Two resistors R1 = 600 +- 3 Ohm and R2 = 300 +- 6 Ohm are connected in series combination. Find the percentage error in net resistance.

Net resistance R = R1 + R2

(In a series combination, resistances get added)

Max error in measurement of R1 = 3 ohm.

Relative error = Max. error/Measured value

Therefore,

Relative error for R1 = 3/600 = 0.005

Max error in measurement of R2 = 6 ohm.

Relative error for R2 = 6/300 = 0.02

When two quantities are added, the respective errors involved also get added.

R = R1 + R2

Measured value of R = 600 + 300 = 900 ohms

Error in measuring R = 3 + 6 = 9 ohm

Relative Error in measurement of R = 9/900 = 0.01

Percentage error = Relative error * 100

or, Percentage error = 0.01* 100

or, Percentage error in measuring R = 1 %