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in a simple electric circuit, in which you have a DC voltage generator, an internal resistence and the resistence of the load, why the maximum power that the generator is able to tranfer to load is when load resistence is equal to internal resistence? I try to think about it: if internal resistence is zero, power developed by load is V*I, where V is the voltage generator and I=V/RL; then if the internal resistence goes up, the power developed by load is V*I, where V minor than the voltage generator and I=V/(RI+RL), which is less then the previous current. So, why in the first case power isn't max?

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If the internal resistance of the source is zero ohms and the load resitance is equal to the internal resistance then the power is infinite. Difficult to get any higher than that.

Otherwise do some sums. If I have a 12v battery with an internal resistance of 6ohms and I connect a

6ohm load resistor to it the power dissipated in the load is 6 watts. Just try any other value of load resistance and see if you can get its dissipation above 6 watts.
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In article , zooeb wrote: :in a simple electric circuit, in which you have a DC voltage :generator, an internal resistence and the resistence of the load, why :the maximum power that the generator is able to tranfer to load is :when load resistence is equal to internal resistence? I try to think :about it: if internal resistence is zero, power developed by load is :V*I, where V is the voltage generator and I=V/RL; then if the internal :resistence goes up, the power developed by load is V*I, where V minor :than the voltage generator and I=V/(RI+RL), which is less then the :previous current. So, why in the first case power isn't max?

Matching the load resistance to the generator resistance for maximum power transfer applies when you have a generator with a known, fixed internal resistance. If you change the characteristics of the generator by lowering its internal resistance, of course you can deliver more power to the load.

```--
Bob Nichols         AT comcast.net I am "rnichols42"```
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But still, if you change the internal resistance of the generator you're going to decrease the power at the load, until the load resistance changes to match the internal resistance of the source.

You can increase the voltage to the load by making its resistance larger than the resistance of the source but you will only get max power transfer when they are equal.

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DM```
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```--
That's not true.  Consider the generator to be a voltage source with a
resistance in series with it and the load to be a resistance to```
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No. The current is infinite. The power delivered to the load is zero, since power is I^2R and R is zero. Anything multiplied by zero is zero.

A zero ohm load is called a short circuit.

```--
Then there's duct tape ...
(Garrison Keillor)```
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You are correct. I did all my calculations based on a constant source resistance and varying the load resistance, if I'd reversed my values I would have seen that.

```--
DM```
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infinite.

Yep. You are certainly corrent that the power dissipated in the resistor would be zero.

A zero ohm load is just that. An unwanted zero ohm load is a f*****g short.

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infinite.

zero.

No Fred. The power is infinite since power is V^2 / R and R is zero. Anything divided by zero is infinite.

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```--
However, since E = IR and the load resistance is zero ohms, the
voltage dropped across it must be zero volts.  Therefore, the power```
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V=IR, hence if R is zero, V is zero.

Except zero.

```--
Then there's duct tape ...
(Garrison Keillor)```
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But not the converse. Sometimes we *want* a short.

```--
Then there's duct tape ...
(Garrison Keillor)```
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V=I*R if R=0 then V=0, i.e. V=R the power deveoped in the resistor is V^2/R but since V=R this becomes (V^2/V=) V or conversely (R^2/R=) R. The watts developed is thus V volts. This conclusively proves that a resistor of zero ohms has a resistance of N volts, where N is the source voltage.

The power developed in the resistor may also be expressed as I^2R. Now since R=V this may be written as I^2V. Since the power is also expressed as V^2/V, equating these two, V^2/V=I^2V which may be re-aranged as I^2=V^2/(V*V). I is thus SQRT(1) which is one. i.e. the current flowing through the resistor equals1 volt independant of the source voltage (provided the source resistance is equal to the supply volts in volts).

I hope this explains it fully for you.

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ROFL!

```--
Then there's duct tape ...
(Garrison Keillor)```

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