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Subject: Re: Subject: Re: Some H340 impressions

Re: Subject: Re: Some H340 impressions

From: Jerry Van Baren <>
Date: Thu, 05 Jan 2006 14:30:42 -0500

Jerry Van Baren wrote:
> Couple of counter-points:
> * 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
  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 non-resistive loads because most non-resistive loads are also not

Received on 2006-01-05

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