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If you need to measure the transformer at full voltage them do this: A typical multimeter has a 10MΩ input resistance. If you connect nine 10MΩ resistors in series and measure the voltage through them the voltage will be reduced by a factor of ten, so the meter will read 180V.
Note: Be very careful with such a high voltage since it's decidedly lethal.
I would never do that without HV probes.
You misunderstand the circuit. The 10MegΩ resistors are connected in series with the 10MegΩ multimeter input impedance. The meter resistor is part of the voltage divider circuit. The drop provided by the DMM is desired (indeed it's a required part of the circuit).10MegΩ is too high and would make drop when DMM is connected, it can be in KΩ range,
just take more care, you can measure it
You misunderstand the circuit. The 10MegΩ resistors are connected in series with the 10MegΩ multimeter input impedance. The meter resistor is part of the voltage divider circuit. The drop provided by the DMM is desired (indeed it's a required part of the circuit).
Resistors in the kΩ range would dissipate a lot of power
what about 100k x10 numbers in series? power discipation would drop to 3.24 W?
also each resistor discipates 0.324?
connect the meter across one of them and measure?
what about the accuracy of the result when 10M ohms are in series with the meter? ( input impedence of the DMM is so accurate?)