C.16.IR:5.1 - One source, two loads and an unidentified influence

Suppose a source has unknown open-circuit voltage E ≥ 0 and finite internal resistance Rs ≥ 0. An ideal meter of known input resistance Rm > 0 reads the connected voltage:

V = E*Rm/(Rs + Rm).

The relation and passive-source assumptions are supplied. The question is whether E is at most 8 V.

A reading of 5 V with Rm = 1 MΩ gives:

E = 5 V*(1 + Rs/(1 MΩ)).

Rs = 0 and E = 5 V fit the reading; Rs = 1 MΩ and E = 10 V also fit it. The threshold question is unresolved. The reading does establish E ≥ 5 V under the stated domain.

Now a meter with Rm = 3 MΩ reads 7.5 V. Assume E and Rs stayed unchanged between the readings and both meters obey the given model. The two equations share those two unknowns. Express E from each equation and equate:

V1*(1 + Rs/R1) = V2*(1 + Rs/R2).

Hence, when the denominator is nonzero,

Rs = (V2 - V1)/(V1/R1 - V2/R2).

The supplied readings give Rs = 1 MΩ and E = 10 V. Substitution recovers both 5 V and 7.5 V. The answer to E ≤ 8 V is now no.

The shared-source condition does work here. If E changes between readings, replacing it with E1 and E2 leaves the original inference unsupported. A second number is useful through the relation connecting its measurement to the first.

For two zero readings under the same model, E = 0 follows while Rs can be any finite nonnegative value. A zero denominator in the derived formula is a reason to return to the original equations. Full recovery of Rs is unnecessary for this voltage result.