In LR circuit, f = 50 Hz, L= 2 H, `epsilon = 5V, R = 1 Omega`, then the energy stored in the inductor is:

Options

A. 25 J
B. 50 J
C. 100 J
D. None of these

Explanation

To find the energy stored in the inductor of an LR circuit, we can follow these steps:

### Step 1: Identify the given values
- Frequency (f) = 50 Hz
- Inductance (L) = 2 H
- EMF (ε) = 5 V
- Resistance (R) = 1 Ω

### 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:
\[ \omega = 2\pi \times 50 = 100\pi \, \text{rad/s} \]

### Step 3: Calculate the inductive reactance (X_L)
The inductive reactance (X_L) is calculated using the formula:
\[ X_L = \omega L \]
Substituting the values of ω and L:
\[ X_L = 100\pi \times 2 = 200\pi \, \Omega \]

### Step 4: Calculate the total impedance (Z)
The total impedance (Z) in an LR circuit is given by:
\[ Z = \sqrt{R^2 + X_L^2} \]
Substituting the values of R and X_L:
\[ Z = \sqrt{1^2 + (200\pi)^2} = \sqrt{1 + 40000\pi^2} \]

### Step 5: Calculate the current (I)
The current (I) can be calculated using Ohm's law:
\[ I = \frac{\epsilon}{Z} \]
Substituting the values:
\[ I = \frac{5}{\sqrt{1 + 40000\pi^2}} \]

### Step 6: Calculate the energy stored in the inductor (U)
The energy stored in the inductor is given by the formula:
\[ U = \frac{1}{2} L I^2 \]
Substituting the values of L and I:
\[ U = \frac{1}{2} \times 2 \times \left(\frac{5}{\sqrt{1 + 40000\pi^2}}\right)^2 \]
This simplifies to:
\[ U = \frac{1}{2} \times 2 \times \frac{25}{1 + 40000\pi^2} = \frac{25}{1 + 40000\pi^2} \]

### Step 7: Final Calculation
To find the numerical value, we can substitute π ≈ 3.14:
\[ U \approx \frac{25}{1 + 40000 \times (3.14)^2} \]
Calculating \( 40000 \times (3.14)^2 \):
\[ 40000 \times 9.8596 \approx 394384 \]
Thus:
\[ U \approx \frac{25}{1 + 394384} \approx \frac{25}{394385} \approx 6.35 \times 10^{-5} \, \text{J} \]

### Conclusion
The energy stored in the inductor is approximately \( 6.35 \times 10^{-5} \, \text{J} \).