Re: cooler

From: ludvil1.edu@mail.cez.cz
Date: 04. 02. 2002, 11:54 CET

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> Prikon P vetraku je uveden na stitku nebo v katalogu. Rekneme,
> ze prikon je 1.92 W. Chces snizit napajeci napeti U1=12 V napr.
> na polovinu, tedy o napeti U2=6 V. Potrebny odpor potom bude
> R=U2/(P/U1) neboli 6/(1.92/12) = 6/0.16 = 37.5 [Ohm].
> pako Paco.

Toto je blbost, odpor (cinny) vetraku je R=U^2/P=144/1,92=75 Ohmu
Pokud dam do serie s timto vetrakem (jehoz odpor jsme spocitali na 75 Ohmu)
dalsi odpor 75 Ohmu, rozdeli se napeti presne na polovinu, tj. 6V bude na
ventilatoru a 6V na odporu.
Ten odpor staci kdyz bude na 0,5W, protoze P=U^2/R=36/75=0,48W
Lubos

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