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Capacitor Charging & Discharging

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Shown on eduMation website at http://msdaif.googlepages.com/... practical demonstration that shows the chraging and discharging modes of capacitors. A 160 mF capacitor was used. The video was directed and shot by Ebtisam Anzoor and presented by Mustafa Daif. We were assisted by our trainees Husain Atiya and Moh'd Abdulla

Channel: Howto & Style
Uploaded: November 30, 1999 at 12:00 am
Author: msdaif

Length: 08:52
Rating: 4.65
Views: 25742

Tags: analysis  Anzoor  Atiya  capacitor  charging  circuit  condensor  Daif  discharging  Ebtisam  edumation  electricity  Husain  Mustafa  

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Video Comments

ljheard (November 30, 1999 at 12:00 am)
How do you discharge a HID capacitor?
xXYewTewbXx (November 30, 1999 at 12:00 am)
Calculate the capacitor current you do supply voltage minus the voltage divided by the resistance i think.
Deshpande112233 (November 30, 1999 at 12:00 am)
i apriciat the effort, and it is useful demonstration for students. the language is very simple and lucid .thanks
xXYewTewbXx (November 30, 1999 at 12:00 am)
a capacitor may also explode if the insulation dries up over time.
drewadc (November 30, 1999 at 12:00 am)
No, whenever a capacitor is charged, it acts like an open circuit. This means that no current will flow in the circuit when a capacitor is charged.
korsez (November 30, 1999 at 12:00 am)
if i conect lamp between capacitor - and - terminal will it light brighter when it is charged?
frizspin175 (November 30, 1999 at 12:00 am)
all capacitors are AC, including this one. the 2nd demonstration of the lamp explains that the capacitor has an equal # of electrons on both leads so it stops supplying power to the lamp. capacitors work with AC current because the current alternates, so there is a constant push and pull of electrons between the two leads on the capacitor so it doesn't become neutral.
zomgerln (November 30, 1999 at 12:00 am)
it is fully charged..K, UBER THAT HEAVY AND STORM THE BLUE TEAM!
redSHIFT69 (November 30, 1999 at 12:00 am)
great MCAT review! thanks :)
msdaif (November 30, 1999 at 12:00 am)
For a load with a resistance R, a power supply with e.m.f of E, the current is I=E/R. For the electromagnet, you need to measure its resistance , say RL, and energizing current, say IL. Then,IL=E/RL. You could send me an email explaining what you want to do and I can help you with the design, free of charge.

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