Series–parallel reduction: Two 1.2 kΩ resistors in series form one branch, and this series branch is placed in parallel with a 3.3 kΩ resistor. What is the total equivalent resistance of the network?

Introduction / Context:Combining series and parallel connections is common in practical circuits. You must correctly identify which resistors add and which combine by reciprocal addition. This exercise walks through a straightforward two-step reduction to arrive at a single equivalent resistance seen by a source.

Given Data / Assumptions:

• Branch A: two resistors of 1.2 kΩ each in series.
• Branch B: one resistor of 3.3 kΩ.
• Branches A and B are connected in parallel.
• Ideal resistors; temperature effects neglected.

Concept / Approach:First, add the series pair in Branch A: R_A = 1.2 kΩ + 1.2 kΩ = 2.4 kΩ. Then, compute the parallel combination with Branch B using R_total = (R_A * R_B) / (R_A + R_B). Convert to ohms if needed and simplify to a tidy decimal or rounded whole-number value.

Step-by-Step Solution:

Verification / Alternative check:Quick bounds: The equivalent must be less than the smaller of 2.4 kΩ and 3.3 kΩ, hence less than 2.4 kΩ. Our result (≈1.389 kΩ) satisfies this and is reasonable.

Why Other Options Are Wrong:

• 138 Ω: An order of magnitude too small; likely from misplacing the decimal.
• 5,700 Ω: That is the sum, not the parallel result.
• 880 Ω: Not produced by any correct combination of the given values.

Common Pitfalls:

• Accidentally adding all three values as if all were in series.
• Forgetting that parallel equivalents are always less than the smallest branch resistance.

Final Answer:1,389 Ω