A coil of `200Omega` resistance and `1.0H` inductance is conneted to an `AC` source of frequency `200//2pi Hz`. Phase angle between potential and current will be

Options

A. `30^(@)`
B. `90^(@)`
C. `45^(@)`
D. `0^(@)`

Answer

To find the phase angle between the potential and current in a coil connected to an AC source, we can follow these steps:

### Step-by-Step Solution

1. **Identify Given Values**:
- Resistance, \( R = 200 \, \Omega \)
- Inductance, \( L = 1.0 \, H \)
- Frequency, \( f = \frac{200}{2\pi} \, Hz \)

2. **Calculate Angular Frequency (\( \omega \))**:
\[ \omega = 2\pi f \]
Since \( f = \frac{200}{2\pi} \):
\[ \omega = 200 \, \text{rad/s} \]

3. **Calculate Inductive Reactance (\( X_L \))**:
The formula for inductive reactance is given by:
\[ X_L = L \cdot \omega \]
Substituting the values:
\[ X_L = 1 \cdot 200 = 200 \, \Omega \]

4. **Determine the Phase Angle (\( \phi \))**:
The phase angle can be found using the tangent function:
\[ \tan \phi = \frac{X_L}{R} \]
Substituting the values of \( X_L \) and \( R \):
\[ \tan \phi = \frac{200}{200} = 1 \]

5. **Calculate \( \phi \)**:
To find \( \phi \), we take the arctangent:
\[ \phi = \tan^{-1}(1) = 45^\circ \]

6. **Conclusion**:
The phase angle between the potential and the current is \( 45^\circ \).

### Final Answer:
The phase angle between the potential and current is \( 45^\circ \).

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