Question

For two resistors R, and Ry, connected in
parallel, the relative error in their equivalent
resistance is (Where R (10.0 + 0.1) 2 and
R2 = (20.0 10.4) 2)

(1) 0.08
(2) 0.05
(3) 0.01
(4) 0.04​

Answer 1

Answer: For two resistors R₁ and R₂ are connected in parallel combination. where R₁ = (10 ± 0.1) Ω and R₂ = (20 ± 0.4) Ω

To find : The equivalent resistance of two resistors.

solution : equivalent resistance of two resistor R₁ and R₂ is given by, 1/R = 1/R₁ + 1/R₂

To find an error in equivalent resistance,

∆R/R² = ∆R₁/R₁² + ∆R₂/R₂²

first find R,

i.e., 1/R = 1/10 + 1/20 = 3/20

⇒R = 20/3 Ω

now ∆R/(20/3)² = 0.1/(10)² + (0.4)/(20)²

⇒∆R = (20/3)² [ 0.1/100 + 0.4/400 ]

⇒∆R = 400/9 [ 0.1/100 + 0.1/100 ]

⇒∆R = 400/9 × 0.2/100

⇒∆R = 0.8/9 ≈ 0.08

Therefore the relative error in their equipment resistance is 0.08 .

Explanation: