Series current from total resistance: Four resistors each of 90 Ω are connected in series across an 18 V DC source. What is the circuit current?

Introduction / Context:
Calculating current in a simple series network is a direct application of Ohm’s law once the total resistance is known. Series connection increases resistance by summation; hence current decreases relative to a single element for the same source voltage.

Given Data / Assumptions:

• Resistors: four identical units of 90 Ω each, in series.
• Source: 18 V DC.
• Ideal components with negligible lead resistance.

Concept / Approach:
Compute total resistance, then apply I = V / R_total. Because resistors are in series, R_total = 90 + 90 + 90 + 90. Compare the resulting current to answer choices, and convert to milliamperes for readability.

Step-by-Step Solution:

Verification / Alternative check:
If a single 90 Ω were across 18 V, current would be 0.2 A. In series with four such resistors, current is reduced by a factor of 4, giving 0.05 A. This matches the calculation.

Why Other Options Are Wrong:
0.2 A is the single-resistor case. 5 A and 20 A are unrealistic for 18 V across hundreds of ohms. 9 mA would imply 2 kΩ total, not 360 Ω.

Common Pitfalls:
Forgetting to sum series resistances or mishandling unit conversions between amperes and milliamperes.

Final Answer:
50 mA