Two-resistor parallel formula — is the total resistance simply the arithmetic sum of the two resistors?

Introduction / Context:
Computing equivalent resistance for two resistors in parallel is a frequent task in circuit design. This question challenges a common misconception that parallel totals add like series totals.

Given Data / Assumptions:

• Two resistors R1 and R2 connected in parallel across the same source.
• Resistors are linear and temperature effects are neglected.

Concept / Approach:
For two resistors in parallel, the correct equivalent resistance is Req = (R1 * R2) / (R1 + R2). More generally, 1/Req = 1/R1 + 1/R2. The arithmetic sum R1 + R2 applies to series connections, not parallel. Because parallel adds conductances, the final equivalent must be less than the smallest individual resistor value.

Step-by-Step Solution:

Verification / Alternative check:
Numerical example: R1 = 6 Ω, R2 = 3 Ω → Req = (6 * 3) / (6 + 3) = 18/9 = 2 Ω; clearly not 9 Ω.

Why Other Options Are Wrong:

• “True” incorrectly transfers the series rule to a parallel situation and would drastically overestimate current.

Common Pitfalls:
Forgetting to invert sums when switching between resistance and conductance; to avoid errors, compute with admittances in parallel networks.

Final Answer:
False