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: The instantaneous power delivered to build up current in an inductor is \(P = Vi\), where the induced voltage is \(V = L\dfrac{di}{dt}\), so \(P = Li\dfrac{di}{dt}\).

Step 2: The total energy stored is this power integrated over time as the current ramps up from \(0\) to its final value \(I\): \(U = \displaystyle\int_0^I Li\, di = L\left[\dfrac{i^2}{2}\right]_0^I = \dfrac{1}{2}LI^2\).

Step 3: Substitute \(L = 32\times10^{-3}\text{ H}\) and \(I=3\text{ A}\): \(U = \dfrac{1}{2}(32\times10^{-3})(3)^2 = \dfrac{1}{2}(32\times10^{-3})(9)\).
\[ U = 0.144\text{ J} = \boxed{0.144\ \text{J}} \]