The resistor is effectively shorting your meter. Briefly connect the capacitors to 120 Volts and measure the current.
Then calculate the capacitance. The error caused by the resistor will be negligible on a starting capacitor that may be manufactured to a 20% tolerance.
Be cautious checking unknown capacitors that may be shorted internally. Testing the capacitor in series with a heating appliance of about 1000 Watts may be safer. Be sure to check the voltage across the capacitor as well as the current. Never assume.
Bill
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"Why not the best?"
Jimmy Carter
Safety. The stored charge may have enough energy to be dangerous or even lethal.
In Canada the Electrical code requires a resistor to reduce the terminal voltage to 50 Volts or less within one minute or less.
Bill
--------------------
"Why not the best?"
Jimmy Carter
You could do it with no risk if you just use the discharge resistor in an RC Circuit where the capacitor is the C. No extra connections, except for the measurement strips.
Use the old discharge formula u = U[sub]pk[/sub]*(1-exp(t/RC)) which says that voltage drops 63.2 % in one RC time constant.
For AC capacitors, you can go with low voltage and there will be no grenade problems and no non-linear problems like you could have with electrolytics.
Charge to 15.8 V DC (for reasons explained later). Remove DC and with the voltage drop. Measure time to 5.8 V and there is your RC time constant.
Value of R should be on the component, or you could measure it. It takes some time Before you get a valid Reading because the capacitor needs to be charge at the same time. Picture shows how to do it.
Cursor 1 is a little "too early", thence the 320 ms dt. Better adjustment would probably give a more correct answer.
Why 15.8 V? Simply because the voltage drop in one time constant then is 10 V. Any other value could, of course, be used. But needs more calculation.
Alternatively you could use a soldering iron to remove the resistor from one terminal, test the capacitor using your meter, then re-attach the resistor.
