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?

 

[Fred Abse]

On Wed, 09 Feb 2005 21:31:19 +0000, R.Lewis wrote:

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.

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

 

[R.Lewis]

"Fred Abse" <excretatauris@cerebrumconfus.it> wrote in message
newsan.2005.02.10.20.56.38.419914@cerebrumconfus.it...

On Wed, 09 Feb 2005 21:31:19 +0000, R.Lewis wrote:

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.

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

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]

Fred Abse wrote:

On Wed, 09 Feb 2005 21:31:19 +0000, R.Lewis wrote:

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.

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

 

[Fred Abse]

On Fri, 11 Feb 2005 00:36:49 +0000, R.Lewis wrote:

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

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

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

 

[Fred Abse]

On Thu, 10 Feb 2005 18:29:00 -0800, redbelly wrote:

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.

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

 

[Robert Nichols]

In article <62b9dcf3.0502091225.1660366e@posting.google.com>,
zooeb <zooeb@libero.it> 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
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.

--
Bob Nichols AT comcast.net I am "rnichols42"

 

[DM]

Robert Nichols wrote:

In article <62b9dcf3.0502091225.1660366e@posting.google.com>,
zooeb <zooeb@libero.it> 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
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.
--
DM

 

[John Fields]

On Thu, 10 Feb 2005 11:49:35 -0800, DM <dm@nospam.vfemail.net> wrote:

Robert Nichols wrote:
In article <62b9dcf3.0502091225.1660366e@posting.google.com>,
zooeb <zooeb@libero.it> 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
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.

---
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:


E˛ 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


1.333˛V
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 E2˛/0.5R = 2
watts, and so will R1.
---

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

 

[DM]

John Fields wrote:

On Thu, 10 Feb 2005 11:49:35 -0800, DM <dm@nospam.vfemail.net> wrote:


Robert Nichols wrote:

In article <62b9dcf3.0502091225.1660366e@posting.google.com>,
zooeb <zooeb@libero.it> 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
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.


---
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:


E˛ 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


1.333˛V
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 E2˛/0.5R = 2
watts, and so will R1.
---


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.

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

 

[John Fields]

On 10 Feb 2005 18:29:00 -0800, "redbelly" <redbelly98@yahoo.com>
wrote:

Fred Abse wrote:
On Wed, 09 Feb 2005 21:31:19 +0000, R.Lewis wrote:

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.


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

---
However, since E = IR and the load resistance is zero ohms, the
voltage dropped across it must be zero volts. Therefore, the power
dissipated in the load will be E˛/R = 0/0 = ???

--
John Fields

 

[R.Lewis]

"Fred Abse" <excretatauris@cerebrumconfus.it> wrote in message
newsan.2005.02.11.21.30.08.834051@cerebrumconfus.it...

On Thu, 10 Feb 2005 18:29:00 -0800, redbelly wrote:


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.

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

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]

On Sat, 12 Feb 2005 14:06:34 +0000, R.Lewis wrote:

"Fred Abse" <excretatauris@cerebrumconfus.it> wrote in message
newsan.2005.02.11.21.30.08.834051@cerebrumconfus.it...
[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!


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