maximum power to load?

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

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.

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.

--- 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 ground, and you'll have this:

E1 | R1 | +----E2 | R2 | GND

A simple voltage divider, where:

E1 is the generator's voltage source R1 is the generator resistance E2 is the generator's output voltage, and R2 is the load resistance

Now, if E1 = 2V R1 = 1R, and R2 = 1R

then

E1R2 2V * 1R E2 = ------- = --------- =1V R1+R2 1r + 1R

and the power being dissipated in R2 will be:

E² 1 P = ---- = --- = 1W R2 1

Now, if we change the generator's internal resistance by _lowering_ it to 0.5 ohm, the voltage across the load will _increase_ to:

and the power the load will dissipate will be

Which is _higher_ than the dissipation with the generator resistance at 1 ohm, but lower than it would be if the load resistance was equal to the generator resistance.

Since we know that if R1 and R2 are equal E2 has to be 1/2 of E1, then with R1 and R2 bothe equal to 0.5 ohms, R2 will dissipate E2²/0.5R = 2 watts, and so will R1.

---

-- John Fields

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.

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.

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.

infinite.

zero.

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

V=IR, hence if R is zero, V is zero.

Except zero.

But not the converse. Sometimes we *want* a short.

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.

ROFL!