NEET Class neet Answered
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Asked by jhajuhi19 | 30 Dec, 2020, 08:36: PM
Expert Answer
Equivalent resistance of R1 = 3 Ω and R2 = 6 Ω is ( R1 R2 ) /( R1 + R2 ) = (18/9 ) Ω = 2 Ω
if percentage error of R1 is 1% , then error in R1 , Δ R1 = .01 × 3 = 0.03 Ω
if percentage error of R2 is 2% , then error in R2 , Δ R2 = .02 × 6 = 0.12 Ω
If R is equivalenet resistance of R1 and R2 , then we have (1/R) = ( 1 / R1 ) + ( 1 / R2 ) ................(1)
By differentiating eqn.(1) , we get , ( ΔR / R2 ) = ( ΔR1 / R12 ) + ( ΔR2 / R22 ) ...................(2)
Hence from eqn.(2) , we get , ΔR = ( R2 / R12 ) ΔR1 + ( R2 / R22 ) ΔR2
Let us substitute values in above expression, then we get , ΔR = ( 22 / 32 ) × 0.03 + ( 22 / 62 ) × 0.12 = 0.0266
Percentage error in parallel combination of R1 and R2 is given as , ( ΔR / R ) × 100 = 1.33 %
Answered by Thiyagarajan K | 30 Dec, 2020, 11:41: PM
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