An AC source of frequency `omega` when fed into a RC series circuit, current is recorded to be l. If now frequency is changed to `(omega)/(4)` (keeping voltage same), the current is found to 1/2. The ratio of reactance to resistance at original frequency `omega` is

Options

A. 2
B. `1/2`
C. `(1)/(sqrt(2))`
D. None of these

Answer

To solve the problem step by step, we will analyze the given information about the RC circuit and the changes in frequency and current.

### Step 1: Understand the Circuit and Given Information
We have an RC series circuit with:
- Initial frequency: \( \omega \)
- Current at frequency \( \omega \): \( I \)
- New frequency: \( \frac{\omega}{4} \)
- Current at frequency \( \frac{\omega}{4} \): \( \frac{I}{2} \)

### Step 2: Write the Expression for Current
The current \( I \) in an RC circuit is given by:
\[ I = \frac{V}{Z} \]
where \( Z \) is the impedance of the circuit.

### Step 3: Calculate Impedance at Frequency \( \omega \)
The impedance \( Z \) at frequency \( \omega \) is given by:
\[ Z_1 = \sqrt{R^2 + X_C^2} \]
where \( X_C \) is the capacitive reactance given by:
\[ X_C = \frac{1}{\omega C} \]
Thus,
\[ Z_1 = \sqrt{R^2 + \left(\frac{1}{\omega C}\right)^2} \]

### Step 4: Write the Expression for Current at Frequency \( \omega \)
Using the impedance, the current at frequency \( \omega \) can be expressed as:
\[ I = \frac{V}{Z_1} = \frac{V}{\sqrt{R^2 + \left(\frac{1}{\omega C}\right)^2}} \]

### Step 5: Calculate Impedance at Frequency \( \frac{\omega}{4} \)
At frequency \( \frac{\omega}{4} \), the capacitive reactance becomes:
\[ X_C' = \frac{1}{\frac{\omega}{4} C} = \frac{4}{\omega C} \]
Thus, the new impedance \( Z_2 \) is:
\[ Z_2 = \sqrt{R^2 + X_C'^2} = \sqrt{R^2 + \left(\frac{4}{\omega C}\right)^2} \]

### Step 6: Write the Expression for Current at Frequency \( \frac{\omega}{4} \)
The current at frequency \( \frac{\omega}{4} \) is:
\[ \frac{I}{2} = \frac{V}{Z_2} = \frac{V}{\sqrt{R^2 + \left(\frac{4}{\omega C}\right)^2}} \]

### Step 7: Set Up the Equation
From the two expressions for current, we have:
\[ \frac{I}{2} = \frac{V}{\sqrt{R^2 + \left(\frac{4}{\omega C}\right)^2}} \]
and
\[ I = \frac{V}{\sqrt{R^2 + \left(\frac{1}{\omega C}\right)^2}} \]

### Step 8: Equate the Two Expressions
We can equate the two expressions for current:
\[ \frac{V}{\sqrt{R^2 + \left(\frac{1}{\omega C}\right)^2}} = 2 \cdot \frac{V}{\sqrt{R^2 + \left(\frac{4}{\omega C}\right)^2}} \]
Cancelling \( V \) and squaring both sides gives:
\[ R^2 + \left(\frac{1}{\omega C}\right)^2 = 4 \left(R^2 + \left(\frac{4}{\omega C}\right)^2\right) \]

### Step 9: Simplify the Equation
Expanding and simplifying leads to:
\[ R^2 + \frac{1}{\omega^2 C^2} = 4R^2 + \frac{64}{\omega^2 C^2} \]
Rearranging gives:
\[ -3R^2 = \frac{63}{\omega^2 C^2} \]
Thus,
\[ 3R^2 = \frac{63}{\omega^2 C^2} \]

### Step 10: Solve for \( \frac{X_C}{R} \)
From \( X_C = \frac{1}{\omega C} \), we can express the ratio:
\[ \frac{X_C}{R} = \frac{1}{R \cdot \omega C} \]
Substituting \( \omega C = \frac{2}{R} \) gives:
\[ \frac{X_C}{R} = \frac{1}{2} \]

### Final Answer
The ratio of reactance to resistance at the original frequency \( \omega \) is:
\[ \frac{X_C}{R} = \frac{1}{2} \]