A coil of inductance `L = 5//8 H` and of resistance `R = 62.8 (Omega)` is connected to the mains alternating voltage of frequency 50 Hz. What can be the capacitance of the capacitor `(in (mu)F)` connected in series with the coil if the power dissipated has to remain unchanged? Take `(pi)^(2) = 10`.

Explanation

To solve the problem step by step, we need to find the capacitance \( C \) that will keep the power dissipated in the circuit unchanged when a capacitor is added in series with the coil.

### Step 1: Write down the given values
- Inductance \( L = \frac{5}{8} \, \text{H} \)
- Resistance \( R = 62.8 \, \Omega \)
- Frequency \( f = 50 \, \text{Hz} \)
- \( \pi^2 = 10 \)

### Step 2: Understand the power dissipation condition
The power dissipated in the circuit must remain unchanged when a capacitor is added in series. This means:
\[ P_1 = P_2 \]
Where \( P_1 \) is the power in the original circuit (with only the inductor and resistor) and \( P_2 \) is the power in the new circuit (with the capacitor added).

### Step 3: Write the expression for power in both cases
The power in the first case (without the capacitor) can be expressed as:
\[ P_1 = \frac{V_{\text{rms}}^2}{Z^2} R \]
Where \( Z \) is the impedance given by:
\[ Z = \sqrt{R^2 + X_L^2} \]
And \( X_L = \omega L = 2 \pi f L \).

### Step 4: Calculate \( X_L \)
First, calculate \( \omega \):
\[ \omega = 2 \pi f = 2 \pi \times 50 = 100 \pi \, \text{rad/s} \]
Now, calculate \( X_L \):
\[ X_L = \omega L = 100 \pi \times \frac{5}{8} = \frac{500 \pi}{8} = 62.5 \pi \, \Omega \]

### Step 5: Write the expression for power in the second case
In the second case (with the capacitor), the impedance is:
\[ Z' = \sqrt{R^2 + (X_C - X_L)^2} \]
Where \( X_C = \frac{1}{\omega C} \).

### Step 6: Set up the equation for unchanged power
Since \( P_1 = P_2 \):
\[ \frac{V_{\text{rms}}^2}{Z^2} R = \frac{V_{\text{rms}}^2}{(Z')^2} R \]
This simplifies to:
\[ Z^2 = (Z')^2 \]

### Step 7: Substitute the expressions for impedance
From the previous steps, we have:
\[ R^2 + X_L^2 = R^2 + (X_C - X_L)^2 \]
Cancelling \( R^2 \) gives:
\[ X_L^2 = (X_C - X_L)^2 \]

### Step 8: Expand and simplify the equation
Expanding the right side:
\[ X_L^2 = X_C^2 - 2X_C X_L + X_L^2 \]
Cancelling \( X_L^2 \) gives:
\[ 0 = X_C^2 - 2X_C X_L \]
Factoring out \( X_C \):
\[ X_C (X_C - 2X_L) = 0 \]
Thus, \( X_C = 2X_L \).

### Step 9: Substitute \( X_L \) and solve for \( C \)
We know:
\[ X_L = 62.5 \pi \, \Omega \implies X_C = 2 \times 62.5 \pi = 125 \pi \, \Omega \]
Using the relationship \( X_C = \frac{1}{\omega C} \):
\[ 125 \pi = \frac{1}{100 \pi C} \]
Rearranging gives:
\[ C = \frac{1}{125 \pi \cdot 100 \pi} = \frac{1}{12500 \pi^2} \]

### Step 10: Substitute \( \pi^2 = 10 \)
\[ C = \frac{1}{12500 \times 10} = \frac{1}{125000} \, \text{F} \]
Converting to microfarads:
\[ C = \frac{1}{125000} \times 10^6 = \frac{10^6}{125000} = 8 \, \mu F \]

Thus, the capacitance \( C \) is:
\[ \boxed{8 \, \mu F} \]