C.16.MR:5.1 - Recover an open-circuit voltage from a loaded indication

The sought value is a source’s open-circuit voltage E. The supplied ideal circuit represents the source by E in series with resistance Rs. Connecting a voltmeter of input resistance Rm completes the circuit. Assume positive resistances, a stable source and a meter whose indication equals its terminal voltage.

The common current I satisfies

E = I*Rs + I*Rm
V = I*Rm

Eliminating I gives

V = E*Rm/(Rs+Rm)
E = V*(1+Rs/Rm)

For Rs=Rm=1 megohm and V=5 volts, the inferred open-circuit voltage is 10 volts. Substitution returns I=5 microamperes and the original 5-volt indication. The two voltage drops explain both the sign and magnitude of the correction.

If the desired quantity were instead the connected terminal voltage, V would already be the value in this ideal model. The measurement question determines which result is needed.

As Rm becomes large relative to Rs, loading becomes small and V approaches E. For a use that needs only a bound on this effect, E-V=V*Rs/Rm supplies it. If Rs/Rm is at most 0.001 and V is 5 volts, the loading correction is at most 0.005 volts. Whether that is negligible depends on the receiving question and the other uncertainties.

If Rs is unknown, the same relation can expose an unresolved contribution. With V=5 volts, Rm=1 megohm and Rs between 0.8 and 1.2 megohms, E lies between 9 and 11 volts under the ideal assumptions. The interval may be enough. A performed measurement additionally accounts for indication and resistance uncertainty and relevant model inadequacy.