In a series LCR circuit, the frequencies at which the current amplitude is`(1)/(sqrt(2))`mes the current amplitude at resonance are` f_(1) and f_(2) (gt f_(1)).` Find the frequency bandwidth of resonance which is defined as `Deltaf=f_(2)-f_(1).` Express your answer in terms of R and L. Assume that resonance frequency `f_(0)gtgtDeltaf`

Answer

To solve the problem of finding the frequency bandwidth of resonance in a series LCR circuit, we will follow these steps:

### Step 1: Identify the Resonance Frequency
The resonance frequency \( f_0 \) for a series LCR circuit is given by the formula:
\[ f_0 = \frac{1}{2\pi\sqrt{LC}} \]

### Step 2: Write the Expression for Current Amplitude
The current amplitude \( I \) at a frequency \( f \) is given by:
\[ I = \frac{V_0}{Z} \]
where \( Z \) is the impedance of the circuit. The impedance \( Z \) for a series LCR circuit is:
\[ Z = \sqrt{R^2 + \left( \omega L - \frac{1}{\omega C} \right)^2} \]
Here, \( \omega = 2\pi f \).

### Step 3: Set Up the Condition for Current Amplitude
At the frequencies \( f_1 \) and \( f_2 \), the current amplitude is given as:
\[ I = \frac{V_0}{Z} = \frac{1}{\sqrt{2}} \cdot \frac{V_0}{R} \]
This leads to:
\[ Z = \sqrt{2} R \]
Thus, we have:
\[ R^2 + \left( \omega L - \frac{1}{\omega C} \right)^2 = 2R^2 \]
This simplifies to:
\[ \left( \omega L - \frac{1}{\omega C} \right)^2 = R^2 \]

### Step 4: Solve for Frequencies \( f_1 \) and \( f_2 \)
Taking the square root gives us two equations:
1. \( \omega L - \frac{1}{\omega C} = R \)
2. \( \omega L - \frac{1}{\omega C} = -R \)

For \( f_1 \):
\[ 2\pi f_1 L - \frac{1}{2\pi f_1 C} = R \quad \text{(Equation 1)} \]
For \( f_2 \):
\[ 2\pi f_2 L - \frac{1}{2\pi f_2 C} = -R \quad \text{(Equation 2)} \]

### Step 5: Manipulate the Equations
From Equation 1:
\[ 2\pi f_1 L = R + \frac{1}{2\pi f_1 C} \]
From Equation 2:
\[ 2\pi f_2 L = -R + \frac{1}{2\pi f_2 C} \]

### Step 6: Find the Bandwidth \( \Delta f \)
To find the bandwidth \( \Delta f = f_2 - f_1 \), we can use the relationships derived from the equations. By manipulating the equations and using the approximation \( f_0 \gg \Delta f \), we can derive:
\[ \Delta f = \frac{R}{2\pi L} \]

### Final Answer
Thus, the frequency bandwidth of resonance \( \Delta f \) is given by:
\[ \Delta f = \frac{R}{2\pi L} \]