How do you set amperage in a simple circuit?
We take current measurements by connecting an ammeter (or a multimeter with the leads and range set to mA or A) in series with the load - not across the load.







simulate this circuit &ndash; Schematic created using CircuitLab

Figure 1. AM1 shows the correct way to measure the current. Typically the voltage will be measured with respect to ground (usually battery negative) so you could measure V1, V2 and V3 to see increasing voltages with respect to ground.

Note that we can also calculate the current if we know the voltage across a resistor of known value. In this case you have a 1 k&Omega; resistor and the voltage drop across that will be 1 V/mA. You should take a current reading (AM1) and a voltage reading (VM1) taking care to select the right probe sockets and range select before powering on the circuit. The current reading and calculated current should agree.



Figure 2. IV curves for various colours of LED. Source: LED IV curves.

To figure out the resistance required for an LED you can use the graph of Figure 2. Select the current you require, trace a line across to the colour of your LED and drop down to the voltage line to see what the forward-voltage, Vf of the LED will be. You seem to be using 0.002 A or 2 mA for yours. 

I'll choose 10 mA as it will be easier to read on the graph and I'm going to assume you have a green LED. Tracing across from 10 mA on the vertical axis we see that the grid line crosses the green curve at 2.0 V. That means your voltage-dropping resistors need to drop 1 V (since you have a 3 V supply). We then calculate \$ R = \frac V I = \frac 1 {0.01} = 100 \ \Omega \$.

Notice that at 2 mA the voltage will still be close to 2.0 V so you can revise the calculation to give \$ R = \frac V I = \frac 1 {0.002} = 500 \ \Omega \$. There's no need to be too exact with this circuit as your battery voltage is going to vary with state of charge.