How to properly measure the output voltage of a boost converter?

Question

Electronics

In order to avoid voltage drop with a multimeter, it is recommended to use a divider, for example, 100 MΩ. I do not understand how the divider will help in this case? Let's say the voltage at the output of the converter is 400v, we connect a divider of 10 10MΩ resistors, measure the voltage across one resistor. The resistor and the multimeter are connected in parallel, their total resistance (if the internal resistance of the multimeter is 10 MΩ) will be 5 MΩ, so the total resistance of the circuit will be 95 MΩ, the current is 400/95000000 = 4.21 μA, respectively. The measured voltage across the resistor will be 4.21 μA * 5MOhm = 21V. Multiplying 21 by 10 we get 210v, but not 400v. What do I think wrong?

Answer 1

Are there such problems with the power of the source that 10 MΩ of its resistance will make significant changes in the output voltage? Or is it not possible to find a multimeter with a range of up to 1000 V?
As an option, put a buffer on the op-amp with an input current of several pA (orders of magnitude less than the current of the measured circuit) on the lower resistor and calmly watch the signal at the output.
The current in the lower resistor is known.
The total voltage will be equal to: current * total divider resistance: 4.21 μA * 95 MΩ = 400 V with rounding errors.
You forgot to take into account that the divider became not ten times after taking into account the resistance of the measuring device.
Multiply the voltage across the resistor by 19.

Answer 2

At 5 MΩ, you have 21V drops, and the rest of the resistors are 10MΩ each, and 42V each will fall on them!
So not 21x10, but 21 + 42x9!