Question

Suppose you want to measure resistances in the range from $\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{\Omega }$to $\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{k\Omega }$using a Wheatstone bridge that has $\frac{{\mathbf{R}}_{\mathbf{2}}}{{\mathbf{R}}_{\mathbf{1}}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{000}$. Over what range should${\mathbf{R}}_{\mathbf{3}}$be adjustable?

Short answer

The range over which the value ${\mathrm{R}}_3$is adjustable is ${\mathrm{R}}_3=5\mathrm{\Omega }-5\mathrm{k\Omega }$.

Step-by-step solution

Concept Introduction

A Wheatstone bridge is an electrical circuit that balances two legs of a bridge circuit, one of which includes the unknown component, to measure an unknown electrical resistance. The circuit's main advantage is its capacity to deliver exceedingly exact readings.

Given Information

• The value range for the resistance in a Wheatstone bridge:${\mathrm{R}}_{\mathrm{x}}=10\mathrm{\Omega }-10\mathrm{k\Omega }$.
• The value of $\frac{{\mathrm{R}}_2}{{\mathrm{R}}_1}=2$.

Range for resistance

Calculate the necessary resistance ${\mathrm{R}}_3$for the range by calculating ${\mathrm{R}}_3$what corresponds ${\mathrm{R}}_{\mathrm{x}}=10\mathrm{\Omega }$to ${\mathrm{R}}_{\mathrm{x}}=10\mathrm{\Omega }$and using the relation,

$\frac{{\mathrm{R}}_{\mathrm{x}}}{{\mathrm{R}}_3}=\frac{{\mathrm{R}}_2}{{\mathrm{R}}_1}$

The calculation for the minimum value –

role="math" localid="1656396469353" $$\begin{gathered} {\mathrm{R}}_3^{\min }={\mathrm{R}}_3^{\min }\left(\frac{{\mathrm{R}}_1}{{\mathrm{R}}_2}\right) \\ =10\mathrm{\Omega }\times \left(\frac{1}{2}\right) \\ =5\mathrm{\Omega } \end{gathered}$$

The calculation for the maximum value,

$$\begin{gathered} {\mathrm{R}}_3^{\max }={\mathrm{R}}_3^{\max }\left(\frac{{\mathrm{R}}_1}{{\mathrm{R}}_2}\right) \\ =10\mathrm{k\Omega }\times \left(\frac{1}{2}\right) \\ =5\mathrm{k\Omega } \end{gathered}$$

Therefore, the resistance range is ${\mathrm{R}}_3=5\mathrm{\Omega }-5\mathrm{k\Omega }$.