I've been working through some of the exercises in the new third edition of "The Art of Electronic", and exercise 1.9 is the first one that I'm not sure of my answer. Here's the exercise:

"The very high internal resistance of digital multimeters, 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 M ohm "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 2 V dc range has 10 M ohm internal resistance, and you have available a dc source of +10V. How can you use what you've got to measure accurately the leakage resistance?"

My thought was to connect everything in series, but I have two problems with this solution:

1. For the leakage resistance varying +-10% around 1000 Mohm, the voltage measured by the DMM only changes by about 20mV (90mV for 1100 M ohm and 110mV for 900 M ohm). Can the DMM detect this small of a voltage change accurately?

2. The formula for the leakage resistance is 10M*((10/VDMM)-1), where the 10M multiplier is the DMM's internal resistance. This means that accurately calculating the leakage resistance depends on how accurately you know the DMM's internal resistance. How accurately do you typically know a DMM's internal resistance?

Is there a better way to solve this problem? Thanks in advance for any help you can give me.