Answer :
Final answer:
The current through a 0.510 H inductor connected to a 59.1 Hz, 474 VAC source is approximately 2.5 A. When the frequency is increased to 97 kHz, the current decreases significantly to about 1.5 mA.
Explanation:
The current through an inductor in a AC circuit can be calculated using the formula: I = V/Z, where V is the voltage, and Z is the reactance. The reactance of an inductor (Z) can be calculated using the formula: Z = 2πfL, where f is the frequency and L is the inductance.
For part a, the reactance is Z = 2π * 59.1 Hz * 0.510 H = 189.3 Ω. Therefore, the current is I = 474 V / 189.3 Ω = 2.5 A (rounded to one decimal place).
For part b, we need to consider frequency in kHz, so the reactance is now Z = 2π * 97000 Hz * 0.510 H = 310717 Ω. Therefore, the current at 97kHz is I = 474 V / 310717 Ω = 0.0015 A, or 1.5 mA when converted to milliamps.
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