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Power dissipation in a resistor

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 = I2 ∙ R             P = V2 / 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 vRMS and iRMS. 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 1Vt/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 1Vt sinusoidal signal is 0.707V. Of course, the approximation is more accurate if you take more samples. Using some math, you can prove that vRMS = A/√2 (for a sinusoidal signal). Using this equation we can calculate the RMS value of the signal of our example: vRMS=1/√2. = 0.707V.

Using the theory above, we can calculate the power dissipation of a 100ohms resistor connected to a 9Vt sinusoidal signal: P=v2RMS/R = A2/2R = 81/200 = 0.401W.

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