Electrical – A DMM low current measurement question

current measurementmultimeter

I'm stuck with a question when reading a textbook:

The question is a situation as follows:


The very high internal resistance of digital multi-meters, in their voltage measuring ranges, can be used to measure extremely low currents (even though the DMM may not offer a low current range explicitly).

Suppose, for example, you want to measure the small current that flows through a 1000 Mohm "leakage" resistance (that term is used to describe a small current that ideally should be absent entirely, for example through the insulation of an underground cable).

You have available a standard DMM, whose 2V DC range has 10 Mohm internal resistance, and you have available a dc source of +10 V.

How can you use what you've got to measure accurately the leakage resistance?


Anyone can understand this question and have an idea?

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Best Answer

This is how I solved; I used voltage divider: 10/(1000M+10M) = Vddm/10M so Vdd = (10 x 10 M) / 1010M = 0.099V

I = (10V - 0.099V)/1000M = 9.901x10^-9A