Series resistors — quick sum check: If 7.3 kΩ, 1.8 kΩ, and 4.9 kΩ resistors are connected in series, is the total resistance equal to 14 kΩ? Verify the claim using series-resistance rules.

Introduction / Context:
Being able to sanity-check series resistor totals quickly is invaluable for design and troubleshooting. Series combinations add directly because the same current flows through each resistor and each contributes its share of voltage drop. This problem asks you to confirm a specific numerical claim for three common E24 values.

Given Data / Assumptions:

• Resistor values: 7.3 kΩ, 1.8 kΩ, 4.9 kΩ.
• Series connection: same current through all three resistors.
• Ideal components; tolerance effects ignored for the conceptual check.

Concept / Approach:
For series resistances, the equivalent resistance is the arithmetic sum: R_total = R1 + R2 + R3. Units should be consistent; since all are in kΩ, we can add directly in kΩ or convert to Ω first and convert back later. The calculation is straightforward but must be performed carefully to avoid simple arithmetic slips.

Step-by-Step Solution:

Verification / Alternative check:
Convert to ohms for confirmation: 7300 + 1800 + 4900 = 14,000 Ω, which equals 14 kΩ. Either way, the arithmetic agrees with the claim.

Why Other Options Are Wrong:

• Incorrect: The sum is exactly 14 kΩ under ideal arithmetic.
• Valid only at 25 °C / analog meter only: Temperature and instrument type do not change the ideal series rule; tolerances might slightly vary measured values but not the nominal total.
• Cannot be determined without current: Equivalent resistance is independent of current magnitude.

Common Pitfalls:
Mistyping values (e.g., 7.3 as 7.03), mixing units (Ω vs kΩ), or mistakenly averaging instead of summing for series networks (averaging applies to some parallel intuition, not series).

Final Answer:
Correct