Question

Two resistors have resistances R1= ( 24+- 0.5)ohm and R2 = (8 +-0.3)ohm . Calculate the absolute error and the percentage relative error in calculating the combination of two resistances when they are in parallel. ​

Answer 1

Answer:

Absolute error=0.006

Percentage Relative error=0.6

Explanation:

The resistors have resistance of 24 ohm and 8 ohm.

The change in resistance is 0.5 and 0.3 ohm respectively.

Relative error for parallel combination of resistors is

= dR/R²

= dR1/R1² + dR2/R2²

= 0.5/(24)² + 0.3/(8)²

= 0.5/ 24*24 + 0.3/ 64

= 0.5/576+0.3/ 64

= 32 + 172.8/ 36,864

=204.8/ 36,864=0.0055

=0.006

Percentage error =Relative error *100

= 0.006* 100 = 0.6