An `AC` voltage is applied to a resistance `R` and an inductance `L` in series. If `R` and the inductive reactance are both equal to `3 Omega`, the phase difference between the applied voltage and the current in the circuit is

Options

A. ` pi `/ 4
B. ` pi ` / 2
C. zero
D. ` pi `/ 6

Explanation

To find the phase difference between the applied voltage and the current in a series circuit containing a resistance \( R \) and an inductance \( L \), we can follow these steps:

### Step 1: Understand the circuit components
In a series circuit with resistance \( R \) and inductive reactance \( X_L \), the total impedance \( Z \) can be expressed as:
\[ Z = \sqrt{R^2 + X_L^2} \]
where \( R \) is the resistance and \( X_L \) is the inductive reactance.

### Step 2: Given values
From the problem, we know:
- \( R = 3 \, \Omega \)
- \( X_L = 3 \, \Omega \)

### Step 3: Calculate the phase angle
The phase angle \( \phi \) between the voltage and the current can be calculated using the formula:
\[ \tan \phi = \frac{X_L}{R} \]
Substituting the given values:
\[ \tan \phi = \frac{3}{3} = 1 \]

### Step 4: Find the angle \( \phi \)
To find \( \phi \), we take the inverse tangent:
\[ \phi = \tan^{-1}(1) \]
The angle whose tangent is 1 is:
\[ \phi = \frac{\pi}{4} \text{ radians} \]

### Conclusion
Thus, the phase difference between the applied voltage and the current in the circuit is:
\[ \phi = \frac{\pi}{4} \text{ radians} \]

### Final Answer
The correct option is \( \frac{\pi}{4} \).

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