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. Zero
B. `(pi)/(6)`
C. `(pi)/(4)`
D. `(pi)/(2)`

Answer

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

### Step 1: Understand the relationship between resistance, inductive reactance, and phase difference
In an R-L circuit, the phase difference \( \phi \) between the voltage and the current can be calculated using the formula:
\[ \tan(\phi) = \frac{X_L}{R} \]
where \( X_L \) is the inductive reactance and \( R \) is the resistance.

### Step 2: Identify the given values
From the problem, we know:
- \( R = 3 \, \Omega \)
- \( X_L = 3 \, \Omega \) (since it is given that the inductive reactance is equal to the resistance)

### Step 3: Substitute the values into the formula
Now, substituting the values into the formula for \( \tan(\phi) \):
\[ \tan(\phi) = \frac{X_L}{R} = \frac{3}{3} = 1 \]

### Step 4: Calculate the phase difference \( \phi \)
To find \( \phi \), we take the arctangent (inverse tangent) of 1:
\[ \phi = \tan^{-1}(1) \]
The value of \( \tan^{-1}(1) \) is known to be:
\[ \phi = \frac{\pi}{4} \, \text{radians} \]

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