Question

A 450 μH inductor is connected across an AC generator that produces a peak voltage of 5.4 V .

a) At what frequency f is the peak current 42 mA ?

b) What is the instantaneous value of the emf at the instant when iL=IL?

Answer 1

a) The peak value of a sinusoidal function is Vpk = (√2)•(Vʀмs) and it makes
no difference which one you work with so long as you are consistent.

The inductive reactance = Xʟ = (2•π) • f • L


Ohm's law: Iʟ = Vʟ ⁄ Xʟ = Vʟ ⁄ [2•π • f • L]

(0.0420 amps-pk) = (5.4 volts-pk) ⁄ [2•π • f • (450 ×10^-6)]

f = 45.5 kHz

b) The current and voltage for a simple one-inductor circuit are

     

For iL = IL, must be equal to 2nπ, where n = 0, 1, 2, …. This means .

Since the source voltage has the expression . Thus, the instantaneous value of the emf at the instant when iL = IL is V_L = 0 V.