The values of two resistors are (8.0±0.3) Ω and (12.0±0.2) Ω. What is the percentage error in the equivalent resistance when they are connected in series and parallel?
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R1= 8.0Ω . dR1 = +- 0.3Ω
R2 = 12.0 . dR2 = +- 0.2Ω.
In series:
Net Resistance = R= 20 Ω.
Net error: 0.2+0.3= 0.5Ω.
% error: 0.5×100/ 20 = 2.5%
In parallel:
1/R = 1/R1 + 1/R2
- dR/R^2 = - dR1/R1^2 - dR2/R2^2.
R = 8×12/(8+12)= 4.8Ω.
dR/R = 4.8 * 0.2/12^2 + 4.8 * 0.3/8^2
= 2.9%
R2 = 12.0 . dR2 = +- 0.2Ω.
In series:
Net Resistance = R= 20 Ω.
Net error: 0.2+0.3= 0.5Ω.
% error: 0.5×100/ 20 = 2.5%
In parallel:
1/R = 1/R1 + 1/R2
- dR/R^2 = - dR1/R1^2 - dR2/R2^2.
R = 8×12/(8+12)= 4.8Ω.
dR/R = 4.8 * 0.2/12^2 + 4.8 * 0.3/8^2
= 2.9%
kvnmurty:
:-))
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