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- Path: sparky!uunet!usc!news.cerf.net!proton!galaxy!watserv.ucr.edu
- From: judson@watserv.ucr.edu (Mike Judson)
- Newsgroups: misc.consumers.house
- Subject: Re: Power and current
- Message-ID: <25427@galaxy.ucr.edu>
- Date: 26 Jan 93 01:39:08 GMT
- References: <C1Bq16.2EA@math.okstate.edu>
- Sender: news@galaxy.ucr.edu
- Reply-To: judson@watserv.ucr.edu
- Lines: 18
- Nntp-Posting-Host: watmon.ucr.edu
-
- In article <C1Bq16.2EA@math.okstate.edu> dunne@math.okstate.edu (Ed Dunne)
- writes:
- > If I have an appliance with a heating element rated at 850 watts, how
- > much current (in amps) is it going to draw in a standard 110V circuit?
-
-
- P*V=I
- thus,
- (110 V) * (.850 Kw) = 93.5 A
-
- --
- You know what would make a good story? Something about a clown
- who makes people happy, but inside he's real sad. Also, he has
- severe diarrhea.
-
- Jack Handy
-
- judson@watserv.ucr.edu
-