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​

Answer 1

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 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 .

Answer 2

Answer:

The correct answer to this question is $0.08$

Explanation:

Given - Two resistors R1, and R2, connected in parallel, the relative error in their equivalent resistance.

To Find - Choose the correct option of the equivalent resistance of two resistors.

The equivalent resistance is the total resistance of all connected devices in parallel or series. The circuit is essentially built in either series or parallel.

Now, according to the question,

The equivalent resistance of two resistors is the R₁ and R₂ is given by, 1/R = 1/R₁ + 1/R₂

Error in equivalent resistance is,

∆R / R² = ∆R₁ / R₁² + ∆R₂/R₂²

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 $0.08$.

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