Two resistors in parallel: A 1.6 kΩ resistor is connected in parallel with a 120 Ω resistor. Which range best describes the total equivalent resistance?
Correct Answer: less than 120 Ω but greater than 100 Ω
Introduction / Context:For two resistors in parallel, the equivalent resistance is always less than the smaller of the two values. Estimating the range quickly is a valuable troubleshooting skill even without a calculator.
Given Data / Assumptions:
- R1 = 1.6 kΩ = 1,600 Ω.
- R2 = 120 Ω.
- Connection: R1 ∥ R2 (parallel).
Concept / Approach:Use R_eq = 1 / (1/R1 + 1/R2). Because one branch is much smaller (120 Ω), the result will be slightly below 120 Ω. A quick computation refines the range.
Step-by-Step Solution:
Compute conductances: 1/R1 = 1/1600 = 0.000625 S; 1/R2 = 1/120 ≈ 0.008333 S.Sum: 0.000625 + 0.008333 = 0.008958 S.Equivalent resistance: R_eq = 1 / 0.008958 ≈ 111.7 Ω.Verification / Alternative check:Product-over-sum shortcut for two in parallel: R_eq = (R1 * R2) / (R1 + R2) = (1600 * 120) / (1720) ≈ 111.6 Ω, matching the detailed calculation.
Why Other Options Are Wrong:
- Greater than 1.6 kΩ or greater than 120 Ω but less than 1.6 kΩ: Parallel equivalent cannot exceed the smallest branch (120 Ω).
- Less than 100 Ω: Too low; calculation yields ~111.7 Ω.
Common Pitfalls:
- Assuming a large difference forces R_eq near the smaller value without checking how close.
- Arithmetic slips when inverting sums of reciprocals.
Final Answer:less than 120 Ω but greater than 100 Ω