Series circuit calculation: what is the total resistance when four resistors 220 Ω, 470 Ω, 1 Ω, and 1.2 Ω are connected in series?

Introduction / Context:
Combining resistors in series is a fundamental task in circuit analysis, appearing in labs, interviews, and exams. In a series connection, the same current flows through every component and the total voltage drop equals the sum of individual drops. This property leads directly to the rule for total resistance.

Given Data / Assumptions:

• Four ideal resistors: 220 Ω, 470 Ω, 1 Ω, and 1.2 Ω.
• All connected in series (no parallel branches).
• Temperature effects and tolerance are neglected.

Concept / Approach:
For series resistors, total resistance is the arithmetic sum of individual resistances: R_total = R1 + R2 + R3 + R4. This follows from Ohm’s law applied to each element and KVL (sum of voltage drops equals source voltage). The units remain ohms.

Step-by-Step Solution:
Write the series sum: R_total = 220 + 470 + 1 + 1.2.Add in steps: 220 + 470 = 690.Add the small values: 690 + 1 = 691; 691 + 1.2 = 692.2.Therefore, R_total = 692.2 Ω.

Verification / Alternative check:
Quick estimate: 220 + 470 ≈ 690; adding roughly 2 more gives ≈ 692 Ω, consistent with the exact 692.2 Ω.

Why Other Options Are Wrong:
844.5 Ω: no basis in series addition.3890 Ω: off by an order of magnitude.1234 Ω: arbitrary; not the sum.None of the above: incorrect because 692.2 Ω is correct.

Common Pitfalls:
Mistaking series with parallel rules; rounding too early; misreading 1.2 Ω as 12 Ω.

Final Answer:
692.2 Ohm