# Electric power correction

The electrical energy is defined as: $$E(t) = Q(t)V(t)$$

The general definition of power: $$P(t)=\frac{dE(t)}{dt}$$ and becomes $$P(t) = \frac{dQ(t)}{dt}V(t) + \frac{dV(t)}{dt}Q(t) = I(t)V(t) + \frac{dV(t)}{dt}Q(t)$$

My question: electrical power is often written as $$P=I(t)V(t)$$. Why is this second term neglected?

For example, the grid has an oscillating voltage, so you cannot neglect $$dV(t)/dt$$. Also, $$Q(t)$$ is different from zero because otherwise $$dQ(t)/dt$$ would also be zero (which is not the case), so why is the power of a resistor then given by only the first term?

You have the logical order backwards. In circuit theory power is primary and then energy is derived from that. So you start with $$P=IV$$ and then from that you can derive $$E=\int P\ dt=\int IV \ dt$$. If $$V$$ is constant then that works out to $$QV$$ but that is a derived result assuming constant $$V$$