A voltage $V_{1}$ is measured 100 times and its average and standard deviation are 100 V and 1.5 V respectively. A second voltage $V_{2}$ , which is independent of $V_{1},$ is measured 200 times and its average and standard deviation are 150 V and 2 V respectively. $V_{3}$ is computed as: $V_{3} = V_{1} + V_{2} .$ Then the standard deviation of $V_{3}$ in volt is
2.5

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Solution

The correct option is A 2.5
$V_{1} = 100 \pm 1.5 V$
$V_{2} = 150 \pm 2 V$
$V_{3} = V_{1} + V_{2}$

Standard deviation in $V_{3}$ will be
$σ_{V_{3}} = \sqrt{{(\frac{\partial V_{3}}{\partial V_{1}})}^{2} σ_{V_{1}}^{2} + (\frac{\partial V_{3}}{\partial V_{2}}) σ_{V_{2}}^{2}}$

$\frac{\partial V_{3}}{\partial V_{1}} = \frac{\partial}{\partial V_{1}} (V_{1} + V_{2}) = 1$

$\frac{\partial V_{3}}{\partial V_{2}} = \frac{\partial}{\partial V_{2}} (V_{1} + V_{2}) = 1$

$∴ σ_{V_{3}} = \sqrt{1^{2} \times (1.5)^{2} + 1^{2} \times 2^{2}}$

$= \sqrt{6.25} = 2.5 V$

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