[pure-silver] Re: Now this realy an Off Topic question..

  • From: "Richard Knoppow" <dickburk@xxxxxxxxxxxxx>
  • To: <pure-silver@xxxxxxxxxxxxx>
  • Date: Fri, 27 Nov 2009 12:48:58 -0800

----- Original Message ----- From: "Ralph W. Lambrecht" <info@xxxxxxxxxxxxxxxx>
To: <pure-silver@xxxxxxxxxxxxx>
Sent: Friday, November 27, 2009 8:51 AM
Subject: [pure-silver] Re: Now this realy an Off Topic question..

I'm not an Electrical Engineer, but I thought this can be roughly
calculated from the battery specification.

Watt = Volt * Amps

Consequently, if it is a 12-Volt battery and the bulb has a power of 24 Watt, it will draw 2 Amps. If the battery has a capacity of 60 Ah,
the bulb will burn 30 hours under ideal conditions.

Doesn't it work this way?


Ralph W. Lambrecht


That is the simplified principal, however real batteries don't work quite that way. The rate at which the current is drawn can affect how much of the rated capacity is available. Since high current increases internal battery heating it reduces the real capacity. Also, the battery internal resistance goes up as it is discharged so that the voltage drops and available maximum current also drops. In other words the ampere/hour ratings for automobile batteries tends to be rather optimistic. The main virtue of the lead-acid batteries commonly used for autos is the capability to deliveer very high currents for short periods. This is needed in starting the car. They will also tolerate more abuse than many other kinds of batteries although the life will be extended if kept properly charged. Usually, the intermittant charging batteries get in use is not the best for long life. Batteries fail for many reasons but the most common one is the accumulation of junk in the space below the plates eventually reaching the plates and shorting them. Battery charge condition can usually be determined by measuring the specific gravity of the electrolyte but the more accurate method is to measure the voltage drop under load.

Richard Knoppow
Los Angeles, CA, USA
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