For a household lamp rated 100 W and supplied from a 120 V AC source (assume ideal resistive behavior at rated conditions), what steady-state current does it draw?

Introduction / Context:
Lamp nameplates list rated power and voltage. From these two values, the operating current can be estimated assuming resistive behavior at steady state. This is a straightforward application of the power relation P = V * I.

Given Data / Assumptions:

• Rated power P = 100 W.
• Supply voltage V ≈ 120 V (RMS).
• Assume the lamp behaves approximately as a resistive load at operating temperature.

Concept / Approach:
For a resistive load, power equals the product of RMS voltage and RMS current. Rearranging gives I = P / V. Although incandescent filaments change resistance with temperature when warming up, the steady-state current at rating follows directly from this formula.

Step-by-Step Solution:
Use I = P / V.I = 100 W / 120 V = 0.833… A.Express in milliamperes: 0.833 A ≈ 833 mA, commonly rounded to 830 mA.Therefore, the lamp draws about 830 mA at 120 V.

Verification / Alternative check:
Compute equivalent resistance: R = V^2 / P = 120^2 / 100 = 144 Ω. Then I = V / R = 120 / 144 = 0.833 A, consistent with the direct calculation.

Why Other Options Are Wrong:

• 1.2 amps or 12 amps: Would imply P = 144 W or 1440 W at 120 V, not 100 W.
• 12000 amps: Physically impossible here; a clear distractor.

Common Pitfalls:

• Confusing watts with volt–amperes or using peak instead of RMS voltage.
• Ignoring warm-up behavior; initial inrush may be higher, but the rating refers to steady state.

Final Answer:
830 mA