Yes. The measurement cannot take place without applying a voltage, and there is always a chance that voltage will remain in the cap at the instant it is disconnected.Is there any chance of leaving them charged ??
Not true! The cap will have whatever voltage was applied at the instant of disconnection. I had a friend in grade school that would touch a capacitor - probably a big, high voltage ceramic from an old TV - to the wall outlet and then zap his friends with it. It would deliver a jolt most of the time.You will not leave a capacitor charged with alternating voltage alone so a test circuit the only applies ac will not normally leave a capacitor charged.
Do you not consider this pretty irresponsible behaviour? He was your friend?I had a friend in grade school that would touch a capacitor - probably a big, high voltage ceramic from an old TV - to the wall outlet and then zap his friends with it. It would deliver a jolt most of the time.
Indeed it would.But the AC voltage applied by a meter would be much smaller.
If 1 volt is applied to a 10μF capacitor in series with a 2Ω resistor, how can about 1.5 volts ever appear across the capacitor? How can more than the applied voltage appear across the capacitor?A 1V signal at 1000Hz is applied to the CUT and a low value resistor in series.
The voltage across the resistor is read with an AC millivoltmeter calibrated in capacitance.
Now the reactance of 10 microfarads at 1kHz is 16Ω so a 2Ω resistor will have about 1 x 2/18 = 100 millivolts full scale across it, so the max possible voltage across the CUT is about 1.5 volts.
You are being sloppy. When you said 1 volt was applied you didn't say that you meant RMS, then later when you described the voltage across the capacitor, you didn't say that you had changed to peak volts. If you are going to change from RMS to peak in midstream you should inform your readers.1volt RMS has a peak of nearly 1.5 volts.
I agree. But I think this is also somewhat uncharacteristic of studiot, so it was probably akin to a typo.You are being sloppy. When you said 1 volt was applied you didn't say that you meant RMS, then later when you described the voltage across the capacitor, you didn't say that you had changed to peak volts. If you are going to change from RMS to peak in midstream you should inform your readers.
How do you get the 12.4?Furthermore, you give an example calculation that makes it appear as if a capacitor having a reactance of 16 ohms in series with a resistor of 2 ohms forms a voltage divider with 2/18 of the applied voltage appearing across the resistor. But actually the voltage across the resistor would be 1 x 1/12.4
Yes, I pressed the wrong button on the calculator.How do you get the 12.4?
That would be 2/24.8 implying that the total impedance has a magnitude of 24.8Ω. But the most it can be is (16+2)Ω=18Ω and the least it can be is (16-2)Ω=14Ω. Furthermore, since we know that one is pure real and the other is pure reactive, we know that the lower limit is the larger of the two, so 16Ω.
Since 16Ω >> 2Ω, the magnitude of the impedance is going to be just a bit over 16Ω, so the divider would be close to 1/8.
This is assuming the 16Ω is correct (and it sounds right since 1/(2τ) is 0.159 and everything else is a simple decimal point shift). Oh, hell, might as well do it. 1000*10u = 1/100, so since that's in the denominator it makes it 15.9Ω. Yep, I agree.