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 parameters
We are given:
- Resistance \( R = 3 \, \Omega \)
- Inductive reactance \( X_L = 3 \, \Omega \)

### Step 2: Use the formula for phase difference
In an RL series circuit, the phase difference \( \phi \) between the voltage and the current can be calculated using the formula:
\[ \tan \phi = \frac{X_L}{R} \]

### Step 3: Substitute the known values
Substituting the given values into the formula:
\[ \tan \phi = \frac{X_L}{R} = \frac{3 \, \Omega}{3 \, \Omega} = 1 \]

### Step 4: Calculate the phase difference
To find \( \phi \), we take the arctangent (inverse tangent) of 1:
\[ \phi = \tan^{-1}(1) \]
The angle whose tangent is 1 is:
\[ \phi = 45^\circ \quad \text{or} \quad \phi = \frac{\pi}{4} \, \text{radians} \]

### Step 5: Conclusion
The phase difference between the applied voltage and the current in the circuit is:
\[ \phi = \frac{\pi}{4} \, \text{radians} \]

### Final Answer
The phase difference is \( \frac{\pi}{4} \) radians.

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