A resistor of 4k ohm with tolerance 10% is connected with a resistor of 6k ohm with tolerance 10% in parallel. What is the tolerance of the parallel combination
Answers
Given info : A resistor of 4k ohm with tolerance 10% is connected with a resistor of 6k ohm with tolerance 10% in parallel.
To find : The tolerance of the parallel combination is...
solution : in parallel combination,
equivalent resistance of two resistors R₁ and R₂ is given by, 1/R = 1/R₁ + 1/R₂
now differentiating both sides,
-dR/R² = -dR₁/R₁² - dR₂/R₂²
⇒dR/R² = dR₁/R₁² + dR₂/R₂²
⇒∆R/R² = ∆R₁/R₁² + ∆R₂/R₂²
⇒∆R/R × 100 = R[∆R₁/R₁² + ∆R₂/R₂²] × 100
⇒% error (tolerance) in R = R[∆R₁/R₁² + ∆R₂/R₂²] × 100
here, R = 1/4k + 1/6k = (3 + 2)/12k = 5/12k
R = 12k/5
∆R₁ = 0.1R₁ = 0.4k , ∆R₂ = 0.6R₂ = 0.6k
now, % error in R = 12k/5 [0.4k/(4k)² + 0.6k/(6k)²] × 100
= 12k/5[0.4/16k² + 0.6k/36k²] × 100
= 12k/5 [1/40k + 1/60k] × 100
= 12k/5[ 5/120k] × 100
= 10%
Therefore the tolerance of parallel combination is 10%