Electrical – How to a multimeter measure current while being parallel to the element

Assumed the current you want to measure is indeed about 10uA and you feel OK with a burden voltage of about 1V and your multimeter has a voltage range of about 1V, then all you have to do is put a resistor is series with the FET you want to test. Most multimeters have a voltage range where 1V is somewhat in the middle of the range. In this case:

$$R_{series} = \dfrac{U_{burden}}{I_{expected}} = \dfrac{1\text{V}}{10\mu\text{A}}=100\text{k}\Omega$$

and measure the voltage across the resistor. The voltage measured is a pretty good indication of the current through the resistor, but it is important to check your meter's impedance. For good results it should be at least 10 times higher than the resistor. In this case the error will be better than 10% (\$\frac{100\text{k}\cdot 100\%}{1\text{M}}=10\%\$).

The leakage current will flow through the resistor and Ohm says:

$$I=\dfrac{U}{R}$$

So the voltage across the resistor is proportional with the current through it. R is known, U is measured, so you can calculate current I.

Two things you can do:

1. Measure the current flowing through the resistor, then calculate the voltage drop across it (Ohms Law).
2. Create a Voltage Follower circuit with a much higher input impedance and measure the output voltage of that.

However, by what you're describing, I'm not sure you're using the resistor right in the first place. Please expand your question with a schematic and a full reasoning behind the use of the resistor.