Question

find the percentage error in the equivalent resistance of two parallel resistance​

Answer 1

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Answer 2

The percentage error in the equivalent resistance is 5%

Explanation:

When the resistors are connected in parallel, the equivalent resistance is given by the formula:

1/R = 1/r₁ + 1/r₂

R = (r₁r₂)/(r₁ + r₂)

On differentiation the above equation, we get,

-1/R² dR = -1/r₁² dr₁ - 1/r₂² dr₂

1/R² dR = 1/r₁² dr₁ + 1/r₂² dr₂

1/R |dR/R| = 1/r₁ |dr₁/r₁| + 1/r₂ |dr₂/r₂|

|dR/R| = R [1/r₁ |dr₁/r₁| + 1/r₂ |dr₂/r₂|]

|dR/R| = (r₁r₂)/(r₁ + r₂) [1/r₁ |dr₁/r₁| + 1/r₂ |dr₂/r₂|]

|dR/R| = (r₂ |dr₁/r₁| + r₁ |dr₂/r₂|)/(r₁ + r₂)

On substituting the values, we get,

|dR/R| = (7 × 0.25/5 + 5 × 0.35/7)/(5 + 7)

|dR/R| = (7 × 0.5 + 5 × 0.5)/(5 + 7)

|dR/R| = (0.35 + 0.25)/(5 + 7)

|dR/R| = (0.6)/(12)

|dR/R| = 0.05

∴ % |dR/R| = 5%