maximum power to load?

[zooeb]

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?

 

[R.Lewis]

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?

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.

 

[Robert Nichols]

: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
revious 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.

 

[DM]

: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
revious 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.

 

[John Fields]

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.

---
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:



2V * 1R
E2 = ----------- = 1.333V
0.5R + 1R


and the power the load will dissipate will be


1.333²V
P = --------- = 1.333W
1R

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.
---

 

[Fred Abse]

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.

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.

 

[DM]

---
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:



2V * 1R
E2 = ----------- = 1.333V
0.5R + 1R


and the power the load will dissipate will be


1.333²V
P = --------- = 1.333W
1R

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.

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.

 

[R.Lewis]

infinite.

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.

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.

 

[redbelly]

infinite.

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.

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

 

[John Fields]

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

 

[Fred Abse]

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

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

Anything divided by zero is infinite.

Except zero.

 

[Fred Abse]

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

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

 

[R.Lewis]

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


Except zero.

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.

 

[Fred Abse]

[quoted text muted]

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!