An inductor of 30 MHz is connected to an alternating current with a frequency of 50 Hz. Determine the inductive reactance of the coil
Initial data: L (inductance of the considered coil) = 30 mH (30 * 10-3 H); ν (AC mains frequency) = 50 Hz.
The inductive resistance of the coil in question is determined by the formula: XL = 2 * Π * ν * L.
Calculation: XL = 2 * 3.14 * 50 * 30 * 10-3 = 9.42 ohms.
Answer: The inductive resistance of the coil in question is 9.42 ohms.
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