Two ammeters A1 and A2 measure the same current and provide readings I1 and I2 respectively. The ammeter error can be characterized as independent zero mean Gaussian random variable of standard deviations σ1 and σ2 respectively. The value of the current is computed as:I = μI1 + (1 - μ)I2The value of μ which gives the lowest standard deviation of I is

Question

TuteeHUB · Accepted Answer

$\frac{{{\rm{\sigma }}_1^2}}{{{\rm{\sigma }}_1^2 + {\rm{\sigma }}_2^2}}$

TuteeHUB · Answer

$\frac{{{\rm{\sigma }}_2^2}}{{{\rm{\sigma }}_1^2 + {\rm{\sigma }}_2^2}}$

TuteeHUB · Answer

$\frac{{{{\rm{\sigma }}_2}}}{{{{\rm{\sigma }}_1} + {{\rm{\sigma }}_2}}}$

TuteeHUB · Answer

$\frac{{{{\rm{\sigma }}_1}}}{{{{\rm{\sigma }}_1} + {{\rm{\sigma }}_2}}}$

1.	Two ammeters A1 and A2 measure the same current and provide readings I1 and I2 respectively. The ammeter error can be characterized as independent zero mean Gaussian random variable of standard deviations σ1 and σ2 respectively. The value of the current is computed as:I = μI1 + (1 - μ)I2The value of μ which gives the lowest standard deviation of I is
A.	\(\frac{{{\rm{\sigma }}_2^2}}{{{\rm{\sigma }}_1^2 + {\rm{\sigma }}_2^2}}\)
B.	\(\frac{{{\rm{\sigma }}_1^2}}{{{\rm{\sigma }}_1^2 + {\rm{\sigma }}_2^2}}\)
C.	\(\frac{{{{\rm{\sigma }}_2}}}{{{{\rm{\sigma }}_1} + {{\rm{\sigma }}_2}}}\)
D.	\(\frac{{{{\rm{\sigma }}_1}}}{{{{\rm{\sigma }}_1} + {{\rm{\sigma }}_2}}}\)
Answer» B. \(\frac{{{\rm{\sigma }}_1^2}}{{{\rm{\sigma }}_1^2 + {\rm{\sigma }}_2^2}}\)

Discussion

No Comment Found

Related MCQs

Reply to Comment