Two identical resistors of resiatance `(2 pm 0.1) Omega` are connected in series. Calculate the effective resistance along with the percentage error.

Explanation

Given `R_(1)=(2pm 0.1)Omega`
`R_(2)=(2pm 0.1)Omega`
`R.= R_(1)+R_(2)` (series connection)
`R.= 2+2= 4 Omega`
`triangleR.= pm(triangleR_(1)+triangleR_(2))`
`triangleR.= pm(0.1+0.1)= pm 0.2`
Percentage error `= pm (0.2)/(4)xx100= pm 5%`
`R.= (4 pm 0.2)Omega= (4 Omega pm 5%)`.