If frequency of R- L circuit is f then impedence will be

Options

A. `sqrt(R^2+(2pifL)^2)`
B. `R^2+ (2pif^2)^2`
C. `sqrt(R^Lpif^2)`
D. `sqrt((R^2 - Lpif^2))`

Explanation

To solve the question regarding the impedance of an R-L circuit at a frequency \( f \), we will follow these steps:

### Step-by-Step Solution:

1. **Understand the Impedance Formula**:
The impedance \( Z \) of an R-L circuit is given by the formula:
\[ Z = \sqrt{R^2 + X_L^2} \]
where \( R \) is the resistance and \( X_L \) is the inductive reactance.

2. **Identify Inductive Reactance**:
The inductive reactance \( X_L \) is defined as:
\[ X_L = \omega L \]
where \( \omega \) is the angular frequency and \( L \) is the inductance.

3. **Relate Angular Frequency to Frequency**:
The angular frequency \( \omega \) can be expressed in terms of the frequency \( f \):
\[ \omega = 2\pi f \]

4. **Substitute Angular Frequency into Inductive Reactance**:
Now, substitute \( \omega \) into the equation for \( X_L \):
\[ X_L = (2\pi f) L \]

5. **Substitute \( X_L \) back into the Impedance Formula**:
Now we can substitute \( X_L \) back into the impedance formula:
\[ Z = \sqrt{R^2 + (2\pi f L)^2} \]

6. **Final Expression for Impedance**:
Thus, the impedance \( Z \) of the R-L circuit at frequency \( f \) is:
\[ Z = \sqrt{R^2 + (2\pi f L)^2} \]