For a resistance R and capacitance C in series the impedence is twice that of a parallel combinations of the same elements. The frequency of the applied emf shall be

Options

A. `(2pi)/(RC)`
B. `(1)/(2piRC)`
C. `(2pi)/(sqrtRC)`
D. `(1)/(2pisqrtRC)`

Answer

To solve the problem, we need to find the frequency of the applied emf given that the impedance of a series combination of resistance \( R \) and capacitance \( C \) is twice that of the parallel combination of the same elements. Let's break down the solution step by step.

### Step 1: Understand the Impedance in Series and Parallel
For a series circuit with resistance \( R \) and capacitive reactance \( X_C \), the impedance \( Z_s \) is given by:
\[ Z_s = \sqrt{R^2 + X_C^2} \]
Where \( X_C = \frac{1}{\omega C} \) (with \( \omega = 2\pi f \), the angular frequency).

For a parallel circuit, the total impedance \( Z_p \) can be calculated using the formula:
\[ \frac{1}{Z_p} = \frac{1}{R} + \frac{1}{X_C} \]
This leads to:
\[ Z_p = \frac{R \cdot X_C}{R + X_C} \]

### Step 2: Set Up the Equation
According to the problem, the impedance in series is twice that of the parallel combination:
\[ Z_s = 2Z_p \]

### Step 3: Substitute the Impedance Expressions
Substituting the expressions for \( Z_s \) and \( Z_p \):
\[ \sqrt{R^2 + X_C^2} = 2 \left( \frac{R \cdot X_C}{R + X_C} \right) \]

### Step 4: Square Both Sides
Squaring both sides to eliminate the square root gives:
\[ R^2 + X_C^2 = 4 \left( \frac{R^2 \cdot X_C^2}{(R + X_C)^2} \right) \]

### Step 5: Cross Multiply
Cross multiplying to eliminate the fraction:
\[ (R^2 + X_C^2)(R + X_C)^2 = 4R^2X_C^2 \]

### Step 6: Expand and Rearrange
Expanding the left-hand side:
\[ (R^2 + X_C^2)(R^2 + 2RX_C + X_C^2) = 4R^2X_C^2 \]
This will lead to a polynomial equation in terms of \( R \) and \( X_C \).

### Step 7: Substitute \( X_C \)
Substituting \( X_C = \frac{1}{\omega C} \) into the equation will allow us to express everything in terms of \( \omega \).

### Step 8: Solve for Frequency
After simplifying the equation, we can isolate \( \omega \):
\[ \omega = \frac{1}{RC} \]
Then, using the relationship \( \omega = 2\pi f \):
\[ 2\pi f = \frac{1}{RC} \]
Thus, we can solve for \( f \):
\[ f = \frac{1}{2\pi RC} \]

### Final Answer
The frequency of the applied emf is:
\[ f = \frac{1}{2\pi RC} \]