Question

An AC source operating at a frequency of 250 Hz has a maximum output voltage of 6.00 V. What is the smallest inductor that can be connected across the source and the rms current remain less than 3.00 mA

Answer 1

Answer:

1.274 H

Explanation:

using

V = XLI...................Equation 1

Where V = voltage, XL = Inductive reactance, I = current.

Make XL the subject of the equation

XL = V/I.............. Equation 2

Given: V = 6.00 V, I = 3.00 mA = 0.003 A

Substitute into equation 2

XL = 6/0.003

XL = 2000 Ω

But,

XL = 2πFL............... Equation 3

Where F = Frequency, L = inductance.

Make L the subject of the equation

L = XL/(2πF).............. Equation 4

Given: F = 250 Hz, XL = 2000 Ω

Constant: π = 3.14

L = 2000/(2×3.14×250)

L = 2000/1570

L = 1.274 H.