If your multimeter does not
measure capacitance, you could measure the capacitive reactance using s 60 Hz low voltage transformer and a resistive voltage divider with your multimeter. A 165 pF capacitor has 16 MΩ reactance at 60 Hz. A 1650 pF capacitor has 1.6 MΩ reactance at 60 Hz. If you connect ten of the "unknown" capacitors in parallel, these reactances will drop to 1.6 MΩ for ten 165 pF capacitors in parallel and to 160 kΩ for ten 1650 pF capacitors in parallel. That gets the capacitive reactance low enough that you can construct a resistive/capacitive voltage divider without the multimeter presenting an excessively low impedance during voltage measurements performed on the divider components.
So, the procedure is this: Connect a 100 kΩ resistor in series with the ten paralleled capacitors and apply low-voltage AC across the series combination. Measure the applied AC voltage and the voltage across the 100 kΩ resistor. If the "unknown" capacitor is 1650 pF the parallel combination will drop about 36% of the AC voltage across the resistor and the remaining 64% across the paralleled capacitors. If instead of 1650 pF capacitors your "unknown" capacitors are only 165 pF, then most of the AC excitation voltage will appear across the paralleled capacitors instead of across the resistor. The problem is to measure this without excessively loading the resistor and obtaining unreliable values.
A typical digital multimeter has an input impedance of one to ten megohms, so if you use the multimeter to measure the AC voltage across the 100 kΩ resistor, it will present an impedance slightly less than 100 kΩ (1 MΩ in parallel with 100 kΩ is 90909 ohms). When you measure the voltage across the 100 kΩ resistor, it will be about 36% of the AC excitation voltage if the ten "unknown" capacitors are each 1650 pF, presenting a reactance of 160 kΩ at 60 Hz AC. In other words you have a voltage divider with 90.9 kΩ resistance (which includes the multimeter input impedance) in series with 160 kΩ of capacitive reactance. The divider ratio (across the resistor) is 90.9/(160+90,9) = 0.36.
If the ten "unknown" capacitors are each 165 pF, the voltage across the 100 kΩ resistor will be much less than the AC excitation voltage and most of the excitation voltage will appear across the ten paralleled 165 pF capacitors. It should be fairly easy to determine which of these two instances is the correct one.
