200 `Omega` resistance and 1H inductance are connected in series with an A.C. circuit. The frequency of the source is `(200)/(2pi)` Hz. Then phase difference in between V and I will be-

Options

A. `30^(@)`
B. `60^(@)`
C. `45^(@)`
D. `90^(@)`

Answer

To solve the problem, we need to find the phase difference (φ) between the voltage (V) and the current (I) in a series RL circuit. Here are the steps to arrive at the solution:

### Step 1: Identify the given values
- Resistance (R) = 200 Ω
- Inductance (L) = 1 H
- Frequency (f) = \( \frac{200}{2\pi} \) Hz

### Step 2: Calculate the angular frequency (ω)
The angular frequency (ω) is given by the formula:
\[ \omega = 2\pi f \]
Substituting the value of f:
\[ \omega = 2\pi \left(\frac{200}{2\pi}\right) = 200 \, \text{rad/s} \]

### Step 3: Calculate the inductive reactance (X_L)
The inductive reactance (X_L) is given by:
\[ X_L = \omega L \]
Substituting the values of ω and L:
\[ X_L = 200 \times 1 = 200 \, \Omega \]

### Step 4: Use the formula for phase difference (φ)
In a series RL circuit, the phase difference (φ) between the voltage and current can be calculated using:
\[ \tan \phi = \frac{X_L}{R} \]
Substituting the values of X_L and R:
\[ \tan \phi = \frac{200}{200} = 1 \]

### Step 5: Calculate φ
To find φ, we take the arctangent of both sides:
\[ \phi = \tan^{-1}(1) \]
We know that:
\[ \tan 45^\circ = 1 \implies \phi = 45^\circ \]

### Conclusion
The phase difference between the voltage and the current in the circuit is:
\[ \phi = 45^\circ \]

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