   Chapter 14.4, Problem 39E

Chapter
Section
Textbook Problem

If R is the total resistance of three resistors, connected in parallel, with resistances R1, R2 R3 then 1 R = 1 R 1 + 1 R 2 + 1 R 3 If the resistances are measured in ohms as R1 = 25 Ω, R2 = 40 Ω, and R3 = 50 Ω. with a possible error of 0.5% in each case, estimate the maximum error in the calculated value of R.

To determine

To estimate: The maximum error in the calculated value of R , if 1R=1R1+1R2+1R3 and R1,R2andR3 are three resistors.

Explanation

Given:

The total resistance of three resistors is R .

1R=1R1+1R2+1R3

Where R is measured in ohms.

The three resistors are R1,R2andR3 .

The value of R1=25Ω .

The value of R2=40Ω .

The value of R3=50Ω .

The possible error is 0.5% .

Calculation:

The given function is, 1R=1R1+1R2+1R3 (1)

Substitute R1=25,R2=40andR3=50 and obtain R as follows,

1R=125+140+150=8+5+4200=17200R=20017

Thus, the value of R=2001711.76 .

The error 0.5% at R1=25,R2=40andR3=50 are computed as follows,

ΔR1=(25)(0.005)=0.125ΔR2=40(0.005)=0.2ΔR3=50(0.005)=0.25

Take the partial derivative with respect to R1 of the equation (1),

R1(1R)=R1(1R1+1R2+1R3)1R2(RR1)=1R12+0+01R2(RR1)=1R12RR1=R2R12

Thus, the value of RR1 is R2R12 .

Take the partial derivative with respect to R2 of the equation (1),

R2(1R)=R2(1R1+1R2+1R3)1R2(RR2)=01R22+01R2(RR2)=1R22RR2=R2R22

Thus, the value of RR2 is R2R22

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