Having edited my initial post to insert the correct values, I decided to perform another experiment.

The gubbins and the

` Fluke 8022B `DMM were connected in parallel to the terminals of the

` Level Measuring Set No. 35A`, which was configured to terminate the circuit with a load of 600 Ohms.. All three items were powered on. The DMM indicated a value of 0.250 V, whilst the LMS indicated a value of -9.9 dBm.

`The power = (0.250)`^{2} x 10^{3} / 600 mW

= 0.1042 mW

That power is equivalent to 10 x log_{10} 0.1042 dBm

= -9.8 dBmI am satisfied that all the observed values are essentially consistent with one another.

The only deduction that I can make is that the

` Rheostat No. 1A `and the

` Level Measuring Set No. 35A `differ slightly in their presentation of 600 Ohms. The former being somewhat smaller and the latter being somewhat greater than the nominal 600 Ohms.

Looking closely at both sets of results, we can see that it is possible to calculate the deviation from the nominal 600 Ohms. We make two assumptions --

- The gubbins was providing the same power in both experiments.
- The arithmetic mean of the two resistances is 600 Ohms.

Let R

_{R} be the resistance of the

` Rheostat No. 1A `when it is set to 600 Ohms.

Let R

_{L} be the resistance of the

` Level Measuring Set No. 35A `when it is set to terminate the circuit with 600 Ohms.

We now substitute R

_{R} & R

_{L} into the respective power calculations and then equate them (assumption 1., above) --

`(0.242)`^{2} x 10^{3} / R_{R} = (0.250)^{2} x 10^{3} / R_{L}Rearranging, we can see --

`R`_{R} = R_{L} x (0.242)^{2} x 10^{3} / ((0.250)^{2} x 10^{3})

= R_{L} x (0.242)^{2} / (0.250)^{2}

= R_{L} x 0.937024 (equation 1)From assumption 2., above, we write --

`(R`_{R} + R_{L}) / 2 = 600Rearranging, we can see --

`R`_{R} = 1200 - R_{L} (equation 2)Substituting equation 2 into equation 1 gives --

`1200 - R`_{L} = R_{L} x 0.937024Rearranging, we can see --

`(R`_{L} x 0.937024) + R_{L} = 1200Thus --

`R`_{L} = 1200 / (1 + 0.937024)

= 619.5 Ohms (equation 3)Substituting equation 3 into equation 2 gives --

`R`_{R} = 1200 - 619.5

= 580.5 OhmsA rather surprising conclusion. But one that fits the facts.