Power from voltage and resistance — corrected stem (Recovery-First): Assume the component is a 1 kΩ resistor (not “1000 W”). With 25 V applied across it, how much power does it dissipate?

Introduction / Context:
The original text likely intended “1000 Ω resistor,” not “1000 W resistance.” Using the Recovery-First policy, we minimally correct the stem to a realistic component value. The problem then becomes a straightforward application of power formulas for resistive elements.

Given Data / Assumptions:

• Resistance R = 1 kΩ = 1000 Ω.
• Applied voltage V = 25 V (DC or RMS AC).
• Ideal resistor; temperature and tolerance effects neglected.

Concept / Approach:
For a resistor, power can be computed as P = V^2 / R, equivalently P = I^2 * R or P = V * I. With V and R given, the V^2 / R form is the most direct. Always check units: volts squared over ohms yields watts.

Step-by-Step Solution:
Compute V^2: 25 * 25 = 625.Divide by R: 625 / 1000 = 0.625.Thus, P = 0.625 W.

Verification / Alternative check:
Find current first: I = V / R = 25 / 1000 = 0.025 A (25 mA). Then P = V * I = 25 * 0.025 = 0.625 W. The two methods agree.

Why Other Options Are Wrong:
(a) 0.24 W results from incorrect arithmetic. (c) 80 W would require much lower resistance or higher voltage. (d) 325 kW is wildly unrealistic in this context. (e) Invalid because a correct option exists.

Common Pitfalls:
Misreading “1000 W” for watts instead of ohms; forgetting unit conversions; squaring voltage but not dividing by resistance.

Final Answer:
0.625 W.