The resistances of two resistors are measured as R1 = (6 ± 0.3) Ω and R2 = (12 ± 0.2) Ω. The maximum percentage error in equivalent resistance when they are connected in parallel is
8.5%
3.9%
6.4%
2.1%
Answers
Explanation:
Given The resistances of two resistors are measured as R1 = (6 ± 0.3) Ω and R2 = (12 ± 0.2) Ω. The maximum percentage error in equivalent resistance when they are connected in parallel is
- Given the value of two resistors R1 = 6 ± 0.3 ohm and R2 = 12 ± 0.2 ohm.
- Now this is connected in parallel and we need to find the total resistance and also the percentage error.
- So 1/R eq = 1/R1 + 1/R2
- So R eq = R1 R2 / R1 + R2
- Now substituting the main values we get
- R eq = 6 x 12 / 6 + 12
- R eq = 4 ohm
- Now we need to find Δ R eq
- Now we have 1/R eq = 1/R1 + 1/R2
- So we have ΔR eq = Δ R1 / R1^2 + Δ R2 / R2^2
- We can write this as ΔR1 (R eq / R1)^2 + Δ R2(R eq / R2)^2
- 0.3 (4 / 6)^2 + 0.2 (4 / 12)^2
- 1.4 / 9
- ΔR eq = 0.155 ohm
- So total resistance will be R eq ± Δ R eq
- (4 ± 0.155) ohm
- Now maximum percentage will be Δ R eq / R eq x 100
- 0.155 / 4 x 100
- 0.0387 x 100
- = 3.9 %
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B) 3.9%
Explanation:
Given: The resistances of two resistors are measured as R1 = (6 ± 0.3) Ω and R2 = (12 ± 0.2) Ω.
Find: The maximum percentage error in equivalent resistance when they are connected in parallel.
Solution:
R1 = 6 Ω
∆R1 = 0.3 Ω
R2 = 12 Ω
∆R2 = 0.2 Ω
Equivalent resistance:
1 / Rp = 1/R1 + 1/R2
1/ Rp = 1/6 + 1/12
1/ Rp = 18 / 72
1/ Rp = 1/4
Therefore Rp = 4
Error in equivalent resistance:
∆Rp / Rp² = ∆R1 / R1² + ∆R2 / R2²
∆Rp / Rp = ∆R1 * Rp / R1² + ∆R2 * Rp/ R2²
∆Rp / Rp = 0.3 * 4 / (0.6)² + 0.2 * 4 / (12)²
∆Rp / Rp = 0.389
Maximum percentage of error in equivalent resistance = 0.389 * 100 = 3.89% = 3.9% approximately
Option B is the answer.