Maximum Power Transfer: MDCAT Physics notes
Maximum Power Transfer for MDCAT: when a source delivers most power to a load, R = r condition, P max = E²/4r, efficiency at matching, and power units.
Power from a source with internal resistance
A source of emf $\varepsilon$ and internal resistance $r$ drives current through a load $R$:
$$I = \frac{\varepsilon}{R + r} \qquad P_{\text{load}} = I^2R = \frac{\varepsilon^2R}{(R + r)^2}$$
If $R$ is very small, the current is large but little power is developed in the load. If $R$ is very large, the current is tiny, so the power is again small. The power in the load is greatest somewhere in between.
Maximum power transfer theorem
The power delivered to the load is maximum when the load resistance equals the internal resistance of the source:
$$R = r$$
Substituting $R = r$:
$$I = \frac{\varepsilon}{2r} \qquad P_{\max} = \left(\frac{\varepsilon}{2r}\right)^2r = \frac{\varepsilon^2}{4r}$$
The same result can be written $P_{\max} = \varepsilon^2/4R$, since $R = r$.
Efficiency at maximum power
At matching, equal power is dissipated in $r$ and in $R$. The source produces $\varepsilon I = \varepsilon^2/2r$, and half of it reaches the load, so the efficiency is only 50%. The terminal voltage at matching is $\varepsilon/2$.
| Load | Current | Power in load | Efficiency |
|---|---|---|---|
| $R \ll r$ | Large | Small | Low |
| $R = r$ | $\varepsilon/2r$ | Maximum, $\varepsilon^2/4r$ | 50% |
| $R \gg r$ | Small | Small | High |
Worked example
A battery of emf 10 V has internal resistance 2 $\Omega$. For maximum power the load must be 2 $\Omega$. Then $I = 10/4 = 2.5$ A and $P_{\max} = 10^2/(4\times2) = 12.5$ W. Check: $I^2R = 6.25\times2 = 12.5$ W.
With a 3 $\Omega$ load instead: $I = 10/5 = 2$ A, $P = 4\times3 = 12$ W, which is less than 12.5 W, as expected.
Where it is used
Matching is used where getting the most power into the load matters more than efficiency, for example matching a loudspeaker to an amplifier or an antenna to a receiver. Power supplies for the mains are designed for high efficiency instead, with $R \gg r$.
Power and its units
Electrical power in general is $P = VI = I^2R = V^2/R$. Volt × ampere = watt, so the product $VI$ measures power.
Key formulas
- $P_{\text{load}} = \varepsilon^2R/(R + r)^2$
- Condition: $R = r$
- $P_{\max} = \varepsilon^2/4r$
- Efficiency at maximum power = 50%
Common MDCAT traps
- The condition is $R = r$, not $R = 2r$ or $R = r/2$, and not $R = 0$.
- $P_{\max} = \varepsilon^2/4r$: the 4 in the denominator is often dropped.
- Maximum power does not mean maximum efficiency; efficiency is only 50%.
- Volt × ampere measures power, not resistance or potential difference.
Quick revision
- Maximum power transfer occurs when load resistance equals source resistance.
- At matching, terminal voltage is half the emf.
- Half the power is wasted inside the source at matching.
- 1 V × 1 A = 1 W.