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