Q.An electric bulb is marked 220 V, 100 W. When it is connected across 110 V, its power is :

Deriving P = VI. Tracking a small positive charge dQ around a simple loop containing a battery of voltage V and a resistor: moving through the battery, the charge gains potential energy $dU=V\,dQ$ (at the expense of the battery's stored chemical energy); moving through the resistor, it loses that same energy to collisions with the resistor's atoms. The rate at which this energy is delivered is the electrical power, $P=dU/dt=V\,dQ/dt$, and since $I=dQ/dt$,

This gives the power delivered by a battery (or delivered TO any electrical device) carrying current I under potential difference V, with SI unit the watt ($1\ \text{W}=1\ \text{J/s}$).

Alternative forms via Ohm's law. Substituting $V=IR$ into $P=VI$ gives $P=I^2R$, and substituting $I=V/R$ instead gives $P=V^2/R$ -- three algebraically equivalent expressions for the power dissipated in a resistor, each convenient in different situations depending on which quantity (I or V) is already known. The $I^2R$ (and $V^2/R$) forms make clear that power depends on the SQUARE of current (or voltage): doubling the current through a fixed resistor quadruples, not merely doubles, the power dissipated. …