Power–voltage relationship: A 75 Ω load is consuming 2 W of power from a DC/AC source. Approximately what is the output voltage of the supply, assuming ideal conditions?

Introduction / Context:
Many practical problems in basic electrical engineering require moving between power, voltage, and resistance. Recognizing the correct formula and applying it cleanly allows you to size power supplies and verify whether a load is being driven correctly. This question targets the relationship among power (P), voltage (V), and resistance (R).

Given Data / Assumptions:

• Load resistance R = 75 Ω.
• Load power P = 2 W (steady state).
• Assume ideal source and purely resistive load (no reactive elements).
• We seek the approximate source/output voltage across the load.

Concept / Approach:

For a resistive load, the power–voltage relation is P = V^2 / R. Rearranging gives V = sqrt(P * R). This formula is valid for DC and for RMS values in AC when the load is purely resistive.

Step-by-Step Solution:

Verification / Alternative check:

Compute power back from 12 V: P = V^2 / R = 144 / 75 = 1.92 W, which is near 2 W. Using 12.25 V gives about 2.0 W exactly, so 12 V is a good approximate choice from the options.

Why Other Options Are Wrong:

120 V would yield P = 120^2 / 75 ≈ 192 W (far too high). 1.2 V gives ≈ 0.019 W (too low). 6 V gives 0.48 W (too low). 9 V gives 1.08 W (still too low).

Common Pitfalls:

Mixing up P = V * I with P = V^2 / R; forgetting to use RMS values for AC; mis-squaring or mis-taking the square root.

Final Answer:

12 V