Choose a series resistor: How much resistance is required to draw 17.6 mA from a 12 V source in a simple DC circuit?

Introduction / Context:
Given a fixed supply voltage and a target current, selecting an appropriate series resistor is a basic design task. This example mirrors LED current setting and bias network sizing in many entry-level circuits.

Given Data / Assumptions:

• Supply voltage V = 12 V.
• Desired current I = 17.6 mA = 0.0176 A.
• Simple series circuit; neglect internal source resistance.

Concept / Approach:
Use R = V / I. Because current is on the order of 10 mA and voltage is about 10 V, the expectation is around hundreds of ohms (V in tens divided by I in hundredths of an ampere).

Step-by-Step Solution:

Verification / Alternative check:
Check current with 680 Ω: I ≈ 12 / 680 ≈ 0.01765 A = 17.65 mA, matching the target within rounding and typical resistor tolerances. Power check: P = I^2 * R ≈ (0.0176)^2 * 680 ≈ 0.21 W; a 0.25 W resistor would be acceptable with modest margin.

Why Other Options Are Wrong:

• 212 Ω: Would draw about 56.6 mA, too high.
• 6.8 kΩ: Would reduce current to about 1.76 mA, too low.
• 68 Ω: Would allow about 176 mA, an order of magnitude too large.

Common Pitfalls:

• Selecting 68 Ω or 6.8 kΩ due to similar digits without checking the prefix.
• Forgetting to evaluate resistor power rating after choosing R.

Final Answer:
680 Ω