A coil of inductance `L=5H` and resistance `R=55Omega` is connected in series to the mains alternating voltage of frequency `f=50Hz` in series. What can be the non-zero capacitance of the capacitor (in `muF`) 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 analyze the circuit and the conditions given in the question.

### Step 1: Understanding the Circuit
We have a coil with inductance \( L = 5 \, H \) and resistance \( R = 55 \, \Omega \) connected in series with an alternating voltage source of frequency \( f = 50 \, Hz \). We need to find the capacitance \( C \) that can be added in series such that the power dissipated remains unchanged.

### Step 2: Calculate the Angular Frequency
The angular frequency \( \omega \) is given by:
\[ \omega = 2\pi f \]
Substituting the given frequency:
\[ \omega = 2\pi \times 50 = 100\pi \, \text{rad/s} \]

### Step 3: Calculate the Impedance of the LR Circuit
The impedance \( Z_1 \) of the LR circuit (without the capacitor) is given by:
\[ Z_1 = \sqrt{R^2 + X_L^2} \]
where \( X_L = \omega L \) is the inductive reactance. First, we calculate \( X_L \):
\[ X_L = \omega L = (100\pi)(5) = 500\pi \, \Omega \]
Now, we can calculate \( Z_1 \):
\[ Z_1 = \sqrt{55^2 + (500\pi)^2} \]

### Step 4: Calculate the Impedance of the LRC Circuit
When a capacitor is added in series, the impedance \( Z_2 \) becomes:
\[ Z_2 = \sqrt{R^2 + (X_L - X_C)^2} \]
where \( X_C = \frac{1}{\omega C} \) is the capacitive reactance.

### Step 5: Power Dissipation Condition
The average power \( P \) in an AC circuit is given by:
\[ P = \frac{V^2 R}{Z^2} \]
Since the power must remain unchanged, we have:
\[ P_1 = P_2 \Rightarrow \frac{V^2 R}{Z_1^2} = \frac{V^2 R}{Z_2^2} \]
This simplifies to:
\[ Z_1^2 = Z_2^2 \]

### Step 6: Equating the Impedances
From the previous steps, we have:
\[ R^2 + (500\pi)^2 = R^2 + (500\pi - \frac{1}{\omega C})^2 \]
Cancelling \( R^2 \) from both sides gives:
\[ (500\pi)^2 = (500\pi - \frac{1}{\omega C})^2 \]

### Step 7: Expand and Simplify
Expanding the right side:
\[ (500\pi)^2 = (500\pi)^2 - 2(500\pi)(\frac{1}{\omega C}) + \left(\frac{1}{\omega C}\right)^2 \]
Cancelling \( (500\pi)^2 \) from both sides:
\[ 0 = -2(500\pi)(\frac{1}{\omega C}) + \left(\frac{1}{\omega C}\right)^2 \]

### Step 8: Rearranging the Equation
Rearranging gives:
\[ \left(\frac{1}{\omega C}\right)^2 = 2(500\pi)(\frac{1}{\omega C}) \]
Let \( x = \frac{1}{\omega C} \):
\[ x^2 - 1000\pi x = 0 \]
Factoring out \( x \):
\[ x(x - 1000\pi) = 0 \]
Thus, \( x = 0 \) or \( x = 1000\pi \).

### Step 9: Solving for Capacitance
Since \( x = \frac{1}{\omega C} \), we have:
\[ \frac{1}{\omega C} = 1000\pi \Rightarrow C = \frac{1}{1000\pi \omega} \]
Substituting \( \omega = 100\pi \):
\[ C = \frac{1}{1000\pi \cdot 100\pi} = \frac{1}{100000\pi^2} \]
Using \( \pi^2 = 10 \):
\[ C = \frac{1}{100000 \cdot 10} = \frac{1}{1000000} = 1 \, \mu F \]

### Final Answer
The non-zero capacitance of the capacitor connected in series with the coil is:
\[ C = 1 \, \mu F \]