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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.

[Don Pearce]

On Thu, 20 Jul 2006 18:26:57 GMT, 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.

The source resistance is three ohms, so as you say, you must make the
load three ohms as well. The voltage across the load is now three
volts (the six open circuit volts have been divided in half by the
equal value resistors). The power in the load is thus three squared
over three (V squared / r), so the power is indeed three watts.

d

--
Pearce Consulting
http://www.pearce.uk.com

[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.