The Ammeter and the Voltmeter — Question 10

Question 10

Four-wire sensing removes force-lead drops but retains a finite-input correction

A regulated total current I enters specimen node A and leaves B through separate force leads. The specimen resistance Rx joins A to B. A voltmeter of resistance RV senses A and B through two leads of resistance rs each. Its displayed voltage Vm is across RV only.

1. Derive Vm/I, accounting for the small current in the sense branch.

2. Find the fractional underestimate of Rx. Evaluate it for Rx=.100 ohm, rs=1.00 ohm and RV=10.0 megohm; explain the ideal-input limit.

Question 10 - Solution

Four-wire sensing removes force-lead drops but retains a finite-input correction

(a) Method and calculation

$$R_m=R_V+2r_s,\quad V_{AB}=I\frac{R_xR_m}{R_x+R_m}.$$$$V_m=V_{AB}\frac{R_V}{R_m}\Rightarrow\frac{V_m}{I}=\frac{R_xR_V}{R_x+R_V+2r_s}.$$

(b) Interpretation and further calculation

$$\epsilon=1-\frac{V_m/I}{R_x}=\frac{R_x+2r_s}{R_V+R_x+2r_s}=\frac{2.10}{10000002.10}\simeq2.10\times10^{-7}.$$$$R_V\to\infty:\quad V_m/I\to R_x.$$Force-lead drops lie outside the sensed nodes, provided the driver maintains I.

Physical check

Separating force and sense connections eliminates the leading contact-voltage error; finite sense current remains calculable.