Parallel resistance estimation: Three resistors of 470 Ω, 220 Ω, and 100 Ω are connected in parallel. What is the approximate total resistance?

Introduction / Context:
Parallel resistor networks are ubiquitous in electronics. The total resistance is always less than the smallest branch resistance, and calculating it correctly is essential for current budgeting and power dissipation analysis.

Given Data / Assumptions:

• Resistors: R1 = 470 Ω, R2 = 220 Ω, R3 = 100 Ω.
• All three in parallel across the same source.
• Assume ideal components (no tolerance or temperature effects for this calculation).

Concept / Approach:

For parallel resistors, use the conductance sum: 1 / R_total = 1 / R1 + 1 / R2 + 1 / R3. The numerical result should be slightly larger than 1/100 Ω^-1 (because additional branches add conductance), giving R_total a bit below 100 Ω and greater than 0 Ω.

Step-by-Step Solution:

Verification / Alternative check:

Sanity: R_total must be less than the smallest branch (100 Ω). A value around 60 Ω is reasonable, since the 100 Ω branch dominates with help from the others adding conductance.

Why Other Options Are Wrong:

470 Ω and 790 Ω exceed the smallest branch, impossible in parallel. 30 Ω is too small for these values. 100 Ω equals the smallest branch and ignores the added paths.

Common Pitfalls:

Adding resistances directly (series rule) instead of adding conductances; calculator rounding errors when inverting the sum.

Final Answer:

60 Ω