Electric power initially can be calculated by the few equations:

$$
begin{equation}
P=I V
end{equation}
$$

$$
begin{equation}
P=I^{2} R
end{equation}
$$

$$
begin{equation}
P=frac{V^{2}}{R}
end{equation}
$$

Following written we can see there are multiple ways to calculate electrical power. All of them are valid and the first one can be used for $textbf{all}$ electrical system, as any electrical system is going to have voltage and current flowing. There second and third can be used when we have information about the resistance.

So, it is a matter of practicality. Even sometimes we can calculate current if we know the voltage and resistance, but the extra calculation is not needed. Because, if we need only power we can directly calculate it from the voltage and resistance.

$$
P=I V
$$

$$
P=I^{2} R
$$

$$
P=frac{V^{2}}{R}
$$

The unit kilowatt-hours is a unit that is measuring $textbf{the energy}$. A common misconception is the electricity bill that is paid when it is often said we pay for the current, but we actually are paying energy.

We know that kilowatt-hour is nothing more than the multiplication of power and time interval, which is exactly the energy, or equal to Joules.

$$
1 mathrm{kWh} =3.6 times 10^{6} mathrm{~J}
$$

Because the charge is present in the electrical field, when it is moving across different electric potential the energy will start to differ (as different force starts to be exerted on the charge). Which is equal to:

$$
text { change in electric potential }=frac{text { change in electric potential energy }}{text { charge }} Rightarrow Delta V=frac{Delta P E}{q}
$$

And this change in potential energy would be equal to the:

$$
boxed{color{#c34632}Delta PE = qV}
$$

We have a change in electrical potential energy, $Delta P E = qV$

When operating in same voltage, the power of a light can be given by

$$
P=frac{V^2}{R}
$$

So we can see, that, in this case the power if inversely proportional to the resistance. Hence light A has less resistance than light B.

Here we find the power dissipated from the electric heater if the resistance of the heater is 25 $Omega$ and it is connected to a 120 V outlet. We use the following relation:

$$
P=frac{V^{2}}{R}
$$

Putting in the numbers we have:

$$
P=frac{(120 mathrm{V})^2}{25 Omega}
$$

which gives the result of:

$$
boxed{color{#c34632}P = 576 mathrm{W}}
$$

$$
P = 576 mathrm{W}
$$

In this problem, knowing the electric power and resistance in the electric system, we calculate how much power is dissipated. A current of 2.1 A is going through an 85 $Omega$ resistor. We calculate the power with the following relation:

$$
P=I^{2} R
$$

Putting in the numbers we have:

$$
P=(2.1 mathrm{A})^2 cdot 85 Omega
$$

We have the result of:

$$
boxed{color{#c34632}P = 374.85 mathrm{W}}
$$