Question

Two resistance r1=50+-0.1,r2=100+-0.1 find effective resistance in series parellel

Answer 1

Answer:

r1=50+-0.1

r2=100+-0.1

in series

r=r1+r2=50+100=150

dr=dr1+dr2=0.1+01=0.2

r=150+-0.2

sqrt(((dr1)^2)+((dr2)^2))=sqrt((0.1^2)+(0.1)^2)=0.1414

r=150+-0.1414

in parallel

1/r=1/r1+1/r2

r=r1r2/(r1+r2)=50×100/(50+100)=33.333

dr/r^2=dr1/r1^2 + dr2/r2^2

dr=r^2(dr1/r1^2 + dr2/r2^2)

   =(33.333^2)((0.1/50^2)+(0.1/100^2))

   =1111.089(5×10^-5)

    0.0556

dr=0.0556 ohm

r=33.333+-0.0556 ohm

Explanation: