Power–voltage–resistance calculation: A 220 Ω resistor dissipates 3 W of power in steady state. What is the voltage across the resistor?

Introduction / Context:
Relating power, voltage, and resistance using Ohm's law and power formulas is routine in circuit design. This problem reinforces the use of P = V^2 / R or equivalently V = √(P * R), which is common when choosing component power ratings and verifying safe operating points.

Given Data / Assumptions:

• Resistance R = 220 Ω.
• Power P = 3 W.
• Resistor assumed ideal; DC or RMS values are equivalent for computation.

Concept / Approach:
Use the power–voltage relation for a resistor: P = V^2 / R. Solving for V gives V = sqrt(P * R). Perform the multiplication and take the square root to obtain the voltage magnitude.

Step-by-Step Solution:

Verification / Alternative check:
Check using I = V / R: I ≈ 25.69 / 220 ≈ 0.1168 A. Power P = I^2 * R ≈ (0.1168)^2 * 220 ≈ 3.0 W. The numbers are consistent, confirming the solution.

Why Other Options Are Wrong:

• 73.3 V and 257 V: Would imply much higher power dissipation than 3 W for a 220 Ω resistor.
• 2.5 V: Far too low; would yield P ≈ 0.028 W.

Common Pitfalls:

• Using P = V * I without substituting I = V / R correctly can lead to algebra mistakes.
• Forgetting to take the square root when solving V^2 = P * R.

Final Answer:
25.7 V