# RL High-Pass Filters

MathML formulaFormula ImageStraight text formula
$f c = R 2πL$ fc = R / 2πL

Where

fc = cut off frequency, Hertz

= that frequency at which output power has fallen by half

= that frequency at which output voltage has fallen by 1/(2^0.5) = 0.707

R = resistance in Ohms

L = inductance in Henrys

The cut off frequency is defined as that frequency at which the output power has fallen by 50%, ie 1/(2^0.5) = 0.707 of the voltage.

I intend to build the circuit below to test my understanding of RL high-pass filters. The 1 Ohm resistor makes little difference to the operation of the circuit, but can be used as a convenient way to measure current (a high input impedance voltmeter connected across the 1 ohm resistor will give a reading in volts that corresponds 1:1 with the current in amps).

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