Question

When 3 A dc is passed through a coil of 32 mH inductance, The energy stored in the coil

• A. 288 J
• B. 0.288 J
• C. 0.144 J
• D. 0.0 J

Hint:

Always convert units to the standard SI system before calculating. Here, "mH" (millihenry) must be converted to "H" (Henry) by multiplying by $10^{-3}$ to ensure the answer is in Joules.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
When an electric current flows through an inductor (coil), it creates a magnetic field. The work done to establish this current is stored in the inductor as magnetic potential energy.

Step 2: Key Formula or Approach:
The energy $U$ stored in an inductor is given by the formula: \[ U = \frac{1}{2} LI^2 \] where $L$ is the inductance and $I$ is the current.

Step 3: Detailed Explanation:
Given: $I = 3\text{ A}$ $L = 32\text{ mH} = 32 \times 10^{-3}\text{ H}$ Placing values in the formula: \[ U = \frac{1}{2} \times (32 \times 10^{-3}) \times (3)^2 \] \[ U = 16 \times 10^{-3} \times 9 \] \[ U = 144 \times 10^{-3} \] \[ U = 0.144\text{ J} \]

Step 4: Final Answer:
The energy stored in the coil is 0.144 J.

Answer 2

The energy stored in an inductor carrying a steady current is given by \( U = \dfrac{1}{2}LI^2 \), with \(L\) converted into Henries. Here \(L = 32\text{ mH} = 32\times10^{-3}\text{ H}\) and \(I = 3\text{ A}\).

\[ U = \frac{1}{2}(32\times10^{-3})(3)^2 = \frac{1}{2}(32\times10^{-3})(9) = 0.144\text{ J} \]

1. 288 J: This value arises if the inductance is mistakenly left in millihenries (32) instead of being converted to henries, and the factor of \(\frac{1}{2}\) is skipped, since \(32\times9=288\).
2. 0.288 J: This is close to the correct method but misses the factor of \(\frac{1}{2}\) in the formula, since \((32\times10^{-3})\times9 = 0.288\), leaving the answer exactly double the true value.
3. 0.144 J: This is exactly what the formula gives when \(L\) is correctly converted to henries and the \(\frac{1}{2}\) factor is applied properly, as shown in the computation above.
4. 0.0 J: This would only be true if the current were zero or the coil had no inductance, but a steady \(3\text{ A}\) through a real \(32\text{ mH}\) inductor definitely stores nonzero magnetic energy.

Applying the formula carefully, with correct unit conversion and the factor of one-half included, gives a clear, unique value.

Therefore, the correct answer is 0.144 J.