JEE Main 2024 Physics question
The resistance R = V/I where V = (200 ± 5) V and I = (20 ± 0.2) A, the percentage error in the measurement of R is-
Answer: (A)
Solution
ΔR/R = ΔV/V + ΔI/I = 5/200 + 0.2/20 = 0.035, i.e. 3.5%.
The resistance R = V/I where V = (200 ± 5) V and I = (20 ± 0.2) A, the percentage error in the measurement of R is-
ΔR/R = ΔV/V + ΔI/I = 5/200 + 0.2/20 = 0.035, i.e. 3.5%.