Answer:
0.71%
Explanation:
We are given that
[tex]R_1=6ohm[/tex]
[tex]R_2=7ohm[/tex]
[tex]R_3=8ohm[/tex]
Error in each resistor=0.05 ohm
[tex]R=R_1+R_2+R_3[/tex]
[tex]R=6+7+8=21 ohm[/tex]
[tex]Total error=3(0.05)=0.15[/tex] ohm
We have to find the maximum possible error in computing R.
Maximum possible error=[tex]\frac{error}{R}\times 100[/tex]
Substitute the values
Maximum possible error=[tex]\frac{0.15}{21}\times 100=0.71[/tex]%
Answer:
Explanation:
R1 = 6 ohm
R2 = 7 ohm
R3 = 8 ohm
error = 0.05 ohm
Let R is the equivalent resistance
[tex]\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}[/tex] ... (1)
[tex]\frac{1}{R}=\frac{1}{6}+\frac{1}{7}+\frac{1}{8}[/tex]
R = 2.3 ohm
The error is ΔR.
Differentiate the equation (1)
[tex]\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}}[/tex]
[tex]\frac{\Delta R }{2.3^2}=\frac{0.05}{6}+\frac{0.05}{7}+\frac{0.05}{8}[/tex]
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.
