A series combination of resistor (R), capacitor (C) is connected to an AC source of angular frequency `omega`. Keeping the voltage same, If the frequency is changed to `Omega/3`, the current becomes half of the original current. Then, the ratio of the capacitance reactance and resistance at the former frequency is

Options

A. `sqrt(0.6)`
B. `sqrt(3)`
C. `sqrt(2)`
D. `sqrt(6)`

Explanation

To solve the problem step by step, we need to analyze the given information and apply the relevant formulas.

### Step 1: Understand the relationship between current, voltage, and impedance
The current in an AC circuit with a resistor (R) and capacitor (C) in series is given by:
\[ I_{rms} = \frac{V_{rms}}{Z} \]
where \( Z \) is the impedance of the circuit.

### Step 2: Write the expression for impedance
The impedance \( Z \) for a resistor and capacitor in series is given by:
\[ Z = \sqrt{R^2 + X_C^2} \]
where \( X_C \) is the capacitive reactance given by:
\[ X_C = \frac{1}{\omega C} \]
Thus, we can rewrite the impedance as:
\[ Z = \sqrt{R^2 + \left(\frac{1}{\omega C}\right)^2} \]

### Step 3: Set up the equations for the two frequencies
Let the original frequency be \( \omega \) and the new frequency be \( \frac{\omega}{3} \). The current at the original frequency is:
\[ I_{rms1} = \frac{V_{rms}}{\sqrt{R^2 + \left(\frac{1}{\omega C}\right)^2}} \]

When the frequency is changed to \( \frac{\omega}{3} \), the current becomes half:
\[ I_{rms2} = \frac{V_{rms}}{\sqrt{R^2 + \left(\frac{1}{\frac{\omega}{3} C}\right)^2}} = \frac{1}{2} I_{rms1} \]

### Step 4: Substitute the expressions for currents
Substituting the expressions for \( I_{rms1} \) and \( I_{rms2} \) gives us:
\[ \frac{V_{rms}}{\sqrt{R^2 + \left(\frac{1}{\frac{\omega}{3} C}\right)^2}} = \frac{1}{2} \cdot \frac{V_{rms}}{\sqrt{R^2 + \left(\frac{1}{\omega C}\right)^2}} \]

### Step 5: Simplify the equation
Cancelling \( V_{rms} \) from both sides and squaring both sides, we get:
\[ \frac{1}{R^2 + \left(\frac{3}{\omega C}\right)^2} = \frac{1}{4} \cdot \frac{1}{R^2 + \left(\frac{1}{\omega C}\right)^2} \]

### Step 6: Cross-multiply and simplify
Cross-multiplying gives:
\[ 4 \left(R^2 + \left(\frac{3}{\omega C}\right)^2\right) = R^2 + \left(\frac{1}{\omega C}\right)^2 \]
Expanding and simplifying:
\[ 4R^2 + \frac{36}{\omega^2 C^2} = R^2 + \frac{1}{\omega^2 C^2} \]
\[ 3R^2 + \frac{36}{\omega^2 C^2} - \frac{1}{\omega^2 C^2} = 0 \]
\[ 3R^2 + \frac{35}{\omega^2 C^2} = 0 \]

### Step 7: Rearranging the equation
Rearranging gives:
\[ 3R^2 = -\frac{35}{\omega^2 C^2} \]
This implies:
\[ \frac{1}{\omega^2 C^2} = \frac{3R^2}{35} \]

### Step 8: Find the ratio of capacitive reactance to resistance
Now, we need to find the ratio \( \frac{X_C}{R} \):
\[ X_C = \frac{1}{\omega C} \]
Substituting \( C \) from the previous equation gives:
\[ \frac{X_C}{R} = \frac{1}{\omega C R} = \sqrt{\frac{3}{35}} \]

### Step 9: Final ratio
Thus, the ratio of capacitive reactance to resistance is:
\[ \frac{X_C}{R} = \sqrt{0.6} \]

### Final Answer
The ratio of the capacitance reactance and resistance at the former frequency is:
\[ \sqrt{0.6} \]