4 \, \text{kHz} \) and \( f_2 = 9 \, \text{kHz} \).

2. **Using the Relationship of Current in an L-C-R Circuit:**
In a series L-C-R circuit, the current \( I \) can be expressed as:
\[ I = \frac{V}{Z} \]
where \( Z \) is the impedance of the circuit. The impedance \( Z \) is given by:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
where \( X_L = \omega L \) (inductive reactance) and \( X_C = \frac{1}{\omega C} \) (capacitive reactance).

3. **Setting Up the Condition for Equal Currents:**
Since the current is the same at both frequencies, we can write:
\[ Z_1 = Z_2 \]
This implies:
\[ \sqrt{R^2 + (X_{L1} - X_{C1})^2} = \sqrt{R^2 + (X_{L2} - X_{C2})^2} \]

4. **Expressing Reactances in Terms of Frequencies:**
We can express the reactances at the two frequencies:
- For \( f_1 = 4 \, \text{kHz} \):
\[ X_{L1} = 2\pi f_1 L, \quad X_{C1} = \frac{1}{2\pi f_1 C} \]
- For \( f_2 = 9 \, \text{kHz} \):
\[ X_{L2} = 2\pi f_2 L, \quad X_{C2} = \frac{1}{2\pi f_2 C} \]

5. **Equating the Impedance Conditions:**
From the condition \( Z_1 = Z_2 \), we can derive:
\[ (X_{L1} - X_{C1})^2 = (X_{L2} - X_{C2})^2 \]
This leads to two possible cases:
\[ X_{L1} - X_{C1} = X_{L2} - X_{C2} \quad \text{or} \quad X_{L1} - X_{C1} = -(X_{L2} - X_{C2}) \]

6. **Finding the Resonant Frequency:**
The maximum current occurs at the resonant frequency \( f_r \), where:
\[ X_L = X_C \]
This gives:
\[ 2\pi f_r L = \frac{1}{2\pi f_r C} \]
Rearranging gives:
\[ f_r = \frac{1}{2\pi \sqrt{LC}} \]

7. **Calculating the Resonant Frequency:**
We know that the product of the two frequencies where the current is the same can be used to find the resonant frequency:
\[ f_r = \sqrt{f_1 \cdot f_2} = \sqrt{4 \, \text{kHz} \cdot 9 \, \text{kHz}} = \sqrt{36} \, \text{kHz} = 6 \, \text{kHz} \]

### Final Answer:
The current in the circuit is maximum at **6 kHz**.

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