Should i measure with a true RMS multimeter? - Page 1
Forum section: electronics / beginners

author: robsims
author_rank: Frequent Contributor
timestamp: November 06, 2022, 01:50:43 pm
content: Hi, i have connected a polypropylene film 408 nF capacitor in series with a 390 ohm resistor on mains AC . Mains AC is 126 volts 60 Hz.I calculated that the voltage drop over the capacitor should be 110.5 Volts and the voltage drop over the resistor should be 6.6 VoltsWhen measuring with my average meter i get a voltage dop of 126.2 Volts over the capacitor, which is higher than expected and a voltage drop over the resistor of 7.46 Volts. Should i measure with a true RMS meter to get better results?

author: bdunham7
author_rank: Super Contributor
timestamp: November 06, 2022, 02:32:32 pm
content: Quote from: robsims on November 06, 2022, 01:50:43 pmNo, you should show us your work on how you calculated what those voltages would be.

author: TimFox
author_rank: Super Contributor
timestamp: November 06, 2022, 02:46:27 pm
content: Remember to display "j" in your calculation of the the voltage divider.

author: wasedadoc
author_rank: Super Contributor
timestamp: November 06, 2022, 03:23:02 pm
content: For sine wave AC a true RMS meter and a meter that is not true RMS should give the same reading.

author: Grandchuck
author_rank: Super Contributor
timestamp: November 06, 2022, 05:24:57 pm
content: All the replies are correct and so are your readings (within reasonable error margins). How are you calculating impedance?

author: wasedadoc
author_rank: Super Contributor
timestamp: November 06, 2022, 05:28:48 pm
content: For a visual solution use a right angle triangle. The two sides forming the right angle are the resistance of the resistor and the impedance of the capacitor. Pythagoras gives you the hypotenuse which is the impedance of the two in series. From that you get the current. Multiply by the resistor to get its voltage drop. Ditto for the capacitor.Gives me 7.54 and 125.77 which are more in accord with your measured values.

author: robsims
author_rank: Frequent Contributor
timestamp: November 06, 2022, 05:58:25 pm
content: Quote from: Grandchuck on November 06, 2022, 05:24:57 pmI calculated impedance of the capacitor with an online capacitive reactance calculator

author: bdunham7
author_rank: Super Contributor
timestamp: November 06, 2022, 06:06:02 pm
content: Quote from: robsims on November 06, 2022, 05:58:25 pmAs I'm sure will be pointed out shortly, that doesn't work because the voltages across the resistor and capacitor are ideally 90 degrees out of phase. However, can you share the actual numbers you obtained and used? The problem is that your two calculated voltages don't even add up to the total voltage.

author: wasedadoc
author_rank: Super Contributor
timestamp: November 06, 2022, 06:06:10 pm
content: Quote from: robsims on November 06, 2022, 05:58:25 pmBut how did you combine that with the resistor to get the total series impedance?

author: wasedadoc
author_rank: Super Contributor
timestamp: November 06, 2022, 06:13:45 pm
content: Quote from: bdunham7 on November 06, 2022, 06:06:02 pmMaybe that is just badly worded. That they don't even add up to the total voltage is not an indicator that something is amiss unless "even" is intended to convey the meaning "at least".

author: bdunham7
author_rank: Super Contributor
timestamp: November 06, 2022, 06:21:53 pm
content: Quote from: wasedadoc on November 06, 2022, 06:13:45 pmPerhaps I've used an idiom that is not universally recognized? Yes, 'don't even add up' as I've written it would imply that the minimum that two voltages across components in series would add up to would be the total voltage across both of them, as is the case when both voltages are perfectly in phase. So before getting into how he can understand how the voltages should add up in his example, I wanted to see his math using whatever incorrect principles he applied.