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Two resistors of resistances R 1 = 100+/- 3 ohm and R2= 200 +/- 4 ohm are connected (a) in series, (b) in parallel. Find the equivalent resistance of the (a) series combination, (b) parallel combination. Use for (a) the realtion R = R1+ R2and for (b) 1/R ' = 1/R1+ 1/R2and ΔR '/R '2= ΔR1/R1 2+ ΔR2/R2 2.

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Answer 1

Answer:

Explanation:

a) The equivalent resistance of series combination

R=R1+R2=(100±3)ohm+(200±4)ohm

=300±7ohm

(b) The equivalent resistance of parallel combination

R'=R1R2R1+R2=2003=66.7ohm

Then, from 1R'=1R1+1R2

we get,

ΔR'R2=ΔR1R21+ΔR2R2

ΔR'=(R'2)ΔR1R21+(R2)ΔR2R22

=(66.7100)23+(66.7200)24

Then, R'=66.7±1.8 ohm

(Here, ΔR is expresed as 1.8 instead of 2 to keep in confirmity with the rules of significant figure).

(c) Error in case of a measured quantity raised to a power

Suppose Z=A2

Then.

ΔZ/Z=(ΔA/A)+(ΔA/A)=2(ΔA/A).

Hence, the relative error in A2 is two times the error in A.

In general, if Z=ApBq/Cr

Then,

ΔZ/Z=p(ΔA/A)+q(ΔB/B)+r(ΔC/C)C

hope u understand