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Formulas/physics/Current Electricity/Maximum Power Transfer

Maximum Power Transfer

Maximum power delivered to external resistance R occurs when R equals internal resistance r. At this condition efficiency is 50% — half the total power is wasted internally.
Class 12
Derivation

Power as a function of R

External power:

Pext(R)=I2R=E2R(R+r)2P_{ext}(R) = I^2R = \frac{\mathcal{E}^2 R}{(R+r)^2}

Maximising by differentiation

dPextdR=E2⋅(R+r)2−R⋅2(R+r)(R+r)4=E2⋅(R+r)−2R(R+r)3=E2⋅r−R(R+r)3\frac{dP_{ext}}{dR} = \mathcal{E}^2\cdot\frac{(R+r)^2 - R\cdot 2(R+r)}{(R+r)^4} = \mathcal{E}^2\cdot\frac{(R+r) - 2R}{(R+r)^3} = \mathcal{E}^2\cdot\frac{r-R}{(R+r)^3}

Setting to zero: r−R=0  ⟹  R=rr - R = 0 \implies R = r

Second derivative is negative at R=rR = r — confirmed maximum.

Maximum power

At R=rR = r:

Pmax=E2⋅r(r+r)2=E2r4r2P_{max} = \frac{\mathcal{E}^2 \cdot r}{(r+r)^2} = \frac{\mathcal{E}^2 r}{4r^2} Pmax=E24rat R=r\boxed{P_{max} = \frac{\mathcal{E}^2}{4r} \quad \text{at } R = r}

Graphical behaviour

PextP_{ext} rises from 0 (at R=0R = 0), peaks at R=rR = r, then falls back toward 0 as R→∞R \to \infty. The curve is asymmetric — it rises steeply for R<rR < r and falls slowly for R>rR > r.

Remember
Maximum power transfer theorem: for a source with internal resistance $r$, maximum power to the load occurs at $R_{load} = r$. This is the principle of impedance matching in AC circuits and antenna design.