Frequency of L-C circuit is `f_(1)`. If a resistance R is also added to it the frequency becomes `f_(2)`. The ratio of `(f_(2))/(f_(1))` will be.

Options

A. `sqrt(1+(R^2C)/(4L))`
B. `sqrt(1-(R^(2)C)/(4L))`
C. `sqrt(1+(R^(2)C)/(L))`
D. `sqrt(1-(R^(2)C)/(L))`

Explanation

To solve the problem step by step, we will derive the frequencies of the LC circuit and the RLC circuit, and then find the ratio of these frequencies.

### Step 1: Frequency of the LC Circuit
The frequency \( f_1 \) of an LC circuit (without resistance) is given by the formula:
\[ f_1 = \frac{1}{2\pi\sqrt{LC}} \]
where \( L \) is the inductance and \( C \) is the capacitance.

### Step 2: Frequency of the RLC Circuit
When a resistance \( R \) is added to the circuit, the frequency \( f_2 \) of the RLC circuit is given by:
\[ f_2 = \frac{1}{2\pi\sqrt{L \left( C - \frac{R^2}{4L^2} \right)}} \]
This formula accounts for the damping effect of the resistance.

### Step 3: Finding the Ratio \( \frac{f_2}{f_1} \)
To find the ratio \( \frac{f_2}{f_1} \), we can substitute the expressions for \( f_1 \) and \( f_2 \):
\[ \frac{f_2}{f_1} = \frac{\frac{1}{2\pi\sqrt{L \left( C - \frac{R^2}{4L^2} \right)}}}{\frac{1}{2\pi\sqrt{LC}}} \]
This simplifies to:
\[ \frac{f_2}{f_1} = \frac{\sqrt{LC}}{\sqrt{L \left( C - \frac{R^2}{4L^2} \right)}} \]

### Step 4: Simplifying the Ratio
Now we can simplify the expression:
\[ \frac{f_2}{f_1} = \sqrt{\frac{LC}{L \left( C - \frac{R^2}{4L^2} \right)}} = \sqrt{\frac{C}{C - \frac{R^2}{4L^2}}} \]

### Step 5: Final Expression
Thus, the final expression for the ratio \( \frac{f_2}{f_1} \) is:
\[ \frac{f_2}{f_1} = \sqrt{1 - \frac{R^2C}{4L}} \]

### Conclusion
The ratio of the frequencies \( \frac{f_2}{f_1} \) when resistance \( R \) is added to the LC circuit is:
\[ \frac{f_2}{f_1} = \sqrt{1 - \frac{R^2C}{4L}} \]