Power from energy–time data: If 12,000 J of energy are used in 400 ms, what is the power delivered?

Introduction / Context:
Converting between energy and power is fundamental in electrical engineering and physics. Short-duration events often involve large instantaneous power despite modest energy totals. This question tests correct unit handling and the formula P = E / t.

Given Data / Assumptions:

• Energy E = 12,000 J.
• Time t = 400 ms = 0.4 s.
• Power is average over the stated interval.

Concept / Approach:
Use the relation Power = Energy / Time. Ensure time is in seconds so that watts (joules per second) are obtained directly. Convert milliseconds to seconds prior to division.

Step-by-Step Solution:

Verification / Alternative check:
Dimensional analysis: J / s = W. Quick estimate: if 12 kJ were used in 1 s → 12 kW; in 0.4 s, the power is higher by factor 1/0.4 = 2.5 → 30 kW, confirming the result.

Why Other Options Are Wrong:

• 30 W and 3 W: Off by factors of 1,000 or 10,000 due to unit conversion errors.
• 300 kW: Too large by a factor of 10; likely from misplacing a decimal when converting time.

Common Pitfalls:

• Treating milliseconds as minutes or forgetting to convert to seconds.
• Confusing energy (J) with power (W) and mixing formulas.

Final Answer:
30 kW