An a.c. 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 frequency of the source is changed to `omega//3` (but maintaining the same voltage), the current in the circuit is found to be halved. calculate the ratio of reactance to resistance at the original frequency `omega`.

Options

A. `sqrt(3/5)`
B. `sqrt(2/5)`
C. `sqrt(1/5)`
D. `sqrt(4/5)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Write down the known relationships
We know that the impedance \( Z \) of a series RC circuit is given by:
\[ Z = \sqrt{R^2 + X_C^2} \]
where \( X_C \) is the capacitive reactance given by:
\[ X_C = \frac{1}{\omega C} \]

### Step 2: Write the current equations for both frequencies
For the first case (frequency \( \omega \)):
\[ V = I \cdot Z \implies V = I \cdot \sqrt{R^2 + X_C^2} \]
Substituting \( X_C \):
\[ V = I \cdot \sqrt{R^2 + \left(\frac{1}{\omega C}\right)^2} \]

For the second case (frequency \( \frac{\omega}{3} \)):
The current is halved, so \( I_2 = \frac{I}{2} \):
\[ V = I_2 \cdot Z_2 \implies V = \frac{I}{2} \cdot \sqrt{R^2 + X_{C2}^2} \]
where \( X_{C2} = \frac{1}{\frac{\omega}{3} C} = \frac{3}{\omega C} \).

### Step 3: Set the equations equal since the voltage remains the same
From the first case:
\[ V = I \cdot \sqrt{R^2 + \left(\frac{1}{\omega C}\right)^2} \]
From the second case:
\[ V = \frac{I}{2} \cdot \sqrt{R^2 + \left(\frac{3}{\omega C}\right)^2} \]

Setting these two equations equal:
\[ I \cdot \sqrt{R^2 + \left(\frac{1}{\omega C}\right)^2} = \frac{I}{2} \cdot \sqrt{R^2 + \left(\frac{3}{\omega C}\right)^2} \]

### Step 4: Cancel \( I \) and simplify
Assuming \( I \neq 0 \):
\[ \sqrt{R^2 + \left(\frac{1}{\omega C}\right)^2} = \frac{1}{2} \cdot \sqrt{R^2 + \left(\frac{3}{\omega C}\right)^2} \]

### Step 5: Square both sides to eliminate the square root
\[ R^2 + \left(\frac{1}{\omega C}\right)^2 = \frac{1}{4} \left(R^2 + \left(\frac{3}{\omega C}\right)^2\right) \]

### Step 6: Expand and rearrange
Multiplying through by 4 to eliminate the fraction:
\[ 4R^2 + \frac{4}{\omega^2 C^2} = R^2 + \frac{9}{\omega^2 C^2} \]
Rearranging gives:
\[ 4R^2 - R^2 = \frac{9}{\omega^2 C^2} - \frac{4}{\omega^2 C^2} \]
\[ 3R^2 = \frac{5}{\omega^2 C^2} \]

### Step 7: Solve for the ratio of reactance to resistance
Recall that \( X_C = \frac{1}{\omega C} \):
\[ X_C^2 = \frac{1}{\omega^2 C^2} \]
Thus,
\[ 3R^2 = 5X_C^2 \implies \frac{X_C^2}{R^2} = \frac{3}{5} \]
Taking the square root:
\[ \frac{X_C}{R} = \sqrt{\frac{3}{5}} \]

### Final Answer
The ratio of reactance to resistance at the original frequency \( \omega \) is:
\[ \frac{X_C}{R} = \sqrt{\frac{3}{5}} \]