# 13.4.1 Derivation of P=VI

Consider a circuit where a lamp is connected across a battery. The EMF of the battery is E and the PD across the lamp is V. From the energy perspective, the battery converts chemical energy to EPE, while the lamp converts EPE to heat and light.

In the battery, the EPE converted per unit charge is E (think $\displaystyle E=\frac{W}{Q}$), and charge per unit time is I (think $\displaystyle I=\frac{Q}{t}$). So the power output of the battery is

$P=EI$

Likewise, in the lamp, the EPE converted per unit charge is V (think $\displaystyle V=\frac{W}{Q}$), and charge per unit time is I (think $\displaystyle I=\frac{Q}{t}$). So the rate of energy conversion in the filament must be

$P=VI$

By substituting $V=IR$(Ohm’s Law), we can also express the formula as

\displaystyle \begin{aligned}P&={{I}^{2}}R\\P&=\frac{{{{V}^{2}}}}{R}\end{aligned}

A word of caution: it is probably better if we present these formula as $P=\Delta V.I$ and $\displaystyle P=\frac{{{{{(\Delta V)}}^{2}}}}{R}$ instead, since V is the potential difference across R. But then you know engineers tend to assume people can tell the difference between V the potential and V the potential difference.

Demonstration

35 W vs 50 W Bulb

Concept Test

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