In a series LCR circuit, impedance Z is same at two frequencies `f_1` and `f_2`. Therefore, the resonant frequency of this circuit is

Options

A. `(f_1+f_2)/2`
B. `(2f_1f_2)/(f_1+f_2)`
C. `sqrt(f_1^2+f_2^2)/2`
D. `sqrt(f_1f_2)`

Answer

To find the resonant frequency \( f_R \) of a series LCR circuit where the impedance \( Z \) is the same at two different frequencies \( f_1 \) and \( f_2 \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Impedance in a Series LCR Circuit**:
The impedance \( Z \) in a series LCR circuit is given by:
\[ Z = \sqrt{R^2 + (X_C - X_L)^2} \]
where \( X_C = \frac{1}{\omega C} \) (capacitive reactance) and \( X_L = \omega L \) (inductive reactance).

2. **Set Up the Equations for Two Frequencies**:
Since the impedance is the same at two frequencies \( f_1 \) and \( f_2 \), we can write:
\[ Z(f_1) = Z(f_2) \]
This leads to:
\[ \sqrt{R^2 + \left(\frac{1}{\omega_1 C} - \omega_1 L\right)^2} = \sqrt{R^2 + \left(\frac{1}{\omega_2 C} - \omega_2 L\right)^2} \]

3. **Square Both Sides**:
Squaring both sides eliminates the square root:
\[ R^2 + \left(\frac{1}{\omega_1 C} - \omega_1 L\right)^2 = R^2 + \left(\frac{1}{\omega_2 C} - \omega_2 L\right)^2 \]
Cancel \( R^2 \) from both sides:
\[ \left(\frac{1}{\omega_1 C} - \omega_1 L\right)^2 = \left(\frac{1}{\omega_2 C} - \omega_2 L\right)^2 \]

4. **Expand Both Sides**:
Expanding both sides gives:
\[ \frac{1}{\omega_1^2 C^2} - 2\frac{1}{\omega_1 C}\omega_1 L + \omega_1^2 L^2 = \frac{1}{\omega_2^2 C^2} - 2\frac{1}{\omega_2 C}\omega_2 L + \omega_2^2 L^2 \]

5. **Rearranging Terms**:
Rearranging the equation leads to a relationship between \( \omega_1 \) and \( \omega_2 \):
\[ \frac{1}{\omega_1^2 C^2} - \frac{1}{\omega_2^2 C^2} = 2\left(\frac{1}{\omega_1 C} - \frac{1}{\omega_2 C}\right)L \]

6. **Express in Terms of Frequencies**:
Substitute \( \omega_1 = 2\pi f_1 \) and \( \omega_2 = 2\pi f_2 \):
\[ \frac{1}{(2\pi f_1)^2 C^2} - \frac{1}{(2\pi f_2)^2 C^2} = 2L\left(\frac{1}{2\pi f_1 C} - \frac{1}{2\pi f_2 C}\right) \]

7. **Simplifying the Expression**:
After simplification, we find:
\[ \frac{1}{f_1 f_2} = \frac{1}{f_R^2} \]
where \( f_R \) is the resonant frequency.

8. **Final Result**:
Therefore, the resonant frequency \( f_R \) is given by:
\[ f_R = \sqrt{f_1 f_2} \]

### Final Answer:
The resonant frequency \( f_R \) of the circuit is:
\[ f_R = \sqrt{f_1 f_2} \]