Electronic – Power rating not equal to voltage rating times current rating

There are many unknown factors. Both power and current will vary with load and there is only one way to find out how they relate: measure them.

One of the most important properties of AC power you neglected is the power factor cos(φ) with inductive loads:

\$P = U \cdot I \cdot \cos(\varphi)\$

\$\begin{align} \cos(\varphi) & = \dfrac{P}{U \cdot I} \\ & = \dfrac{1300\text{W}}{120\text{V} \cdot 12\text{A}} \\ & \approx \boxed{0.9} \end{align}\$

which sounds about what I'd expect for a vacuum cleaner.

Check this Wikipedia article on electric power for more background details.

Ohm's law $$ 1: V(t) = I(t)R $$

Instantaneous power dissipation is product of voltage and current $$ 2: P(t) = V(t)I(t)\\ $$

Substitute 1 into 2 to get instantaneous power through a resistor in terms of voltage or current: $$ 3: P(t) = I^2(t)R = \frac{V^2(t)}{R}\\ $$

Average power is definitionally the integral of instantaneous power over a period, divided by that period. Substitute 3 into that to get average power in terms of voltage and current. $$ 4: P_{avg}=\frac{\int_0^T{P(t)dt}}{T}=\frac{R\int_0^T{I^2(t)dt}}{T}=\frac{\int_0^T{V^2(t)dt}}{RT}\\ $$

Definition of RMS current $$ 5: I_{RMS}=\sqrt{\frac{\int_0^T{I^2(t)dt}}{T}}\\ $$ Square both sides $$ 6: I_{RMS}^2 =\frac{\int_0^T{I^2(t)dt}}{T}\\ $$ Multiply by R to find equation 4 for average power $$ 7: I_{RMS}^2R =\frac{R\int_0^T{I^2(t)dt}}{T}=P_{avg}\\ $$ Definition of RMS voltage $$ 8: V_{RMS}=\sqrt{\frac{\int_0^T{V^2(t)dt}}{T}}\\ $$ Square both sides $$ 9: V_{RMS}^2=\frac{\int_0^T{V^2(t)dt}}{T}\\ $$ Divide by R to find equation 4 for average power $$ 10: \frac{V_{RMS}^2}{R}=\frac{\int_0^T{V^2(t)dt}}{RT}=P_{avg}\\ $$ Multiply expressions 7 and 10 for average power $$ 11: P_{avg}^2=V_{RMS}^2I_{RMS}^2\\ $$ Square root of both sides $$ 12: P_{avg} = V_{RMS}I_{RMS}\\ $$ Q.E.D.