[" If voltage "V=100+-5" volt and current "],[I=10+-0.2A," the percentage error in "],[" resistance "R" is "]

If voltage V=(100pm5) V and current I=(10pm0.2) A, the percentage error in resistance R is

If voltage V=(100pm5) V and current I=(10pm0.2) A, the percentage error in resistance R is

Relation between voltage, current and resistance is given by R Measured value of voltage is (50 ± 2) volt, Measured value of current is (20 ±0.2) A. Percentage error in resistance measurement is?

[" According to ohms law "V=iR" ,"],[" where "V" is potential "],[" difference,"l=" electric current "],[" and "R=" resistance.If "],[V=(220+-11)" volt and "],[i=(10+-0.1)" ampere and "],[" percentage error in the "],[" measurement of "R" is "n%" then "],[" find the value of "n" ? "]

The resistance R = Vil. If V= (100 +- 5) volt and I = (10 +- 0.2) A. Find the percentage error in R.

The resistance R=(V)/(I) where V=(200+-5)V and I=(20+-0.2)A ,the percentage error in the measurement of "R" is :

The resistance R=V/I Where V = (100 p5) V and I=(10 pm0.2) . The percentage error in R is

To determine an unknown resistance R in the Iaboratory, readings of potentiaI difference V and eIectric current I are taken. From these readings percentage error is caIcuIated and V and I are obtained as V= (100 pm 5) volt I = (10 pm 0.2) amp. The percentage error in the value of R determined from these V and I will be