C.16.RM:5.2 - Reduce loading enough to settle the question

A source has open-circuit voltage E and internal resistance Rs between 0.8 and 1.2 MΩ. A meter with input resistance Rm reads:

V = E*Rm/(Rs + Rm), hence E = V*(1 + Rs/Rm).

With Rm = 1 MΩ and an ideal reading V = 5 V, the compatible bound is 9 ≤ E ≤ 11 V. It leaves the question E ≤ 10.5 V unresolved.

One option is to determine Rs more closely and correct the loaded reading. Another is to reduce loading. The second option is available here: change to a meter with Rm = 100 MΩ. Assume the same passive-source model and resistance range apply to the new measurement. If its ideal reading is 9.9 V, then:

9.9*(1 + 0.8/100) ≤ E ≤ 9.9*(1 + 1.2/100),

so 9.9792 ≤ E ≤ 10.0188 V. The bound settles E ≤ 10.5 V while Rs remains unresolved. The repair earned its place by reducing the unknown resistance’s influence on this conclusion.

Restore a supplied reading-error bound of ±0.1 V for the new meter. The positive factors give:

9.8*1.008 ≤ E ≤ 10.0*1.012,

or 9.8784 ≤ E ≤ 10.12 V. The same decision remains settled. A more demanding threshold could make the error bound consequential again. Meter range, voltage stability and the loading model belong to the physical premises supporting this use.