Question

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

Answer 1

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.

Answer 2

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=$\frac{0.15}{21}\times 100=0.71$%