Electronic – Trouble understanding why multimeter gives different resistance readings

On the 10A range, your multimeter probably won't have great accuracy with low current, but your measurement sounds reasonable. 0.17 to 0.21 A, or 170 to 210 mA.

Presumably your air pump is a small DC motor, which will have different current measurements at startup, and during running conditions with different loads.

To calculate resistance, you use Ohm's law:

$$R = E / I$$

If the voltage (E) is 3, and the current (I) is 0.17 A, then:

$$\frac{3}{0.17} = 17.6\Omega$$

Therefore, your resistance calculation seems correct. Just keep in mind that a DC motor will measure different resistances (currents) depending on what it is doing, due to induction/reactance.

As you've discovered, an electric motor is not well modeled as a resistor, and as such doesn't obey Ohm's law.

A better model for a DC electric motor is there is some resistance in series with a variable voltage source.

Additionally, a battery has some internal resistance, which can be modeled as a series resistor*. A PC power supply also can use this same model, but the series resistance is likely to be smaller. The system then looks like:

^{simulate this circuit – Schematic created using CircuitLab}

We can explain why in the first case your measured voltage is less than the no-load battery voltage because we have a voltage divider. Doing some math,

\begin{align} V_{emf} = V_+ - I R_m\\ R_s = \frac{V_{bat} - V_+}{I} \end{align}

You measured \$R_m = 3.5 \Omega\$, \$I = 0.19 A\$, and \$V_+ = 2.9V\$, so \$V_{emf} = 2.24 V\$ and \$R_s = 1.47 \Omega\$.

In the second case, \$V_+ = 4.92V\$ and \$I = 0.28A\$. Thus: \$V_{emf} = 3.94 V\$ and \$R_s = 0.43 \Omega\$.

Notice that \$V_{emf}\$ is different between the two. This is because \$V_{emf}\$ is roughly linearly proportional to how fast the motor is spinning. You should have observed the motor spinning faster when hooked to the 5V supply.

Additionally, how multi-meters measure current is by introducing a series shunt resistance and measuring the voltage across this resistor. This further complicates the analysis, so the measured current and load voltage are not exactly correlated. It's more difficult to do this analysis, but is possible if you know the series shunt resistance. This is sometimes quoted as a "burden voltage" at a rated test current and you can use Ohm's law to recover the shunt resistance.

It is possible to reconstruct what the measured load voltage should be with just a single meter, but it requires more information on how \$V_{emf}\$ behaves which is beyond the scope of this answer.

If you set your meter to the largest current range this will use the smallest shunt resistance, you can minimize the impact of having the meter in series at the cost of losing a bit of accuracy.

*note: Batteries don't have a constant internal resistance, but this is a reasonable approximation. It depends on a ton of factors including but not limited to stored energy, temperature, and load.