<html><body><div style="font-family: Arial; font-size: 12pt; color: #000000"><div>Ah, yeah. Didn't catch that (although it's probably a single 6V battery with 4 internal 1.5V cells). </div><div><br></div><div>1.5V would not be sufficient to power a radio signal suitable for even a portable phone.<br></div><div><br></div><div>Bob<br></div><div><br></div><hr id="zwchr"><div style="color:#000;font-weight:normal;font-style:normal;text-decoration:none;font-family:Helvetica,Arial,sans-serif;font-size:12pt;"><br><div><br></div>
<div><div style="font-family: Calibri,sans-serif; font-size: 11pt;">Hi<br>Presumably there are 4 1.5 Volt cells in the phone that are connected in series. 4 * 1.5 = 6, voila.<br><div><br></div>Per<br><div><br></div></div></div><hr><div><br></div>
The mAH is a rating of how much current the battery can supply for
how long:<br>
<br>
'milli Amp Hour'<br>
<br>
Get the highest rated you can find/afford; though the rating on the
Chinese knock-offs--I think most of them are--seems to be
overstated. The load--how much current will be used--is determined
by the circuit on the phone.<br>
<br>
The statement on the charger makes no sense; applying 6V to a 1.5V
battery would cook it in short order (putting 48V on a car battery
would boil the electrolyte; creating a major mess and possibly an
explosion). I suspect it's a nominal rating; i.e. it will put out
150mAH at 6v, but only needs about, oh, 2V or less to charge the
batteries and would supply 450mAH at that voltage.<br>
<br>
E (voltage) = I (current) * R (resistance). Plug in any 2 known
values--and use a little algebra if necessary--and you'll get the
third.<br>
<br>
Bob<br>
<br>
<br>
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