[apa]

Your mistake is in assuming that the 4.8v is still across the load when
the load has changed from 12 ohms to 3 ohms. When the load changes to 3
ohms in this case, the voltage across that load will change to 3v.



wrote:
I'm trying to learn a little more about basic electronics and am using
Success In Electronics by Tom Duncan. There's a diagram, which I've
posted at http://homepage.eircom.net/~gerfmcc/circuit.html.
It just shows a battery, 12 ohm resistor, and a switch in series, with
a voltmeter connected across the battery terminals. The text reads:

In the figure, V is a high resistance voltmeter. It reads 6V when the
switch is open and 4.8V when the switch is closed.
a) Calculate the internal resistance of the battery.

That part is okay, I got this:

The current in the circuit -
I = V / R
I = 4.8 / 12
I = .4A
Then -
e.m.f = useful voltage + lost voltage
6 = 4.8 + I * r
1.2 = .4r

So the internal resistance of the battery is 3 ohms.


--------------------------------------------------------------------------------
b) What value of resistor must replace the 12 ohm resistor to give
maximum power output? What will be that power output?

I thought it was necessary to match the load resistance to the source
resistance for maximum power transfer, so a 3 ohm resistor would be
required here. But that's probably wrong, because the answer at the
back of the book says Power output is 3 watts.

Thanks for any help.