Hello,i need some help solving this one. i'm kind of lost of which formula to apply. Thanks so much for your help. Answer Key (B) A 45-mH inductor is connected to an ac source of emf with a frequency of 400 Hz and a maximum emf of 20 V. The maximum current is:A) 0B) .018 AC) 1.1 AD) 360 AE) 2300 A
The reactance (XL) of an inductor is given by: XL 2πfL where f is the frequency in Hz and L is the inductance in henrys. XL 2π * 400 * 45/1000 XL 113 ohms Now it is just Ohm's Law i v / XL i 20 / 113 i 0.177 A i ≈ 0.18 A { I suspect a typo for answer (B) it should be 0.18 A }
