A 0.2 H inductor is connected across an AC power supply . The current in the circuit increases from 0 to 3.4 A. Calculate the energy stored in magnetic field of inductor during that period.

Answer

To calculate the energy stored in the magnetic field of an inductor when the current changes, we can use the formula for energy stored in an inductor:

\[ U = \frac{1}{2} L I^2 \]

where:
- \( U \) is the energy stored in joules,
- \( L \) is the inductance in henries,
- \( I \) is the current in amperes.

### Step-by-Step Solution:

1. **Identify the given values:**
- Inductance \( L = 0.2 \, \text{H} \)
- Initial current \( I_{\text{initial}} = 0 \, \text{A} \)
- Final current \( I_{\text{final}} = 3.4 \, \text{A} \)

2. **Calculate the initial energy stored in the inductor:**
- Using the formula \( U = \frac{1}{2} L I^2 \):
\[ U_{\text{initial}} = \frac{1}{2} \times 0.2 \times (0)^2 = 0 \, \text{J} \]

3. **Calculate the final energy stored in the inductor:**
- Now, substitute \( I = 3.4 \, \text{A} \):
\[ U_{\text{final}} = \frac{1}{2} \times 0.2 \times (3.4)^2 \]
- Calculate \( (3.4)^2 = 11.56 \):
\[ U_{\text{final}} = \frac{1}{2} \times 0.2 \times 11.56 = 0.1 \times 11.56 = 1.156 \, \text{J} \]

4. **Calculate the change in energy stored in the inductor:**
- The change in energy \( \Delta U \) is given by:
\[ \Delta U = U_{\text{final}} - U_{\text{initial}} = 1.156 \, \text{J} - 0 \, \text{J} = 1.156 \, \text{J} \]

5. **Conclusion:**
- The energy stored in the magnetic field of the inductor when the current increases from 0 A to 3.4 A is \( 1.156 \, \text{J} \).