A 50 Hz AC signal is applied in a circuit of inductance of `(1//pi)`H and resistance `2100Omega.` The impedance offered by the circuit is

Options

A. `1500 Omega`
B. `1700 Omega`
C. `2102 Omega`
D. `2500 Omega`

Explanation

To find the impedance offered by the circuit, we can use the formula for impedance in an R-L circuit:

\[ Z = \sqrt{R^2 + X_L^2} \]

Where:
- \( Z \) is the impedance,
- \( R \) is the resistance,
- \( X_L \) is the inductive reactance, which can be calculated using the formula \( X_L = \omega L \).

### Step 1: Identify the given values
- Frequency \( f = 50 \) Hz
- Inductance \( L = \frac{1}{\pi} \) H
- Resistance \( R = 2100 \, \Omega \)

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

### Step 3: Calculate the inductive reactance \( X_L \)
Using the formula for inductive reactance:
\[ X_L = \omega L \]
Substituting the values of \( \omega \) and \( L \):
\[ X_L = (100\pi) \left(\frac{1}{\pi}\right) = 100 \, \Omega \]

### Step 4: Substitute \( R \) and \( X_L \) into the impedance formula
Now we can substitute \( R \) and \( X_L \) into the impedance formula:
\[ Z = \sqrt{R^2 + X_L^2} \]
\[ Z = \sqrt{(2100)^2 + (100)^2} \]

### Step 5: Calculate \( R^2 \) and \( X_L^2 \)
Calculating \( R^2 \) and \( X_L^2 \):
\[ R^2 = 2100^2 = 4410000 \]
\[ X_L^2 = 100^2 = 10000 \]

### Step 6: Add \( R^2 \) and \( X_L^2 \)
Now, add the two values:
\[ R^2 + X_L^2 = 4410000 + 10000 = 4420000 \]

### Step 7: Calculate the impedance \( Z \)
Now take the square root to find \( Z \):
\[ Z = \sqrt{4420000} \approx 2102 \, \Omega \]

### Final Answer
The impedance offered by the circuit is approximately \( 2102 \, \Omega \).

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