Series circuit calculation (Ohm’s law): three resistors of 220 Ω, 100 Ω, and 180 Ω are connected in series across a 12 V DC source. What is the resulting circuit current (express your choice in appropriate units)?

Introduction / Context:
When components are connected in series, the same current flows through each element and the total resistance is the arithmetic sum of all series resistances. This problem tests correct application of Ohm’s law and proper unit handling to compute current in a simple DC series network.

Given Data / Assumptions:

• Resistors in series: R1 = 220 ohm, R2 = 100 ohm, R3 = 180 ohm.
• Applied source: V = 12 V DC (ideal source).
• Temperature effects and tolerance are ignored; use nominal values.
• No additional parasitic elements or measurement loading assumed.

Concept / Approach:
For series connections, total resistance is R_total = R1 + R2 + R3. The circuit current is given by Ohm’s law: I = V / R_total. Keeping track of ohms and volts ensures current is in amperes; convert to milliamperes if useful for readability.

Step-by-Step Solution:

Verification / Alternative check:
As a quick sense-check, with about 500 ohm across 12 V, current should be a bit under 25 mA (since 12/480 = 25 mA). Our exact 24 mA result is consistent.

Why Other Options Are Wrong:
41.67 A and 500 A: unrealistically large; would require milliohm-level loads, not hundreds of ohms.
120 mA: corresponds to about 100 ohm at 12 V; not our total resistance.
2.4 mA: off by a factor of 10; would imply 5 kohm, not 500 ohm.

Common Pitfalls:
Adding series resistances correctly but then dividing incorrectly; forgetting to convert A to mA; mistakenly using parallel formulas (1/R_total = sum(1/R)).

Final Answer:
24 mA