Let's calculate the power dissipation in a 100 ohms resistor connected to a 9V battery.

The voltage across the resistor will be 9V. The current is 9V/100ohms = 90mA. So the power dissipation will be: 9V ∙ 90mA = 810mW.

It is very important to calcuate the power dissipation in the components in your design. A regular resistor has a maximum dissipation rating of 0.25W (= 250mW). If you would have used such a resistor in the example above, it would have blown. A 1W resistor is a good choise.

Since it's so important, let's create an equation with which we can easily calculate the power dissipation in a resistor. We know:

(1)** P = V ∙
I** (2)** V
= I ∙ R** (3)** I = V / R**

Substituting (2) in (1) and (3) in (1) respectively results in:

**P = I ^{2}
∙
R P
= V^{2} / R**

With these equations you can easily calculate the power dissipation
when you connect a DC voltage source to a resistor. But what will the
power dissipation be if you connect an AC voltage source to a resistor? In
that case, simply substitute V and I by the so called RMS values
v_{RMS} and i_{RMS}. RMS stands
for Root Mean Square. The RMS value is defined as the DC equivalent that
provides the same power as the original waveform. Let's approximate the
RMS value of a 1V_{t}/1Hz sinusoidal signal: v =
sin(2∙π∙f∙t) = sin(2∙π∙t). We take 4 samples: at 0s, 0.25s, 0.5s and at
0.75s. The values are 0, 1, 0, and -1. Next, calculate the square of each
value: 0, 1, 0, and 1. The mean value of these squares is (0 + 1 + 0 +
1)/4 = 2/4 = 0.5. Finally, calculate the square root of the mean of the
squares: √0.5 = 0.707V. So the approximated RMS value of a
1V_{t} sinusoidal signal is 0.707V. Of course, the
approximation is more accurate if you take more samples. Using some math, you can prove that
v_{RMS} = A/√2 (for a sinusoidal signal). Using this
equation we can calculate the RMS value of the signal of our example:
v_{RMS}=1/√2. = 0.707V.

Using the theory above, we can calculate the power dissipation of a
100ohms resistor connected to a 9V_{t} sinusoidal
signal: P=v^{2}_{RMS}/R =
A^{2}/2R = 81/200 = 0.401W.