A capacitor is connected to a 2 mH inductor in series with an AC source. Calculate the value of maximum current in the circuit if maximum energy stored in inductor is `28 xx10^(-6)` J.

Answer

To find the maximum current in the circuit with a capacitor and a 2 mH inductor connected in series with an AC source, given that the maximum energy stored in the inductor is \( 28 \times 10^{-6} \) J, we can use the formula for the energy stored in an inductor:

\[ U = \frac{1}{2} L I_{\text{max}}^2 \]

where:
- \( U \) is the maximum energy stored in the inductor,
- \( L \) is the inductance,
- \( I_{\text{max}} \) is the maximum current.

### Step-by-Step Solution:

**Step 1: Write down the formula for energy stored in the inductor.**

\[ U = \frac{1}{2} L I_{\text{max}}^2 \]

**Step 2: Rearrange the formula to solve for \( I_{\text{max}} \).**

\[ I_{\text{max}}^2 = \frac{2U}{L} \]

\[ I_{\text{max}} = \sqrt{\frac{2U}{L}} \]

**Step 3: Substitute the given values into the formula.**

Given:
- \( U = 28 \times 10^{-6} \) J
- \( L = 2 \) mH = \( 2 \times 10^{-3} \) H

Substituting these values into the equation:

\[ I_{\text{max}} = \sqrt{\frac{2 \times (28 \times 10^{-6})}{2 \times 10^{-3}}} \]

**Step 4: Simplify the expression.**

Calculating the numerator:

\[ 2 \times (28 \times 10^{-6}) = 56 \times 10^{-6} \]

Now, substituting back into the equation:

\[ I_{\text{max}} = \sqrt{\frac{56 \times 10^{-6}}{2 \times 10^{-3}}} \]

\[ I_{\text{max}} = \sqrt{\frac{56 \times 10^{-6}}{2 \times 10^{-3}}} = \sqrt{28 \times 10^{-3}} = \sqrt{0.028} \]

**Step 5: Calculate the square root.**

\[ I_{\text{max}} = \sqrt{0.028} \approx 0.167 \text{ A} \]

### Final Answer:

The maximum current in the circuit is approximately \( 0.167 \) A.

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