The resistance of a coil for DC is `5Omega` . In case of AC, the resistance will

Options

A. remain `5Omega`
B. decrease
C. increase
D. be zero

Explanation

To solve the problem, we need to analyze the difference between resistance in a direct current (DC) circuit and an alternating current (AC) circuit, particularly focusing on a coil (inductor).

### Step-by-Step Solution:

1. **Understanding Resistance in DC:**
- The resistance of the coil for DC is given as \( R = 5 \, \Omega \).

2. **Understanding Resistance in AC:**
- In an AC circuit, the concept of resistance is replaced by impedance (\( Z \)). Impedance takes into account both resistance and reactance (which arises from inductors and capacitors).

3. **Impedance Formula:**
- The impedance \( Z \) in an AC circuit can be calculated using the formula:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
- Where \( X_L \) is the inductive reactance and \( X_C \) is the capacitive reactance.

4. **Identifying Components:**
- In this case, we are dealing with a coil (inductor), so we only need to consider the inductive reactance \( X_L \).
- The inductive reactance is given by:
\[ X_L = \omega L \]
- Here, \( \omega \) is the angular frequency and \( L \) is the inductance of the coil.

5. **Substituting Values:**
- Since \( X_C \) is not present in this problem (as we only have an inductor), we can simplify the impedance formula to:
\[ Z = \sqrt{R^2 + X_L^2} \]
- Substituting \( R = 5 \, \Omega \):
\[ Z = \sqrt{(5)^2 + (\omega L)^2} \]

6. **Analyzing the Result:**
- Since \( X_L \) (inductive reactance) is always positive when there is an inductor, it follows that:
\[ Z = \sqrt{25 + (\omega L)^2} > 5 \]
- This indicates that the impedance \( Z \) in the AC circuit is greater than the resistance \( R \) in the DC circuit.

7. **Conclusion:**
- Therefore, the resistance in the case of AC will effectively increase due to the presence of inductive reactance.

### Final Answer:
The resistance of the coil in case of AC will be greater than \( 5 \, \Omega \).