Recovered data for solvability — a series network has three resistors R1, R2, and R3 with total equivalent resistance R_total = 600 kΩ. If R1 = 100 kΩ and R3 = 300 kΩ, what is the value of R2?

Difficulty: Easy

Correct Answer: 200 kΩ

Explanation:


Introduction / Context:
When analyzing series circuits, the equivalent resistance is simply the sum of individual resistances. If the total is known along with all but one element, you can directly compute the missing value. This mirrors many troubleshooting scenarios where you measure total and some components but one reading is unavailable.


Given Data / Assumptions:

  • Series network: R_total = R1 + R2 + R3.
  • R_total = 600 kΩ, R1 = 100 kΩ, R3 = 300 kΩ.
  • Ideal resistors; temperature effects ignored.


Concept / Approach:
Use the series-sum formula and solve for the unknown. This is an algebraic rearrangement requiring no advanced methods. Ensuring units are consistent (all in kΩ) simplifies the arithmetic and avoids conversion errors.


Step-by-Step Solution:

Write the sum: R_total = R1 + R2 + R3.Substitute known values: 600 = 100 + R2 + 300 (all in kΩ).Combine known terms: 600 = R2 + 400.Solve for the unknown: R2 = 600 − 400 = 200 kΩ.


Verification / Alternative check:
Back-substitute: 100 + 200 + 300 = 600 kΩ, which matches the specified total, confirming the calculation.


Why Other Options Are Wrong:

100 kΩ or 300 kΩ: would sum to 500 kΩ or 700 kΩ, inconsistent with the given total.“More information is needed”: not true; series-sum provides a unique solution.50 kΩ: totals 450 kΩ, not 600 kΩ.


Common Pitfalls:
Mishandling units (Ω vs kΩ); forgetting that only series, not parallel, uses direct addition; accidentally subtracting the wrong way around.


Final Answer:
200 kΩ

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