Question

the resistance r=v/I,where v=200+-10v and i=20+-0.4a determine the maximum possible percentage error in r
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Answer 1

Answer :

• Percentage error in Resistance is equal to 7 %.

Given :

Potential difference is given as : ( 200 ± 10 ) V , so

• V = 200 ; Δ V = 10

Current is given as : ( 20 ± 0.4 ) A , so

• I = 20 ; Δ I = 0.4

To find :

• Percentage error in its Resistance ( R )

Knowledge required :

• Ohm's law

   Resistance = P.d. / current

  or     R = V / I

• Error of a quotient

For any quantity Z = A / B , where measured values of A and B are A ± Δ A and B ± Δ B, then relative error in Z will be given by,

    Δ Z / Z = Δ A / A  +  Δ B / B

• Calculating percentage error

When relative error is expressed in Percentage, it is called percentage error

   percentage error = ( Δ Z / Z ) × 100 %

Solution :

Since, R = V / I

therefore, finding relative error in Resistance

→ Δ R / R = Δ V / V  +  Δ I / I

→ Δ R / R = (10) / (200) + (0.4) / (20)

→ Δ R / R = ( 10 + 4 ) / ( 200 )

→ Δ R / R = ( 14 ) / ( 200 )

→ Δ R / R = 14 / 200  = 7 / 100

Now,

Calculating percentage error in Density

→ Percentage error in R = ( Δ R / R ) × 100 %

→ Percentage error in R = ( 7 / 100 ) × 100 %

→  Percentage error in R = 7 %

therefore,

• Percentage error in Density is equal to 2.5 % .