Question

For two resistors R1, and R2, connected in parallel, the relative error in their equivalent resistance is (Where R1 = (10.0 +- 0.1)ohm and R 2= (20.0 +-0.4)ohm) (1) 0.08 (2) 0.05 (3) 0.01 (4) 0.04



solution.​

Answer 1

Answer:

(3) 0.01

is the right answer

Answer 2

Hence, the relative error in their equivalent resistance is (1) 0.08 Ω.

Given: Two resistors R1, and R2, are connected in parallel.

R1 = (10.0 ± 0.1)ohm

R2 = (20.0 ± 0.4)ohm)

To Find: the relative error in their equivalent resistance

Solution:

We can solve the question using the formula of equivalent resistance of two resistors which is represented as,

ΔR/R² = ΔR1/R1² + ΔR2/R2²                                              .... (1)

Where R1 = 10 Ω , R2 = 20 Ω , ΔR1 = 0.1 Ω , ΔR2 = 0.4 Ω

Now we must find the value of R

Since the resistors are in parallel so equivalent resistance is,

1/R = 1/R1 + 1/R2

     = 1/10 + 1/20

     = 3/20 Ω

R = 20/3 Ω

Putting the respective values in (1) we get,

ΔR/ (20/3)² = 0.1/10² + 0.4/20²

ΔR = 400/9 [0.1/100 + 0.4/400]

ΔR = 400/9 × 0.2/100

ΔR = 0.8/9 Ω

ΔR = 0.08 Ω

Hence, the relative error in their equivalent resistance is (1) 0.08 Ω.

#SPJ2