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Amazing. People who KNOW they don't know answering questions with a straight face.
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To the answer:
The issue is how to most efficiently transfer *maximum* power into the load from a fixed impedance source. If there is a transmission line inolved, then that, too, must be impedance matched. When this is the case, the *maximum power* in the load will always be 50% of the generated power, so 50% efficiency. Always.
So how is it that your calculations show more power in the load when the impedance is not matched?
Because you calculated absolute power in the load and adjusted the source impedance down. Usually, it is the source that is fixed and you must impedance match the load. If you compare a fixed source to various loads, you will find that maximum power in the load occurs when Rl = Rg.
When you compare the efficiencies, you will see that the power in the impedance matched scenario is 50% of the generated power, and always will be.
In the unmatched example, the efficiency is 80%. Much better for power generation, right?
Except for the fact that as efficiency goes up, the magnitude of the power to the load decreases.
So, you still get maximum power to the load when impedances are matched, though not with the greatest efficiency.
Here's a graph of power vs. efficiency. See that max power is at 50% efficiency.
http://upload.wikimedia.org/wikipedia/co...
Let's use your 1 Ohm example:
E^2/Rt = 6050 and half of that is 3025. That's higher than your unmatched example.
Obviously, if you drop Rl below 1 Ohm, more power is disspated in the source.
If you go above 1 Ohm, the efficiency increases but the actual magnitude of the power drops.
Let's use 1.2 Ohms as an example:
E^2/Rt = 5500 and 83% of that is disspated in the load = 3000 Watts, which is 25 Watts less than impedance matched.
Nathan:
One should not compare the short circuit power with the quiescent power since the load is not part of the formula for maximum power generated when that is done.
In the impedance matched case, E^2/Rt = 1512.5, of which half (756.25) will be in the load and half will be dissipated in the generator. 50% efficiency, not 100%.
In the unmatched case: E^2/Rt = 2420, and 4/5 (80%) of that is in the load, not 64%.
Maximum power transfer and maximum efficiency are not the same things, as you can see.
And all of this only applies if the source impedance cannot be adjusted. If it can, then make it as low as possible for power to the load approaching 100% efficiency, but it won't be the maximum power.
You have 2 voltage sources with the same voltage, but the internal resistances are 1 and 4 ohms. Both sources will deliver maximum power to a load that has the same resistance as the respective internal resistance. However, the 1 ohm source will deliver much more power to a 1 ohm load than a 4 ohm source and a 4 ohm load.
Yes! In broadcast Engineering training, one of the basics is that a current source, (battery, generator, whatever), till ALWAYS have maximum power output when the load resistance is equal to the internal resistance of the source.
Want to "play" a little more? Figure it out with higher and LOWER resistance than the internal (source) resistance for the load. You will find that the power will peak when R-sub-l exactly equals R-sub-i If you get a different answer, your math is faulty. Sorry.
Good luck and keep studying.
God Bless.
For a generator the maximum developed power is Vg^2/Rg.
The maximum available power under matched condition (RL = Rg) is Vg^2/(4*Rg).
For generator A:
maximum developed power = 110^2/1 = 110^2 = 12,100 W
maximum available power = 12,100/(4*Rg) = 12,100/4 = 3,025 W
with a load, RL = 4:
IL = Vg/(Rg + RL) = 110/(1 + 4) = 22 A
Load power, PL = (IL^2)*RL = (22^2)*4 = 484*4 = 1,936 W
Efficiency = PL/(maximum available power) = 1,936/3,025 = 0.64 ... (64%)
For generator B:
maximum developed power = 110^2/4 = (110^2)/4 = 12,100/4 = 3,025 W
maximum available power = 12,100/(4*Rg) = 12,100/(4*4) = 12,100/16 = 756.25 W
with a load, RL = 4:
IL = Vg/(Rg + RL) = 110/(4 + 4) = 110/8 = 13.75 A
Load power, PL = (IL^2)*RL = 13.752^2)*4 = 189.0625*4 = 756.25 W
Efficiency = PL/(maximum available power) = 746.25/746.25 = 1 ... (100%)
Generator A delivers more power but is only 64% efficient with respect to available power.
Generator B delivers less power but is 100% efficient with respect to available power.
If generator A had a 1 Ohm load (matched condition Rg = RL) the load power, PL, would = 3,025 W, 56.25% more load power (at 100% efficiency) instead of 1,936 W (at 64% efficiency). So generator A is giving up 1,089 W because of the load mismatch.
Matching gives the maximum POWER TRANSFER (100% efficiency with respect to available power) not necessarily the maximum load power.
Your assumption is correct, just not for maximum POWER TRANSFER..
TomG : My analysis is correct.
Yes ur assumption is correct.We must always try to reduce the internal resistance as min as possible for the better output.
see impedance matching is good
but in ideal condition internal resistance assumed to be zero i/p resistance is assumed high and o/p resistance is assumed to be zero.......
Assume that I have two set of Generators. One high internal impedance (High-in), another one with lower internal impedance (Low-in).
Set A:
Internal Impedance , Ria = 1 Ohm
Generated Voltage, Vga = 110 V
Set B:
Internal Impedance , Rib = 4 Ohm
Generated Voltage, Vgb = 110 V
Connect a 4 Ohm loading resistor (Rload) to both generator respectively. Set A should give more output Power to the load.
Set A calculation:
Ia=110V/(1Ohm+4Ohm)
= 22A
Pa=I^2xRload
= 22 x 22 x 4
=1936W
Set B Calculation (Impedance matching):
Ib=110V/(4Ohm+4Ohm)
= 13.75A
Pb=I^2xRload
= 13.75 x 13.75 x 4
= 756.25 W
Pa>Pb. Impedance matching suppose to deliver the optimum power to the output load for a given internal resistance generator. However generator with lower internal impedance will always be able to produce higher output power compare to the generator that having higher internal impedance. Is my assumption correct?
