Repair for solvability — two resistors of 12 kΩ and 8 kΩ are connected in parallel. What is the total (equivalent) resistance of the circuit?

Difficulty: Easy

Correct Answer: 4.8 kΩ

Explanation:


Introduction / Context:
When resistors are in parallel, their equivalent resistance is less than the smallest branch resistance. Calculating with the reciprocal formula gives precise values used in design checks.


Given Data / Assumptions:

  • R1 = 12 kΩ, R2 = 8 kΩ, connected in parallel.
  • Ideal resistors and wiring.


Concept / Approach:
Use the parallel formula: 1/Req = 1/R1 + 1/R2. For two branches it is often convenient to use the product-over-sum shortcut: Req = (R1 * R2) / (R1 + R2).


Step-by-Step Solution:

1) Compute numerator: R1 * R2 = 12 kΩ * 8 kΩ = 96 (kΩ)^2.2) Compute denominator: R1 + R2 = 20 kΩ.3) Divide: Req = 96 / 20 = 4.8 kΩ.4) Sanity check: 4.8 kΩ is less than 8 kΩ, as required for parallel.


Verification / Alternative check:
Reciprocal method: 1/Req = 1/12 + 1/8 = 0.0833 + 0.125 = 0.2083 (1/kΩ) → Req ≈ 4.8 kΩ.


Why Other Options Are Wrong:

Greater than 8 kΩ / greater than 30 kΩ / 58 kΩ: contradict the rule that parallel equivalent is less than the smallest branch.2.0 kΩ: much too low for these values.


Common Pitfalls:
Adding resistances directly (series rule) instead of using reciprocal sum for parallel.


Final Answer:
4.8 kΩ

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