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For two resistance R1 and R2, connected in parallel, the relative error in their equivalent resistance is (where R1=(10.0 -0.1)ohm and R2=(20.0 -0.4)ohm?
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For two resistance R1 and R2, connected in parallel, the relative erro...
Explanation:
When two resistors R1 and R2 are connected in parallel, their equivalent resistance is given by the formula:
1/R = 1/R1 + 1/R2
where R is the equivalent resistance.

Relative Error:
The relative error in the equivalent resistance is given by the formula:
Relative Error = (ΔR/R) × 100%
where ΔR is the absolute error in the equivalent resistance and R is the true value of the equivalent resistance.

Calculating the Equivalent Resistance:
Using the formula for the equivalent resistance, we get:
1/R = 1/[(10.0 - 0.1)ohm] + 1/[(20.0 - 0.4)ohm]
Simplifying, we get:
R = 6.89 ohm

Calculating the Absolute Error:
The absolute error in R1 and R2 can be calculated as follows:
ΔR1 = 0.1 ohm
ΔR2 = 0.4 ohm

Using the formula for the absolute error in the equivalent resistance, we get:
ΔR = (R2/R) × ΔR1 + (R1/R) × ΔR2
Substituting the values, we get:
ΔR = (20.0/(6.89)) × (0.1) + (10.0/(6.89)) × (0.4)
Simplifying, we get:
ΔR = 0.85 ohm

Calculating the Relative Error:
Using the formula for the relative error, we get:
Relative Error = (ΔR/R) × 100%
Substituting the values, we get:
Relative Error = (0.85/6.89) × 100%
Simplifying, we get:
Relative Error = 12.34%

Therefore, the relative error in the equivalent resistance is 12.34%.
Community Answer
For two resistance R1 and R2, connected in parallel, the relative erro...
For two resistance R1 and R2, connected in parallel, the relative error in their equivalent resistance is (where R1=(10.0+-0.1)ohm and R2=(20.0+-0.4)ohm
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For two resistance R1 and R2, connected in parallel, the relative error in their equivalent resistance is (where R1=(10.0 -0.1)ohm and R2=(20.0 -0.4)ohm?
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