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IrinaVladis [17]
3 years ago
6

The resistance R produced by wiring resistors of R1, R2, and R3 ohms in parallel can be calculated from the formula If R1, R2, a

nd R3 are measured to be 6 ohms, 7 ohms, and 8 ohms respectively, and if these measurements are accurate to within 0.05 ohms, estimate the maximum possible error in computing R.
Physics
2 answers:
Elena-2011 [213]3 years ago
7 0

Answer:

Explanation:

R1 = 6 ohm

R2 = 7 ohm

R3 = 8 ohm

error = 0.05 ohm

Let R is the equivalent resistance

\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}    ... (1)

\frac{1}{R}=\frac{1}{6}+\frac{1}{7}+\frac{1}{8}

R = 2.3 ohm

The error is ΔR.

Differentiate the equation (1)

\frac{\Delta R }{R^2}=\frac{\Delta R_{1}}{R_{1}^{2}}+\frac{\Delta R_{2}}{R_{2}^{2}}+\frac{\Delta R_{3}}{R_{3}^{2}}

\frac{\Delta R }{2.3^2}=\frac{0.05}{6}+\frac{0.05}{7}+\frac{0.05}{8}

ΔR = 2.3 x 2.3 (8.33 + 71.4 + 6.25) x 10^-3

ΔR = 0.115

So, the maximum possible error is 0.115.

SIZIF [17.4K]3 years ago
6 0

Answer:

0.71%

Explanation:

We are given that

R_1=6ohm

R_2=7ohm

R_3=8ohm

Error in each resistor=0.05 ohm

R=R_1+R_2+R_3

R=6+7+8=21 ohm

Total error=3(0.05)=0.15 ohm

We have to find the maximum possible error in computing R.

Maximum possible error=\frac{error}{R}\times 100

Substitute the values

Maximum possible error=\frac{0.15}{21}\times 100=0.71%

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