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{{Short description|Theorem in electrical engineering}}
In ], the '''maximum power (transfer) theorem''' states that to obtain ''maximum'' power from a source with a ''fixed'' internal ] the resistance of the load must be made the same as that of the source.
In ], the '''maximum power transfer theorem''' states that, to obtain ''maximum'' external ] from a power source with ], the ] of the ] must equal the ] as viewed from its output terminals. ] published the maximum power (transfer) theorem around 1840; it is also referred to as "'''Jacobi's law'''".<ref>{{citation |url=https://books.google.com/books?id=dKVbT-ZmdDwC&pg=PA406 |title=Dynamo-Electric Machinery; A Manual for Students of Electrotechnics |author=Thompson Phillips |publisher=BiblioBazaar, LLC | isbn=978-1-110-35104-6 |date=2009-05-30}}</ref>


The ] results in maximum ''power'' transfer from the power source to the load, but not maximum '']'' of useful power out of total power consumed. If the load resistance is made larger than the source resistance, then efficiency increases (since a higher percentage of the source power is transferred to the load), but the ''magnitude'' of the load power decreases (since the total circuit resistance increases).<ref name="harrison2013">{{Cite journal|last=Harrison|first=Mark | date=2013-02-22 | title=Physical collisions and the maximum power theorem: an analogy between mechanical and electrical situations | journal=Physics Education|volume=48|issue=2|pages=207–211|doi=10.1088/0031-9120/48/2/207|bibcode=2013PhyEd..48..207H |s2cid=120330420 |issn=0031-9120}}</ref> If the load resistance is made smaller than the source resistance, then efficiency decreases (since most of the power ends up being dissipated in the source). Although the total power dissipated increases (due to a lower total resistance), the amount dissipated in the load decreases.
The ] applies only when the ] is fixed. If the source resistance were variable (but the load resistance fixed), maximum power would be transferred to the load simply by setting the source resistance to zero. Raising the source impedance to match the load would, in this case, ''reduce'' power transfer. This is the case when driving a load such as a ] with a modern amplifier. In this case, the load presented by the loudspeaker is fixed (typically, 8 ohms for home audio) and maximum power occurs with an ] connection. This type of connection also serves to maximize control of the speaker cone (due to high ]), which serves to lower ].


The theorem states how to choose (so as to maximize power transfer) the load resistance, once the source resistance is given. It is a common misconception to apply the theorem in the opposite scenario. It does ''not'' say how to choose the source resistance for a given load resistance. In fact, the source resistance that maximizes power transfer from a voltage source is always zero (the hypothetical ]), regardless of the value of the load resistance.
== Maximising power transfer versus power efficiency ==
The theorem was originally misunderstood (notably by ]) to imply that a system consisting of an electric motor driven by a battery could not be more than 50% efficient, since the power lost as heat in the battery would always be equal to the power delivered to the motor. In ] this assumption was shown to be false by either ] or his colleague ], who realized that the theorem could be sidestepped by making the resistance of the source (whether a battery or a ]) close to zero. Using this new understanding, they obtained an efficiency of about 90%, and proved that the electric motor was a practical alternative to the ].


The theorem can be extended to ] circuits that include ], and states that maximum power transfer occurs when the load ] is equal to the ] of the source impedance.
]


The mathematics of the theorem also applies to other physical interactions, such as:<ref name="harrison2013" /><ref>{{Cite journal |last=Atkin |first=Keith |date=2013-08-22 |title=Energy transfer and a recurring mathematical function |journal=Physics Education |volume=48 |issue=5 |pages=616–620 |doi=10.1088/0031-9120/48/5/616 |bibcode=2013PhyEd..48..616A |s2cid=122189586 |issn=0031-9120}}</ref>
The condition of maximum power transfer does not result in maximum ]. If we define the efficiency <math>\eta</math> as the ratio of power dissipated by the load to power developed by the source, then it is straightforward to calculate from the above circuit diagram that


* mechanical collisions between two objects,
:<math>
* the sharing of charge between two capacitors,
\eta = {R_\mathrm{load} \over {R_\mathrm{load} + R_\mathrm{source} } } = { 1 \over { 1 + { R_\mathrm{source} \over R_\mathrm{load} } } }
* liquid flow between two cylinders,
</math>
* the transmission and reflection of light at the boundary between two media.


== Maximizing power transfer versus power efficiency ==
Consider three particular cases:
]
The theorem was originally misunderstood (notably by ]<ref>{{Cite web |last=Magnetics |first=Triad |title=Understanding the Maximum Power Theorem |url=https://info.triadmagnetics.com/blog/maximum-power-theorem |access-date=2022-06-08 |website=info.triadmagnetics.com |language=en-us}}</ref>) to imply that a system consisting of an electric motor driven by a ] could not be more than 50% ], since the power dissipated as heat in the battery would always be equal to the power delivered to the motor when the impedances were matched.


In 1880 this assumption was shown to be false by either ] or his colleague ], who realized that maximum efficiency was not the same as maximum power transfer.


To achieve maximum efficiency, the resistance of the source (whether a battery or a ]) could be (or should be) made as close to zero as possible. Using this new understanding, they obtained an efficiency of about 90%, and proved that the ] was a practical alternative to the ].]
* If <math>R_\mathrm{load}=R_\mathrm{source}</math>, then <math>\eta=0.5</math>.
The efficiency {{math|''η''}} is the ratio of the power dissipated by the load resistance {{math|''R''<sub>L</sub>}} to the total power dissipated by the circuit (which includes the voltage ] of {{math|''R''<sub>S</sub>}} as well as {{math|''R''<sub>L</sub>}}):
* If <math>R_\mathrm{load}=\infty</math>, then <math>\eta=1</math>.
* If <math>R_\mathrm{load}=0</math>, then <math>\eta=0</math>.


<math display="block">\eta = \frac{P_\mathrm{L}}{P_\mathrm{Total}} = \frac{I^2 \cdot R_\mathrm{L}}{I^2 \cdot (R_\mathrm{L} + R_\mathrm{S})} = \frac{R_\mathrm{L}}{R_\mathrm{L} + R_\mathrm{S}} = \frac{1}{1 + R_\mathrm{S} / R_\mathrm{L}} \, .</math>


Consider three particular cases (note that voltage sources must have some resistance):
The efficiency is only 50% when maximum power transfer is achieved, but approaches 100% as the load resistance approaches infinity (though the total power level tends towards zero). When the load resistance is zero, all the power is consumed inside the source (the power dissipated in a ] is zero) so the efficiency is zero.
* If <math>R_\mathrm{L}/R_\mathrm{S} \to 0</math>, then <math>\eta \to 0.</math> Efficiency approaches 0% if the load resistance approaches zero (a ]), since all power is consumed in the source and no power is consumed in the short.
* If <math>R_\mathrm{L}/R_\mathrm{S} = 1</math>, then <math>\eta = \tfrac{1}{2}.</math> Efficiency is only 50% if the load resistance equals the source resistance (which is the condition of maximum power transfer).
* If <math>R_\mathrm{L}/R_\mathrm{S} \to \infty</math>, then <math>\eta \to 1.</math> Efficiency approaches 100% if the load resistance approaches infinity (though the total power level tends towards zero) or if the source resistance approaches zero. Using a large ratio is called ].


==Impedance matching== ==Impedance matching==
{{main|impedance matching}}
A related concept is reflectionless ]. In ], ]s, and other ], there is often a requirement to match the ] (such as a transmitter) to the ] (such as an ]) to avoid reflections in the ].
A related concept is reflectionless ].


In ] frequency ]s, and other ], there is often a requirement to match the ] (at the transmitter) to the ] (such as an ]) to avoid reflections in the ].
==In reactive circuits==
The theorem also applies where the source and/or load are not totally resistive. This invokes a refinement of the theorem which says that any reactive components of source and load should be of equal magnitude but opposite phase. (''See below for a derivation.'') This means that the source and load impedances should be '']s'' of each other. In the case of purely resistive circuits, the two concepts are identical. However, physically realizable sources and loads are not usually totally resistive, having some inductive or capacitive components, and so practical applications of this theorem, under the name of complex conjugate impedance matching, do, in fact, exist.


== Calculus-based proof for purely resistive circuits ==
If the source is totally inductive (capacitive), then a totally capacitive (inductive) load, in the absence of resistive losses, would receive 100% of the energy from the source but send it back after a quarter cycle. The resultant circuit is nothing other than a resonant ] in which the energy continues to oscillate to and fro. This is called ]. ](where an inductive reactance is used to "balance out" a capacitive one), is essentially the same idea as ] matching although it is done for entirely different reasons.
In the simplified model of powering a load with resistance {{math|''R''<sub>L</sub>}} by a source with voltage {{math|''V''}} and ] {{math|''R''<sub>S</sub>}}, then by ] the resulting current {{math|''I''}} is simply the source voltage divided by the total circuit resistance:
<math display="block">I = \frac{V}{R_\mathrm{S} + R_\mathrm{L}}.</math>


The power {{math|''P''<sub>L</sub>}} dissipated in the load is the square of the current multiplied by the resistance:
For a fixed reactive ''source'', the maximum power theorem maximizes the real power (P) delivered to the load by complex conjugate matching the load to the source.
<math display="block">P_\mathrm{L} = I^2 R_\mathrm{L} = \left(\frac{V}{R_\mathrm{S} + R_\mathrm{L}}\right)^2 R_\mathrm{L} = \frac{V^2}{R_\mathrm{S}^2 / R_\mathrm{L} + 2 R_\mathrm{S} + R_\mathrm{L}}.</math>


The value of {{math|''R''<sub>L</sub>}} for which this expression is a maximum could be calculated by differentiating it, but it is easier to calculate the value of {{math|''R''<sub>L</sub>}} for which the denominator:
For a fixed reactive ''load'', power factor correction minimizes the ] (S) (and unnecessary current) conducted by the transmission lines, while maintaining the same amount of real power transfer. This is done by adding a reactance to the load to balance out the load's own reactance, changing the reactive load impedance into a resistive load impedance.
<math display="block">R_\mathrm{S}^2 / R_\mathrm{L} + 2 R_\mathrm{S} + R_\mathrm{L}</math>
is a minimum. The result will be the same in either case. Differentiating the denominator with respect to {{math|''R''<sub>L</sub>}}:
<math display="block">\frac{d}{d R_\mathrm{L}} \left(R_\mathrm{S}^2 / R_\mathrm{L} + 2R_\mathrm{S} + R_\mathrm{L}\right) = -R_\mathrm{S}^2 / R_\mathrm{L}^2 + 1.</math>


For a maximum or minimum, the first derivative is zero, so
== Proof of theorem for resistive circuits ==
<math display="block">R_\mathrm{S}^2 / R_\mathrm{L}^2 = 1</math>
]
or
<math display="block">R_\mathrm{L} = \pm R_\mathrm{S}.</math>


In practical resistive circuits, {{math|''R''<sub>S</sub>}} and {{math|''R''<sub>L</sub>}} are both positive, so the positive sign in the above is the correct solution.
In the diagram opposite, power is being transferred from the source, with voltage <math>V</math> and fixed ] <math>R_\mathrm{S}</math>, to a load with resistance <math>R_\mathrm{L}</math>, resulting in a current <math>I</math>. <math>I</math> is simply the source voltage divided by the total circuit resistance:


To find out whether this solution is a minimum or a maximum, the denominator expression is differentiated again:
:<math>
I = {V \over R_\mathrm{S} + R_\mathrm{L}}
</math>


<math display="block">\frac{d^2}{dR_\mathrm{L}^2} \left( {R_\mathrm{S}^2 / R_\mathrm{L} + 2 R_\mathrm{S} + R_\mathrm{L}} \right) = {2 R_\mathrm{S}^2} / {R_\mathrm{L}^3}. </math>
The power <math>P_\mathrm{L}</math> dissipated in the load is the square of the current multiplied by the resistance:


This is always positive for positive values of <math>R_\mathrm{S} </math> and <math>R_\mathrm{L} </math>, showing that the denominator is a minimum, and the power is therefore a maximum, when:
:<math>
<math display="block">R_\mathrm{S} = R_\mathrm{L}. </math>
P_\mathrm{L} = I^2 R_\mathrm{L} = {{ \left( {V \over {R_\mathrm{S} + R_\mathrm{L}}} \right) }^2} R_\mathrm{L} = {{V^2} \over {R_\mathrm{S}^2 / R_\mathrm{L} + 2R_\mathrm{S} + R_\mathrm{L}}}
</math>


The above proof assumes fixed source resistance <math>R_\mathrm{S} </math>. When the source resistance can be varied, power transferred to the load can be increased by reducing <math>R_\textrm{S}</math>. For example, a 100 Volt source with an <math>R_\textrm{S}</math> of <math>10\,\Omega</math> will deliver 250 watts of power to a <math>10\,\Omega</math> load; reducing <math>R_\textrm{S}</math> to <math>0\,\Omega</math> increases the power delivered to 1000 watts.
We could calculate the value of <math>R_\mathrm{L}</math> for which this expression is a maximum, but it is easier to calculate the value of <math>R_\mathrm{L}</math> for which the denominator


Note that this shows that maximum power transfer can also be interpreted as the load voltage being equal to one-half of the Thevenin voltage equivalent of the source.<ref>{{Cite web|url=http://www.electronics-tutorial.net/dccircuits/maximum-power-transfer-theorem/index.html|title = Basic Electronics Tutorials and Revision for Freshers to Advanced Learners}}</ref>
:<math>
R_\mathrm{S}^2 / R_\mathrm{L} + 2R_\mathrm{S} + R_\mathrm{L}
</math>


==In reactive circuits==
is a minimum. The result will be the same in either case. Differentiating with respect to <math>R_\mathrm{L}</math>:
The power transfer theorem also applies when the source and/or load are not purely resistive.


A refinement of the maximum power theorem says that any reactive components of source and load should be of equal magnitude but opposite sign. (''See below for a derivation.'')
:<math>
* This means that the source and load impedances should be '']s'' of each other.
{d\over{dR_\mathrm{L}}} \left( {R_\mathrm{S}^2 / R_\mathrm{L} + 2R_\mathrm{S} + R_\mathrm{L}} \right) = -R_\mathrm{S}^2 / R_\mathrm{L}^2+1
* In the case of purely resistive circuits, the two concepts are identical.
</math>


Physically realizable sources and loads are not usually purely resistive, having some inductive or capacitive components, and so practical applications of this theorem, under the name of complex conjugate impedance matching, do, in fact, exist.
For a maximum or minimum, the first derivative is zero, so


If the source is totally inductive (capacitive), then a totally capacitive (inductive) load, in the absence of resistive losses, would receive 100% of the energy from the source but send it back after a quarter cycle.
:<math>
{R_\mathrm{S}^2 / R_\mathrm{L}^2} = 1
</math>


The resultant circuit is nothing other than a resonant ] in which the energy continues to oscillate to and fro. This oscillation is called ].
or


] (where an inductive reactance is used to "balance out" a capacitive one), is essentially the same idea as complex conjugate impedance matching although it is done for entirely different reasons.
:<math>R_\mathrm{L} = \pm R_\mathrm{S}</math>


For a fixed reactive ''source'', the maximum power theorem maximizes the real power (P) delivered to the load by complex conjugate matching the load to the source.
In practical resistive circuits, <math>R_\mathrm{S}</math> and <math>R_\mathrm{L}</math> are both positive. To find out whether this solution is a minimum or a maximum, we must differentiate again:


For a fixed reactive ''load'', power factor correction minimizes the ] (S) (and unnecessary current) conducted by the transmission lines, while maintaining the same amount of real power transfer.
:<math>{{d^2} \over {dR_\mathrm{L}^2}} \left( {R_\mathrm{S}^2 / R_\mathrm{L} + 2 R_\mathrm{S} + R_\mathrm{L}} \right) = {2 R_\mathrm{S}^2} / {R_\mathrm{L}^3}</math>


This is done by adding a reactance to the load to balance out the load's own reactance, changing the reactive load impedance into a resistive load impedance.
This is positive for positive values of <math>R_\mathrm{S}</math> and <math>R_\mathrm{L}</math>, showing that the denominator is a minimum, and the power is therefore a maximum, when


=== Proof ===
:<math>R_\mathrm{S} = R_\mathrm{L}</math>


]
== For reactive impedances ==


In this diagram, ] is being transferred from the source, with ] magnitude of voltage <math>|V_\text{S}|</math> (positive peak voltage) and fixed ] <math>Z_\text{S}</math> (S for source), to a load with impedance <math>Z_\text{L}</math> (L for load), resulting in a (positive) magnitude <math>|I|</math> of the current phasor <math>I</math>. This magnitude <math>|I|</math> results from dividing the magnitude of the source voltage by the magnitude of the total circuit impedance:
]
<math display="block">|I| = {|V_\text{S}| \over |Z_\text{S} + Z_\text{L}|} .</math>


The average power <math>P_\text{L}</math> dissipated in the load is the square of the current multiplied by the resistive portion (the real part) <math>R_\text{L}</math> of the load impedance <math>Z_\text{L}</math>:
In this diagram, ] is being transferred from the source, with ] magnitude voltage <math>|V_\mathrm{S}|</math> (peak voltage) and fixed ] <math>Z_\mathrm{S}</math>, to a load with impedance <math>Z_\mathrm{L}</math>, resulting in a phasor magnitude current <math>|I|</math>. <math>|I|</math> is simply the source voltage divided by the total circuit impedance:
<math display="block">\begin{align}
P_\text{L} & = I_\text{rms}^2 R_\text{L} = {1 \over 2} |I|^2 R_\text{L}\\
& = {1 \over 2} \left( {|V_\text{S}| \over |Z_\text{S} + Z_\text{L}|} \right)^2 R_\text{L} = {1 \over 2}{ |V_\text{S}|^2 R_\text{L} \over (R_\text{S} + R_\text{L})^2 + (X_\text{S} + X_\text{L})^2},
\end{align}</math>
where <math>R_\text{S}</math> and <math>R_\text{L}</math> denote the resistances, that is the real parts, and <math>X_\text{S}</math> and <math>X_\text{L}</math> denote the reactances, that is the imaginary parts, of respectively the source and load impedances <math>Z_\text{S}</math> and <math>Z_\text{L}</math>.


To determine, for a given source, the voltage <math>V_\text{S}</math> and the impedance <math>Z_\text{S},</math> the value of the load impedance <math>Z_\text{L}, </math> for which this expression for the power yields a maximum, one first finds, for each fixed positive value of <math>R_\text{L}</math>, the value of the reactive term <math>X_\text{L}</math> for which the denominator:
:<math>
|I| = { |V_\mathrm{S}| \over Z_\mathrm{S} + Z_\mathrm{L} } <math display="block">(R_\text{S} + R_\text{L})^2 + (X_\text{S} + X_\text{L})^2 </math>
is a minimum. Since reactances can be negative, this is achieved by adapting the load reactance to:
</math>
<math display="block">X_\text{L} = -X_\text{S}.</math>


This reduces the above equation to:
The average power <math>P_\mathrm{L}</math> dissipated in the load is the square of the current multiplied by the resistive portion of the load impedance:
<math display="block">P_\text{L} = \frac 1 2 \frac{|V_\text{S}|^2 R_\text{L}}{(R_\text{S} + R_\text{L})^2}</math>
and it remains to find the value of <math>R_\text{L}</math> which maximizes this expression. This problem has the same form as in the purely resistive case, and the maximizing condition therefore is <math>R_\text{L} = R_\text{S}.</math>


The two maximizing conditions:
{|
|-
|<math>P_\mathrm{L}</math>
|<math>= I_\mathrm{rms}^2 R_\mathrm{L} = {1 \over 2} |I|^2 R_\mathrm{L} = {1 \over 2} {{\left( {|V_\mathrm{S}| \over {Z_\mathrm{S} + Z_\mathrm{L}}} \right)}^2} R_\mathrm{L}</math>
|-
|
|<math>= {1 \over 2}{{|V_\mathrm{S}|^2 R_\mathrm{L}}\over{(R_\mathrm{S} + R_\mathrm{L})^2 + (X_\mathrm{S} + X_\mathrm{L})^2}}</math>
|}


*<math>R_\text{L} = R_\text{S}</math>
We could calculate the values of <math>R_\mathrm{L}</math> and <math>X_\mathrm{L}</math> (since <math>V_\mathrm{S}</math>, <math>R_\mathrm{S}</math>, and <math>X_\mathrm{S}</math> are fixed) for which this expression is a maximum, but it is easier to calculate the values for which the denominator
*<math>X_\text{L} = -X_\text{S}</math>


describe the ] of the source impedance, denoted by <math>^*,</math> and thus can be concisely combined to:
:<math>
<math display="block">Z_\text{L} = Z_\text{S}^*.</math>
(R_\mathrm{S} + R_\mathrm{L})^2 + (X_\mathrm{S} + X_\mathrm{L})^2
</math>


==See also==
is a minimum. The result will be the same in either case. Since reactances can be negative, the reactive term is easily minimized to zero by making
*]

:<math>
X_\mathrm{L} = -X_\mathrm{S}
</math>

The power equation is now reduced to:

:<math>
P_\mathrm{L} = {1 \over 2}{{|V_\mathrm{S}|^2 R_\mathrm{L}}\over{(R_\mathrm{S} + R_\mathrm{L})^2}}
</math>

This is the same form as above, and the maximizing condition <math>R_\mathrm{L} = R_\mathrm{S}</math> can be found in the same way.

The combination of conditions

*<math>R_\mathrm{L} = R_\mathrm{S}</math>
*<math>X_\mathrm{L} = -X_\mathrm{S}</math>

can be concisely written with a ] (the *) as:


==Notes==
:<math>Z_\mathrm{S} = Z_\mathrm{L}^*</math>
<references/>


==References== ==References==
* H.W. Jackson (1959) Introduction to Electronic Circuits, Prentice-Hall. *H.W. Jackson (1959) Introduction to Electronic Circuits, Prentice-Hall.


== External links == == External links ==
* (]) taken from
*
* (]) taken from
* *
*
* - unconfirmed claim that theorem was discovered by Moritz Jacobi
* MH Jacobi Biographical notes


]
==See also==
]

]
] ]

]

Latest revision as of 02:54, 30 November 2024

Theorem in electrical engineering

In electrical engineering, the maximum power transfer theorem states that, to obtain maximum external power from a power source with internal resistance, the resistance of the load must equal the resistance of the source as viewed from its output terminals. Moritz von Jacobi published the maximum power (transfer) theorem around 1840; it is also referred to as "Jacobi's law".

The theorem results in maximum power transfer from the power source to the load, but not maximum efficiency of useful power out of total power consumed. If the load resistance is made larger than the source resistance, then efficiency increases (since a higher percentage of the source power is transferred to the load), but the magnitude of the load power decreases (since the total circuit resistance increases). If the load resistance is made smaller than the source resistance, then efficiency decreases (since most of the power ends up being dissipated in the source). Although the total power dissipated increases (due to a lower total resistance), the amount dissipated in the load decreases.

The theorem states how to choose (so as to maximize power transfer) the load resistance, once the source resistance is given. It is a common misconception to apply the theorem in the opposite scenario. It does not say how to choose the source resistance for a given load resistance. In fact, the source resistance that maximizes power transfer from a voltage source is always zero (the hypothetical ideal voltage source), regardless of the value of the load resistance.

The theorem can be extended to alternating current circuits that include reactance, and states that maximum power transfer occurs when the load impedance is equal to the complex conjugate of the source impedance.

The mathematics of the theorem also applies to other physical interactions, such as:

  • mechanical collisions between two objects,
  • the sharing of charge between two capacitors,
  • liquid flow between two cylinders,
  • the transmission and reflection of light at the boundary between two media.

Maximizing power transfer versus power efficiency

Simplified model for powering a load with resistance RL by a source with voltage VS and resistance RS.

The theorem was originally misunderstood (notably by Joule) to imply that a system consisting of an electric motor driven by a battery could not be more than 50% efficient, since the power dissipated as heat in the battery would always be equal to the power delivered to the motor when the impedances were matched.

In 1880 this assumption was shown to be false by either Edison or his colleague Francis Robbins Upton, who realized that maximum efficiency was not the same as maximum power transfer.

To achieve maximum efficiency, the resistance of the source (whether a battery or a dynamo) could be (or should be) made as close to zero as possible. Using this new understanding, they obtained an efficiency of about 90%, and proved that the electric motor was a practical alternative to the heat engine.

The red curve shows the power in the load, normalized relative to its maximum possible. The dark blue curve shows the efficiency η.

The efficiency η is the ratio of the power dissipated by the load resistance RL to the total power dissipated by the circuit (which includes the voltage source's resistance of RS as well as RL):

η = P L P T o t a l = I 2 R L I 2 ( R L + R S ) = R L R L + R S = 1 1 + R S / R L . {\displaystyle \eta ={\frac {P_{\mathrm {L} }}{P_{\mathrm {Total} }}}={\frac {I^{2}\cdot R_{\mathrm {L} }}{I^{2}\cdot (R_{\mathrm {L} }+R_{\mathrm {S} })}}={\frac {R_{\mathrm {L} }}{R_{\mathrm {L} }+R_{\mathrm {S} }}}={\frac {1}{1+R_{\mathrm {S} }/R_{\mathrm {L} }}}\,.}

Consider three particular cases (note that voltage sources must have some resistance):

  • If R L / R S 0 {\displaystyle R_{\mathrm {L} }/R_{\mathrm {S} }\to 0} , then η 0. {\displaystyle \eta \to 0.} Efficiency approaches 0% if the load resistance approaches zero (a short circuit), since all power is consumed in the source and no power is consumed in the short.
  • If R L / R S = 1 {\displaystyle R_{\mathrm {L} }/R_{\mathrm {S} }=1} , then η = 1 2 . {\displaystyle \eta ={\tfrac {1}{2}}.} Efficiency is only 50% if the load resistance equals the source resistance (which is the condition of maximum power transfer).
  • If R L / R S {\displaystyle R_{\mathrm {L} }/R_{\mathrm {S} }\to \infty } , then η 1. {\displaystyle \eta \to 1.} Efficiency approaches 100% if the load resistance approaches infinity (though the total power level tends towards zero) or if the source resistance approaches zero. Using a large ratio is called impedance bridging.

Impedance matching

Main article: impedance matching

A related concept is reflectionless impedance matching.

In radio frequency transmission lines, and other electronics, there is often a requirement to match the source impedance (at the transmitter) to the load impedance (such as an antenna) to avoid reflections in the transmission line.

Calculus-based proof for purely resistive circuits

In the simplified model of powering a load with resistance RL by a source with voltage V and source resistance RS, then by Ohm's law the resulting current I is simply the source voltage divided by the total circuit resistance: I = V R S + R L . {\displaystyle I={\frac {V}{R_{\mathrm {S} }+R_{\mathrm {L} }}}.}

The power PL dissipated in the load is the square of the current multiplied by the resistance: P L = I 2 R L = ( V R S + R L ) 2 R L = V 2 R S 2 / R L + 2 R S + R L . {\displaystyle P_{\mathrm {L} }=I^{2}R_{\mathrm {L} }=\left({\frac {V}{R_{\mathrm {S} }+R_{\mathrm {L} }}}\right)^{2}R_{\mathrm {L} }={\frac {V^{2}}{R_{\mathrm {S} }^{2}/R_{\mathrm {L} }+2R_{\mathrm {S} }+R_{\mathrm {L} }}}.}

The value of RL for which this expression is a maximum could be calculated by differentiating it, but it is easier to calculate the value of RL for which the denominator: R S 2 / R L + 2 R S + R L {\displaystyle R_{\mathrm {S} }^{2}/R_{\mathrm {L} }+2R_{\mathrm {S} }+R_{\mathrm {L} }} is a minimum. The result will be the same in either case. Differentiating the denominator with respect to RL: d d R L ( R S 2 / R L + 2 R S + R L ) = R S 2 / R L 2 + 1. {\displaystyle {\frac {d}{dR_{\mathrm {L} }}}\left(R_{\mathrm {S} }^{2}/R_{\mathrm {L} }+2R_{\mathrm {S} }+R_{\mathrm {L} }\right)=-R_{\mathrm {S} }^{2}/R_{\mathrm {L} }^{2}+1.}

For a maximum or minimum, the first derivative is zero, so R S 2 / R L 2 = 1 {\displaystyle R_{\mathrm {S} }^{2}/R_{\mathrm {L} }^{2}=1} or R L = ± R S . {\displaystyle R_{\mathrm {L} }=\pm R_{\mathrm {S} }.}

In practical resistive circuits, RS and RL are both positive, so the positive sign in the above is the correct solution.

To find out whether this solution is a minimum or a maximum, the denominator expression is differentiated again:

d 2 d R L 2 ( R S 2 / R L + 2 R S + R L ) = 2 R S 2 / R L 3 . {\displaystyle {\frac {d^{2}}{dR_{\mathrm {L} }^{2}}}\left({R_{\mathrm {S} }^{2}/R_{\mathrm {L} }+2R_{\mathrm {S} }+R_{\mathrm {L} }}\right)={2R_{\mathrm {S} }^{2}}/{R_{\mathrm {L} }^{3}}.}

This is always positive for positive values of R S {\displaystyle R_{\mathrm {S} }} and R L {\displaystyle R_{\mathrm {L} }} , showing that the denominator is a minimum, and the power is therefore a maximum, when: R S = R L . {\displaystyle R_{\mathrm {S} }=R_{\mathrm {L} }.}

The above proof assumes fixed source resistance R S {\displaystyle R_{\mathrm {S} }} . When the source resistance can be varied, power transferred to the load can be increased by reducing R S {\displaystyle R_{\textrm {S}}} . For example, a 100 Volt source with an R S {\displaystyle R_{\textrm {S}}} of 10 Ω {\displaystyle 10\,\Omega } will deliver 250 watts of power to a 10 Ω {\displaystyle 10\,\Omega } load; reducing R S {\displaystyle R_{\textrm {S}}} to 0 Ω {\displaystyle 0\,\Omega } increases the power delivered to 1000 watts.

Note that this shows that maximum power transfer can also be interpreted as the load voltage being equal to one-half of the Thevenin voltage equivalent of the source.

In reactive circuits

The power transfer theorem also applies when the source and/or load are not purely resistive.

A refinement of the maximum power theorem says that any reactive components of source and load should be of equal magnitude but opposite sign. (See below for a derivation.)

  • This means that the source and load impedances should be complex conjugates of each other.
  • In the case of purely resistive circuits, the two concepts are identical.

Physically realizable sources and loads are not usually purely resistive, having some inductive or capacitive components, and so practical applications of this theorem, under the name of complex conjugate impedance matching, do, in fact, exist.

If the source is totally inductive (capacitive), then a totally capacitive (inductive) load, in the absence of resistive losses, would receive 100% of the energy from the source but send it back after a quarter cycle.

The resultant circuit is nothing other than a resonant LC circuit in which the energy continues to oscillate to and fro. This oscillation is called reactive power.

Power factor correction (where an inductive reactance is used to "balance out" a capacitive one), is essentially the same idea as complex conjugate impedance matching although it is done for entirely different reasons.

For a fixed reactive source, the maximum power theorem maximizes the real power (P) delivered to the load by complex conjugate matching the load to the source.

For a fixed reactive load, power factor correction minimizes the apparent power (S) (and unnecessary current) conducted by the transmission lines, while maintaining the same amount of real power transfer.

This is done by adding a reactance to the load to balance out the load's own reactance, changing the reactive load impedance into a resistive load impedance.

Proof

source and load impedance diagram
source and load impedance diagram

In this diagram, AC power is being transferred from the source, with phasor magnitude of voltage | V S | {\displaystyle |V_{\text{S}}|} (positive peak voltage) and fixed source impedance Z S {\displaystyle Z_{\text{S}}} (S for source), to a load with impedance Z L {\displaystyle Z_{\text{L}}} (L for load), resulting in a (positive) magnitude | I | {\displaystyle |I|} of the current phasor I {\displaystyle I} . This magnitude | I | {\displaystyle |I|} results from dividing the magnitude of the source voltage by the magnitude of the total circuit impedance: | I | = | V S | | Z S + Z L | . {\displaystyle |I|={|V_{\text{S}}| \over |Z_{\text{S}}+Z_{\text{L}}|}.}

The average power P L {\displaystyle P_{\text{L}}} dissipated in the load is the square of the current multiplied by the resistive portion (the real part) R L {\displaystyle R_{\text{L}}} of the load impedance Z L {\displaystyle Z_{\text{L}}} : P L = I rms 2 R L = 1 2 | I | 2 R L = 1 2 ( | V S | | Z S + Z L | ) 2 R L = 1 2 | V S | 2 R L ( R S + R L ) 2 + ( X S + X L ) 2 , {\displaystyle {\begin{aligned}P_{\text{L}}&=I_{\text{rms}}^{2}R_{\text{L}}={1 \over 2}|I|^{2}R_{\text{L}}\\&={1 \over 2}\left({|V_{\text{S}}| \over |Z_{\text{S}}+Z_{\text{L}}|}\right)^{2}R_{\text{L}}={1 \over 2}{|V_{\text{S}}|^{2}R_{\text{L}} \over (R_{\text{S}}+R_{\text{L}})^{2}+(X_{\text{S}}+X_{\text{L}})^{2}},\end{aligned}}} where R S {\displaystyle R_{\text{S}}} and R L {\displaystyle R_{\text{L}}} denote the resistances, that is the real parts, and X S {\displaystyle X_{\text{S}}} and X L {\displaystyle X_{\text{L}}} denote the reactances, that is the imaginary parts, of respectively the source and load impedances Z S {\displaystyle Z_{\text{S}}} and Z L {\displaystyle Z_{\text{L}}} .

To determine, for a given source, the voltage V S {\displaystyle V_{\text{S}}} and the impedance Z S , {\displaystyle Z_{\text{S}},} the value of the load impedance Z L , {\displaystyle Z_{\text{L}},} for which this expression for the power yields a maximum, one first finds, for each fixed positive value of R L {\displaystyle R_{\text{L}}} , the value of the reactive term X L {\displaystyle X_{\text{L}}} for which the denominator: ( R S + R L ) 2 + ( X S + X L ) 2 {\displaystyle (R_{\text{S}}+R_{\text{L}})^{2}+(X_{\text{S}}+X_{\text{L}})^{2}} is a minimum. Since reactances can be negative, this is achieved by adapting the load reactance to: X L = X S . {\displaystyle X_{\text{L}}=-X_{\text{S}}.}

This reduces the above equation to: P L = 1 2 | V S | 2 R L ( R S + R L ) 2 {\displaystyle P_{\text{L}}={\frac {1}{2}}{\frac {|V_{\text{S}}|^{2}R_{\text{L}}}{(R_{\text{S}}+R_{\text{L}})^{2}}}} and it remains to find the value of R L {\displaystyle R_{\text{L}}} which maximizes this expression. This problem has the same form as in the purely resistive case, and the maximizing condition therefore is R L = R S . {\displaystyle R_{\text{L}}=R_{\text{S}}.}

The two maximizing conditions:

  • R L = R S {\displaystyle R_{\text{L}}=R_{\text{S}}}
  • X L = X S {\displaystyle X_{\text{L}}=-X_{\text{S}}}

describe the complex conjugate of the source impedance, denoted by , {\displaystyle ^{*},} and thus can be concisely combined to: Z L = Z S . {\displaystyle Z_{\text{L}}=Z_{\text{S}}^{*}.}

See also

Notes

  1. Thompson Phillips (2009-05-30), Dynamo-Electric Machinery; A Manual for Students of Electrotechnics, BiblioBazaar, LLC, ISBN 978-1-110-35104-6
  2. ^ Harrison, Mark (2013-02-22). "Physical collisions and the maximum power theorem: an analogy between mechanical and electrical situations". Physics Education. 48 (2): 207–211. Bibcode:2013PhyEd..48..207H. doi:10.1088/0031-9120/48/2/207. ISSN 0031-9120. S2CID 120330420.
  3. Atkin, Keith (2013-08-22). "Energy transfer and a recurring mathematical function". Physics Education. 48 (5): 616–620. Bibcode:2013PhyEd..48..616A. doi:10.1088/0031-9120/48/5/616. ISSN 0031-9120. S2CID 122189586.
  4. Magnetics, Triad. "Understanding the Maximum Power Theorem". info.triadmagnetics.com. Retrieved 2022-06-08.
  5. "Basic Electronics Tutorials and Revision for Freshers to Advanced Learners".

References

  • H.W. Jackson (1959) Introduction to Electronic Circuits, Prentice-Hall.

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