Maximum Power Transfer Theorem

The Maximum Power Transfer Theorem affirms that for maximum power to be transferred from a source to a load, the resistance of the load matches to the resistance of the source. This theorem is important in electrical engineering as it helps in designing circuits where power efficiency is critical, such as in audio systems, communication devices, and power electronics. The theorem is particularly useful in matching impedances in AC circuits and analyzing power efficiency in various systems.

1.0Statement of Maximum Power Transfer Theorem

This theorem states that output power will be maximum if external resistance connected to the source matches to the internal resistance of the source.

2.0Formula And Circuit Diagram For Maximum Power Transfer Theorem

Circuit having battery of emf E having internal resistance r be connected across the load R. Let I be the current flowing through the resistance R then power dissipated in the form of heat energy.

3.0Proof of Maximum Power Transfer Theorem

$P=I^{2}R$

$I=\frac{E}{R+r}$

$P=\frac{E^2}{(R+r{)}^2}$………….(1)

Output Power will be maximum if

$\frac{dP}{dR}=0$

$\frac{d}{dR}\left[\frac{E^{2}R}{(R+r{)}^2}\right]=0$

$\frac{E^{2}d}{dR}\left[\frac{R}{(R+r{)}^2}\right]$

$\frac{E^{2}d}{dR}\left[R(R+r{)}^{-2}\right]=0$

$E^2\left[(R+r{)}^{-2}\times 1-2R(R+r{)}^{-3}\right]=0$

$E^2\left[\frac{1}{(R+r{)}^2}-\frac{2R}{(R+r{)}^3}\right]=0$            Since $E^2\ne 0$

$\frac{1}{(R+r{)}^2}-\frac{2R}{(R+r{)}^3}=0$

$\frac{1}{(R+r{)}^2}=\frac{2R}{(R+r{)}^3}$

$1=\frac{2R}{(R+r)}$

$R+r=2R$

$R=r$

$\Rightarrow$ For maximum transfer of power the external resistance connected to the source should be equal to the internal resistance of the source R=r

$P_{Max}=\frac{E^{2}R}{(R+r{)}^2}=\frac{E^{2}R}{4R^2}$

$P_{\text{Max }}=\frac{E^2}{4R}$

Illustration: A 20 V battery of internal resistance of 4 is connected to a rheostat. Find the value of the current drawn from the battery to produce maximum power in the rheostat.

Solution: Rate of Heat Produced

$P=I^{2}R={\left(\frac{EMF}{\text{ Total Resistance }}\right)}^2\times R={\left(\frac{20}{R+r}\right)}^2\times R$

For P to be maximum, $\frac{dP}{dR}=0$

$\frac{d}{dR}\left[{\left(\frac{20}{R+4}\right)}^{2}R\right]=0$

$\frac{d}{dR}\left[R\times (R+4{)}^{-2}\right]=0$

$\frac{2R}{R+4}=1$

$R=4\Omega$

Current Drawn 

$I=\frac{20}{R+4}=\frac{20}{4+4}=2.5\mathrm{A}$

Rate of heat produced is maximum when external resistance matches to the internal resistance of the battery.  (R=r)

4.0Advantage of Maximum Power Transfer Theorem

1. The theorem is widely used in designing circuits for maximum energy transfer, particularly in audio, communication, and RF systems.
2. In power systems, understanding the load-resistance matching helps in the effective distribution of power from sources to various loads.
3. The Maximum Power Transfer Theorem is crucial in impedance matching for AC circuits, where maximum power is transferred when the load impedance matches the source impedance. This is particularly important in radio frequency (RF) and audio applications, such as antenna design and speaker matching.

5.0Limitation of Maximum Power Transfer Theorem

• The efficiency of power transfer (R=r) is only 50%. This means that half of the power is dissipated in the source resistance​ as heat, and only the remaining 50% reaches the load. 

• Source having battery or generator of EMF E having internal resistance r, this source is connected with load resistance R known as load resistance.
• If current I flowing through the R and V be the voltage across the resistance, then power delivered to the load resistance, output power of the source is given as, $P_{\text{out }}=VI$
• Since the battery has an internal resistance (r),there is a potential drop across the internal resistance =I r

V=E-I r

$P_{\text{out }}=(E-Ir)I$

$P_{\text{out }}=\left(EI-I^{2}r\right)$

Here E I is the power supplied by the source or battery.

• Input Power $P_{in}=EI$

$\Rightarrow I^{2}r$ is the difference between input and output powers and is dissipated as heat in the source itself.

• Efficiency of the source is given as,

$\eta =\frac{\text{ Output Power }}{\text{ Input Power }}=\frac{VI}{IE}=\frac{V}{E}$

$V=IR=\frac{ER}{(R+r)}\left(\therefore I=\frac{E}{R+r}\right)$

$\eta =\frac{ER}{E(R+r)}=\frac{R}{(R+r)}$

$\eta =\frac{V}{E}=\frac{R}{(R+r)}$

A source or a battery delivers maximum power when R=r 

$\%\eta =\frac{R}{2R}\times 100\%=50\%$

Efficiency of a source delivering maximum power is 50 %.

6.0Sample Questions On Maximum Power Transfer Theorem

Q-1.State the condition for maximum power transfer theorem?

Solution: When the load resistance equals the source resistance, the power delivered to the load is maximized $\left(R_L=R_S\right)$

Q-2.If the source resistance is 50 Ω, but the load resistance is much smaller or larger than 50 Ω, what happens to the power transferred to the load?

Solution:

• When $R_L$ ​ is much smaller than $R_S$​, most of the voltage is dropped across the source resistance, and less power is transferred to the load.
• When $R_L$ ​ is much larger than $R_S$, the current through the load is very small, again leading to lower power transfer.
• The power transferred to the load will be less than the maximum power that can be transferred when $\left(R_L=R_S\right).$