Power dissipation across a fixed 120 V source: which series combination of 220 Ω resistors (one, two, three, or four in series) dissipates the greatest total power?

Introduction / Context:
Total power drawn from a constant-voltage source depends on the equivalent resistance. This question checks your understanding of how power changes when series resistance increases while the supply voltage is held constant.

Given Data / Assumptions:

• Source voltage: 120 V (constant).
• Each resistor value: 220 Ω.
• Series connection only; no tolerance or heating effects considered.

Concept / Approach:

For a fixed voltage source, total power P_total equals V^2 / R_total. Therefore, smaller R_total yields larger P_total. In series, resistances add: R_total = n * 220 Ω, where n is the number of resistors in series.

Step-by-Step Solution:

Verification / Alternative check:

Numerical example: P_1 = 120^2 / 220 ≈ 65.45 W; P_2 ≈ 32.73 W; P_3 ≈ 24.55 W; P_4 ≈ 16.36 W. The trend confirms the reasoning.

Why Other Options Are Wrong:

Adding more resistors increases R_total and therefore reduces total power drawn from the same source voltage.

Common Pitfalls:

Confusing constant-current with constant-voltage cases; incorrectly averaging resistances instead of summing for series.

Final Answer:

one 220 Ω resistor