From: Mark Lawton (markslawton@hotmail.com)
Date: Sat May 25 2002 - 09:03:10 PDT
Date: Sat, 25 May 2002 09:03:10 -0700 Subject: re: battery questions From: Mark Lawton <markslawton@hotmail.com> Message-ID: <B91503CE.23F0%markslawton@hotmail.com>
Geoff,
The wires do indeed have some resistance. The battery also is not an ideal
voltage source. Ideal would mean that it has no internal resistance and the
voltage is constant.
Rather the battery has internal resistance which causes a voltage drop
inside of the battery itself. You can think of the battery as an ideal
voltage source in series with a resistor.
So, there can not be an instantaneous discharge because the current is
limited by the internal resistance and the resistance of the wires.
Also, as the battery discharges the stored charge becomes depleted. This
decreases the voltage which in turn decreases the rate of discharge. It
acts like a capacitor and the voltage = Vinitial * e^(-t/constant)
The second question is the same as the first. The bulb does not discharge
the battery immediately because the bulb has resistance and the current is
limited by ohms law V=IR.
-Mark
>
> One of my brilliant students came up with some questions that I don't
> understand.
>
> If you directly connect the terminals of a battery with a wire, it
> will start to shortcircuit and heat up. Why, though, doesn't the
> battery immediately discharge when it's shortcircuited? Is it because
> the wire has a resistance and therefore can only carry a certain
> amount of amperage at once?
>
> Here's another question: if you hook up a battery to a small
> lightbulb, it can power the lightbulb for hours, until the battery
> runs dead. Why doesn't the battery immediately "try" to discharge in
> a massive burst of current, and burn out either the wires or the
> lightbulb filament?
>
> In other words, it seems like something is controlling the rate of
> discharge of the battery.
>
> Both these questions came up through studying electrochemistry, not
> through the physics side of e/m.
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