Formulas/physics/Current Electricity/Efficiency of a Cell

Efficiency of a Cell

Fraction of total power delivered to external circuit. Maximum power transfer (η = 50%) occurs when R = r, but maximum efficiency requires R ≫ r.

Class 12

Derivation

For a cell (EMF $\mathcal{E}$ , internal resistance $r$ ) connected to external resistance $R$ , current $I = \mathcal{E}/(R+r)$ .

Total power generated by the cell:

P_{total} = \mathcal{E}I = \frac{\mathcal{E}^2}{R+r}

Power delivered to external circuit:

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

Efficiency:

\eta = \frac{P_{ext}}{P_{total}} = \frac{I^2R}{\mathcal{E}I} = \frac{IR}{\mathcal{E}} = \frac{IR}{I(R+r)}

\boxed{\eta = \frac{R}{R+r} = 1 - \frac{Ir}{\mathcal{E}}}

Efficiency vs power transfer

Condition	$\eta$	$P_{ext}$
$R \to \infty$	$\to 1$ (100%)	$\to 0$
$R = r$	50%	Maximum
$R \to 0$	$\to 0$	$\to 0$

Maximum efficiency ( $\eta \to 1$ ) requires $R \gg r$ but gives low power. Maximum power transfer occurs at $R = r$ but efficiency is only 50%. These two objectives are fundamentally in conflict.

Note

In power engineering (generators, transmission), maximum efficiency ($R \gg r$) is the goal — wasting 50% of generated power is unacceptable. In signal electronics and communications, maximum power transfer ($R = r$, impedance matching) is preferred.

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