Discussion
The solution for the unknown resistance for a Wheatstone Bridge is \(R_{4} = \frac{R_{2}R_{3}}{R_{1}}\)
Where, R1 = 100 ± 0.5% Ω
R2 = 1000 ± 0.5% Ω
R3 = 842 ± 0.5% Ω
What will be the limiting error in percent for Resistance R4 ?
Answer: D
The limiting error of \(R_{4}\) is 1.5%.
When a result is a product or quotient of quantities, the limiting (worst-case) relative errors add, whichever side of the fraction each quantity is on.
\[ \begin{aligned} \frac{\delta R_{4}}{R_{4}} &= \frac{\delta R_{1}}{R_{1}} + \frac{\delta R_{2}}{R_{2}} + \frac{\delta R_{3}}{R_{3}} \\ &= 0.5 + 0.5 + 0.5 = 1.5\% \end{aligned} \]
The nominal value is \(R_{4} = 1000 \times 842/100 = 8420\ \Omega\), so it lies within \(8420 \pm 126.3\ \Omega\). 0.50% counts only one resistor. 1.6% and 0.16% do not follow from adding three 0.5% errors.
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