An ac source of angular frequency `omega` is fed across a resistor R and a capacitor C in series. The current registered is I. If now the freqency of source is chaged to `(omega)//3` (but maintainging the same voltage), the current in the circuit is found to be halved. Calculate the ration of hte reactance to resistance at the original frequency `omega`.

Explanation

To solve the problem, we need to analyze the behavior of the circuit with a resistor \( R \) and a capacitor \( C \) in series when the frequency of the AC source changes. Let's break it down step-by-step:

### Step 1: Write the expression for current at frequency \( \omega \)

The current \( I \) in the circuit can be expressed using Ohm's law for AC circuits. The impedance \( Z \) of the circuit is given by:

\[ Z = \sqrt{R^2 + X_C^2} \]

where \( X_C \) is the capacitive reactance, defined as:

\[ X_C = \frac{1}{C \omega} \]

Thus, the current \( I \) can be expressed as:

\[ I = \frac{V}{Z} = \frac{V}{\sqrt{R^2 + \left(\frac{1}{C \omega}\right)^2}} \]

### Step 2: Write the expression for current at frequency \( \frac{\omega}{3} \)

When the frequency is changed to \( \frac{\omega}{3} \), the new capacitive reactance \( X_C' \) becomes:

\[ X_C' = \frac{1}{C \left(\frac{\omega}{3}\right)} = \frac{3}{C \omega} \]

The new impedance \( Z' \) is:

\[ Z' = \sqrt{R^2 + \left(\frac{3}{C \omega}\right)^2} \]

The new current \( I' \) is given as half of the original current \( I \):

\[ I' = \frac{V}{Z'} = \frac{V}{\sqrt{R^2 + \left(\frac{3}{C \omega}\right)^2}} = \frac{I}{2} \]

### Step 3: Set up the equations

From the first equation, we have:

\[ I = \frac{V}{\sqrt{R^2 + \left(\frac{1}{C \omega}\right)^2}} \]

From the second equation, we have:

\[ \frac{I}{2} = \frac{V}{\sqrt{R^2 + \left(\frac{3}{C \omega}\right)^2}} \]

### Step 4: Divide the equations

Now, we can divide the two equations:

\[ \frac{I}{\frac{I}{2}} = \frac{V}{\sqrt{R^2 + \left(\frac{3}{C \omega}\right)^2}} \cdot \frac{\sqrt{R^2 + \left(\frac{1}{C \omega}\right)^2}}{V} \]

This simplifies to:

\[ 2 = \frac{\sqrt{R^2 + \left(\frac{1}{C \omega}\right)^2}}{\sqrt{R^2 + \left(\frac{3}{C \omega}\right)^2}} \]

### Step 5: Square both sides

Squaring both sides gives:

\[ 4 = \frac{R^2 + \left(\frac{1}{C \omega}\right)^2}{R^2 + \left(\frac{3}{C \omega}\right)^2} \]

### Step 6: Cross-multiply and simplify

Cross-multiplying leads to:

\[ 4(R^2 + \left(\frac{3}{C \omega}\right)^2) = R^2 + \left(\frac{1}{C \omega}\right)^2 \]

Expanding and simplifying gives:

\[ 4R^2 + \frac{36}{C^2 \omega^2} = R^2 + \frac{1}{C^2 \omega^2} \]

\[ 3R^2 = \frac{1 - 36}{C^2 \omega^2} \]

### Step 7: Solve for the ratio \( \frac{X_C}{R} \)

Rearranging gives:

\[ \frac{1}{C^2 \omega^2} = \frac{3}{5R^2} \]

This can be rewritten as:

\[ X_C^2 = \frac{3}{5} R^2 \]

Taking the square root:

\[ X_C = \sqrt{\frac{3}{5}} R \]

### Final Result

Thus, the ratio of the reactance to resistance at the original frequency \( \omega \) is:

\[ \frac{X_C}{R} = \sqrt{\frac{3}{5}} \]