Reply to "Insulated Rail Relay Based Sensor"

stan2004OGR Forum Member

For DC or low-frequency (60 Hz), the power dissipation in an inductor is I^2 x R just as it would be for a resistor. As GRJ says, the inductor's resistance is 0.45 ohms. So if current is 0.7A, power is 0.7A x 0.7A x 0.45 Ohms = 0.22 Watts.

An ideal inductor has 0 resistance (and hence dissipates 0 power) so in the example given in addition to stealing 0.22 Watts from the output, the real-world inductor drops the voltage going into the regulator by V = I x R or 0.3 Volts (0.7A x 0.45 Ohms).

3/28/175:07 PM

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