Practical shunt metering.

Assuming you're going to get the math from quick and ready programs, here's a practical overview that should help you make it stick.

E=IR If ten amps are flowing through one ohm of resistance, then the voltage drop across that resistor will be 10 volts. 10 times l equals 10.

A shunt of one ohm may be too large for accurate readings at 12 volts, let's say, so a practical shunt would be .1 ohms. At that, ten amps would drop one volt. (.01 ohms would be even more accurate, interfering even less with the circuit. But let's stay with .1 ohm.)

From there you can set up the microamp meter to read full scale at one volt. Again, at one volt through what resistance will cause one microamp to flow? 1 divided by .000001 equals 1,000,000 ohms.

So. a one megohm resistor in series with the negligible resistance of a microamp meter will read full scale at one volt.

If one amp is flowing through the shunt, the voltage drop will be 1 times .1. .1 volt diverted through the metering resistor will read:
.1 divided by 1,000,000 or .0000001 amp. One tenth of a micro-amp.

From this you can see that the metering resistor should be at least 20 times larger than the shunt so as not to divert too much current. In this example, it should be extremely accurate. Except that .1 ohm will be in series with the load and will drop the total current of the circuit proportionately. Still...

The simple answer to your question is: Ohm's Law.