Two resistance are given as R 1 = ( 10 ± 0 . 5 )   Ω and R 2 = ( 15 ± 0 . 5 )  …

Two resistance are given as R1=(10±0.5) Ω and R2=(15±0.5) Ω. The percentage error in the measurement of equivalent resistance when they are connected in parallel is

  1. 6.33
  2. 2.33
  3. 5.33
  4. 4.33

Solution

For a parallel combination the equivalent resistance is 

1R=1R1+1R2   ...(i)

A general expression for error can be written as x±x

Differentiating equation (i)

dRR2=dR1R12+dR2R22   ...(ii)

The equivalent resistance with $R_1 = 10 \, \Omega$ and $R_2 = 15 \, \Omega$ is

1R=110+115

R=15025=6Ω

Substituting the above value in equation (ii)

dR=360.5100+0.515×15=0.26

R=6±0.26

Hence, the percentage error is 

dRR×100=0.266×100=4.33%

Asked in: JEE Main 2023 (06 Apr Shift 1)

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