Question

An inductor is connected to an AC power supply having a maximum output voltage of 3.00 V at a frequency of 280 Hz. What inductance is needed to keep the rms current less than 2.50 mA? (Provide the minimum value.)

Answer 1

Answer:

L = 0.48 H

Explanation:

let L be the inductance, Irms be the rms current, Vrms be the rms voltage and Vmax  be the maximum voltage and XL be the be the reactance of the inductor.

Vrms = Vmax/(√2)

         = (3.00)/(√2)

         = 2.121 V

then:

XL = Vrms/I  

     = (2.121)/(2.50×10^-3)

     = 848.528 V/A

that is L = XL/(2×π×f)

              = (848.528)/(2×π×(280))

              = 0.482 H

Therefore, the inductance needed to kepp the rms current less than 2.50mA is 0.482 H.