Question

Two resisters of R1 = 100±3 Ω and R2 = 200±4 Ω are connected on series. Find the maximum absolute error, relative error and percentage error in the equivalent resistance of the combination

Answer 1

Solution

Maximum absolute error

$\Delta R=\Delta R_1+\Delta R_2$

$\Delta R=3+4=7 \Omega$

Relative error

$\frac{\Delta R}{R}=\frac{\Delta R_1} {R_1} +\frac{\Delta R_2} {R_2}$

$\frac{\Delta R}{R}=\frac{3} {100} +\frac{4} { 200}=0.07$

Percentage error

$\frac{\Delta R}{R}\times 100=(\frac{\Delta R_1} {R_1} +\frac{\Delta R_2} {R_2}) \times100$

Percentage error =$7\%$