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If R1 = (6±0.2)Ω and R2= (8±0.6)Ω then find. 1.total resistance if it is connected in parallel. 2. maximum %age error in resistance.?
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If R1 = (6±0.2)Ω and R2= (8±0.6)Ω then find. 1.total resistance if it ...
Let the equivalent resistance be R. Then,

1/R = 1/6 + 1/8 => R = 24/7 Ω = 3.42 Ω

1/R = 1/R1 + 1/R2

Now differentiate throughout,

dR/R² = dR1/R1² + dR2/R2²

dR = (3.42)²{(0.2/36) + (0.6/64)}

dR = 0.17

% error in resistance = dR/R × 100 = 4.9 %

R = (3.42 ± 0.17) Ω
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If R1 = (6±0.2)Ω and R2= (8±0.6)Ω then find. 1.total resistance if it is connected in parallel. 2. maximum %age error in resistance.?
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