
Rockbox mail archiveSubject: Re: Subject: Re: Some H340 impressionsRe: Subject: Re: Some H340 impressions
From: Jerry Van Baren <gerald.vanbaren_at_smithsaerospace.com>
Date: Thu, 05 Jan 2006 14:30:42 0500 Jerry Van Baren wrote: > Couple of counterpoints: > * Power consumption is current times voltage: > P = IE > If you are using a resistor: > E = IR, I = E/R > Therefore, _in a resistor_, power is quadratically related to current > (given a constant voltage) or voltage (given a contant current): > P = I^2 R > P = E^2 / R > If your piece of equipment isn't a resistor, the above may not be true > (but it tends to be quadratic). A perverse counter example is a zener > diode where the diode fixes the voltage (E) and thus the power > dissipated by the zener is linearly related to the current through the > diode. Sorry to reply to myself, but the above discussion of power dissipaged by a resistor is poor. Since resistors are linear devices, in order to double the voltage through a given resistor, you must also double the current (or vice versa). Since you are doubling both voltage and current, the power is quadratic: P = I x E 4P = 2I x 2E 9P = 3I x 3E This is only true for a constant resistance. This generally is not true for nonresistive loads because most nonresistive loads are also not linear. gvb Received on 20060105 Page was last modified "Sat May 23 08:12:40 2020" The Rockbox Crew 