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?

Difficulty: Easy

Correct Answer: 1,389 Ω

Explanation:


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:

Series sum: R_A = 2.4 kΩ.Parallel with 3.3 kΩ: R_total = (2.4 kΩ * 3.3 kΩ) / (2.4 kΩ + 3.3 kΩ).Compute numerator: 2.4 * 3.3 = 7.92 (kΩ^2); denominator: 5.7 kΩ.R_total = 7.92 / 5.7 kΩ ≈ 1.38947 kΩ ≈ 1,389 Ω.


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

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