Question

the potential difference across the end of wire has been measured to be (100+_5)and current in the wire as (10+_0.2)A.what is the percentage error in the computed resistance of the wire

Answer 1

v=(100+-5)a

I=(10+-0.2)a

R=V/I =100/10=10Ω

ΔR/R=+_[ ΔV/V+ ΔI/I]

=+_[5/100+0.2/10]

=+_[7/100]

=+_0.07

ΔR=R*0.07

=+_0.07*10

=0.7 ohms

Percentage error ΔR/R*100=+_0.07*100=+_7 percent