Question

A 18 mH inductor is connected across an AC generator that produces a peak voltage of 11.0 V .

What is the peak current through the inductor if the emf frequency is 100 kHz?

What is the peak current through the inductor if the emf frequency is 100 Hz?

Answer 1

We can solve this problem by using formula of inductive tractable which is

X_L=2πfL.......................(1)

Where: ƒ is the Frequency and L is the Inductance of the Coil and 2πƒ = ω.

Now here in first problem f=100kHz=100×10³Hz

L=18mHz=18/1000=0.018Hz

Now substitute all the value in (1)

X_L=2×3.14×100×10³×18×10^-3=11304ohm

Now peak current through indicator = peak voltage /inductive rectance

Therefore I=V/XL

I=11/11304=0.000973

I=9.73×10⁴A

Now for second question we will do same process but for different value of f

Therefore X_L=2πfL.......................(2)

Where L=18×10^-3Hz, f=100Hz

X_L=2×3.14×100×0.018

X_L=11.304ohm

Now peak current is

I=V/X_L

I=11/11.304=0.973A