The Ammeter and the Voltmeter — Question 8

Question 8

A higher voltmeter range trades reduced loading against a larger absolute specification

An analog voltmeter has input resistance 10.0 kilohm per volt of full-scale range and a maximum indication error of .5 percent of full scale. A source has E=10.0 V and r=20.0 kilohm. Compare its 10.0 V and 100 V ranges.

1. Find the loaded voltage on each range.

2. Bound the worst absolute difference between indicated voltage and E by adding loading error and the full-scale error allowance. Which range has the smaller bound?

Question 8 - Solution

A higher voltmeter range trades reduced loading against a larger absolute specification

(a) Method and calculation

$$R_{10}=100\,\mathrm{k\Omega},\quad R_{100}=1.00\,\mathrm{M\Omega}.$$$$V_{10}=10\frac{100}{120}=25/3\,\mathrm V,\quad V_{100}=10\frac{1000}{1020}=500/51\,\mathrm V.$$

(b) Interpretation and further calculation

$$B_{10}=10-25/3+.005(10)=103/60\,\mathrm V\simeq1.72\,\mathrm V.$$$$B_{100}=10-500/51+.005(100)=71/102\,\mathrm V\simeq.696\,\mathrm V.$$The 100 V range has the smaller specified worst-case bound here.

Physical check

Range choice depends on both circuit loading and the instrument error model; nominal resolution alone is insufficient.