Find the missing series element: A series circuit has three resistors. Two resistors are 1.2 kΩ each, and the total resistance is 12 kΩ. What is the value of the third resistor?

Difficulty: Easy

Correct Answer: is 9,600 Ω

Explanation:


Introduction / Context:
Solving for an unknown in a series network is a direct application of the additive property of resistances. This is common when back-calculating a missing component from a measured total or when designing a target equivalent resistance from available parts.


Given Data / Assumptions:

  • Total series resistance: 12 kΩ.
  • Two known resistors: 1.2 kΩ and 1.2 kΩ.
  • All are simple linear resistors in series.


Concept / Approach:
For series resistors, R_total = R1 + R2 + R3. Rearrange to find the unknown: R3 = R_total − (R1 + R2). Keep units consistent, then convert back to a convenient expression (ohms or kilo-ohms) as needed.


Step-by-Step Solution:

Sum the knowns: R1 + R2 = 1.2 kΩ + 1.2 kΩ = 2.4 kΩ.Compute the unknown: R3 = 12 kΩ − 2.4 kΩ = 9.6 kΩ.Express in ohms if preferred: 9.6 kΩ = 9600 Ω.


Verification / Alternative check:
Add them back: 1.2 kΩ + 1.2 kΩ + 9.6 kΩ = 12.0 kΩ exactly. The arithmetic closes perfectly, confirming the result.


Why Other Options Are Wrong:

  • is 960 Ω or is 1.2 kΩ: Far too small; their sum with the two 1.2 kΩ parts would not reach 12 kΩ.
  • cannot be determined: Incorrect; the series sum uniquely determines the missing value.


Common Pitfalls:

  • Mixing kilo-ohms and ohms without conversion, causing factor-of-1000 mistakes.
  • Subtracting only one of the known resistors from the total instead of both.


Final Answer:
is 9,600 Ω

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