|
|
> I thought the voltage of a battery only decreases if you try to draw
> current faster than the chemical processes inside the battery can restore
> it? (I.e., it's a flaw parculiar to chemical batteries.)
No, it decreases the instant start to draw any current. Maybe trying to
teach you about complex AC circuit theory and power electronics was a bad
starting point :-)
When you connect a resistance, R, to a battery, the circuit you have made is
identical to a true, fixed voltage source (say 12V), with two resistances
connected in series. One of the resistances is R, the one you connected,
the other, called the internal resistance, Ri, is representing the resistive
parts of the battery that you have no control over.
You can analyse this circuit easily. You have a voltage, V of say 12 V, and
a combined resistance of R+Ri. So the current is found by Ohm's Law, and is
equal to 12 / (R + Ri).
Now that you know the current in the circuit, you can work out the voltage
across each resistor, again using Ohm's law.
For the resistor you just connected, R, the voltage across it will be given
by I * R. You know I because we just calculated it, so the voltage is 12*R
/ (R + Ri).
If Ri is very small compared to R, then this is approximately just 12 volts.
But when R gets small enough the voltage can no longer be approximated by 12
volts. And when R gets very much smaller than Ri, the voltage essentially
becomes zero.
If the voltage R *is* zero, ie you have connected a thick wire or
super-conductor between the terminals, then the voltage will be given by
12*0/(0+Ri), ie zero volts. *But* the current is still defined by 12 /
(R+Ri), so that simply becomes 12/Ri. This is the maximum current a battery
can deliver.
This explains why on old cars the radio switches off momentarily as you
start your car - due to the huge current drawn from the starter motor the
battery voltage drops, often to as low as 6 or 8 V. Also when charging the
battery the opposite happens, and the voltage often goes up to 13 or 14
volts.
I'll leave it as an exercise to work out the maximum power the battery can
output...
Post a reply to this message
|
|