A capacitor of `10 (mu)F` and an inductor of 1 H are joined in series. An ac of 50 Hz is applied to this combination. What is the impedance of the combination?

Options

A. `(5(pi^(2)-5))/(pi) Omega`
B. `(10(10-pi^(2)))/(pi) Omega`
C. `(10(pi^(2)-5))/(pi) Omega`
D. `(5(10-pi^(2)))/(pi) Omega`

Explanation

To solve the problem of finding the impedance of a series combination of a capacitor and an inductor connected to an AC source, we will follow these steps:

### Step 1: Calculate the inductive reactance (X_L)
The inductive reactance \(X_L\) is given by the formula:
\[ X_L = \omega L \]
where \(\omega = 2\pi f\) and \(L\) is the inductance.

Given:
- \(f = 50 \, \text{Hz}\)
- \(L = 1 \, \text{H}\)

Calculating \(\omega\):
\[ \omega = 2\pi \times 50 = 100\pi \, \text{rad/s} \]

Now substituting into the formula for \(X_L\):
\[ X_L = 100\pi \times 1 = 100\pi \, \Omega \]

### Step 2: Calculate the capacitive reactance (X_C)
The capacitive reactance \(X_C\) is given by the formula:
\[ X_C = \frac{1}{\omega C} \]

Given:
- \(C = 10 \, \mu\text{F} = 10 \times 10^{-6} \, \text{F}\)

Substituting the values:
\[ X_C = \frac{1}{2\pi \times 50 \times 10 \times 10^{-6}} \]
Calculating:
\[ X_C = \frac{1}{100\pi \times 10^{-6}} = \frac{1000}{\pi} \, \Omega \]

### Step 3: Calculate the total impedance (Z)
The total impedance \(Z\) of a series circuit containing an inductor and a capacitor is given by:
\[ Z = \sqrt{(X_L - X_C)^2} \]

Substituting the values of \(X_L\) and \(X_C\):
\[ Z = \sqrt{(100\pi - \frac{1000}{\pi})^2} \]

To simplify:
1. Find a common denominator:
\[ Z = \sqrt{(100\pi^2 - 1000)^2 / \pi^2} \]
2. Thus:
\[ Z = \frac{1}{\pi} \sqrt{(100\pi^2 - 1000)^2} \]

### Final Result
The impedance \(Z\) can be expressed as:
\[ Z = 100 \left(10 - \frac{\pi^2}{100}\right) \, \Omega \]