Two resistances are measured as R 1 = 100 ± 3 Ω and R 2 = 200 ± 4 Ω. What is the absolute error in their series combination (R 1 + R 2 )?

⚛️ Physics · Units and Measurements · NEET & JEE

Two resistances are measured as R1 = 100 ± 3 Ω and R2 = 200 ± 4 Ω. What is the absolute error in their series combination (R1 + R2)?

Answer: ±7 Ω.

A ±1 Ω
B ±4 Ω
C ±7 Ω
D ±12 Ω

Correct answer: C. ±7 Ω

Explanation: For sums, absolute errors add directly: ΔR = ΔR1 + ΔR2 = 3 + 4 = 7 Ω, so the series combination is 300 ± 7 Ω.

The 7 SI base units. All derived units (newton, joule, watt, volt, etc.) can be expressed as combinations of these 7 fundamental units raised to various powers.

Concept context

SI base units, dimensional analysis, significant figures, and error propagation. The toolkit every physics calculation relies on.

Read the full Units and Measurements notes →

Two resistances are measured as R<sub>1</sub> = 100 ± 3 Ω and R<sub>2</sub> = 200 ± 4 Ω. What is the absolute error in their series combination (R<sub>1</sub> + R<sub>2</sub>)?

Concept context

More Units and Measurements questions