When you prove a voltage to be induced using a Simpson why does the input impeadance on the different voltage scales produce different readings for the amount of voltage?

Answer 1

One possibility is that the accuracy of the Simpson is different on the different scales. Another (more probable) possibility is that the impedance of the Simpson on the different scales is sufficiently different so as to affect the reading. This is a common issue with low impedance multi-meters. Lets say you are using a typical Simpson meter with 20,000 Ohms per Volt. On a three volt scale, that means the meter itself has an impedance of 60,000 Ohms. On a 60 volt scale, however the meter has an impedance of 1,200,000 Ohms. Depending on the circuit impedance, that can have a significant impact on the final reading, which must be taken into consideration. Look at the equation for parallel resistance: RT = R1R2 / (R1+R2). If the meter impedance changes the circuit impedance by more than, say, 5%, that is going to affect the observed value. (You pick the percent limit - it depends on the situation.) Even for the case with a high impedance meter, say a 10,000,000 Ohm Digital Multi-meter, impedance must be considered if the circuit impedance is high enough. (I have a WWVB receiver that requires a 1,000,000,000 Ohm voltmeter to correctly measure the AGC voltage - no ordinary digital multimeter will suffice.)This does not mean that you have to spend lots of money on a high performance, high impedance, meter. You simply have to consider what the impedance of the meter is going to do to the circuit, and calculate that impact, before you state the results.