If you plug an inductor with inductance L in your house outlet (120 V), would it dissipate any power? I know the expression for power dissipated by an inductor is (L*i*di/dt) but how would I get the current i?Thanks
actually there will be some power dissipated because we cannot make perfect inductor - at least without using special materials and going to temperatures near absolute zero (superconductivity). this means that any normal inductor will behave as series connection of resistance and inductance. resistance is what will cause losses (heat). inductance would store energy and cause phase shift. you can solve for current using different methods. the most accurate way is to integrate and get time response. in most cases this is overkill and simple phasor analysis is enough. in some cases even resistance is neglected (usually this is rather coarse approximation). p(t)integral v(t)*i(t)*dt
By using the ohms law I E/ XL 120/2pi *f*L 1/pi*L
If you connect an ideal inductor to 120 volt ac outlet the inductor will not dissipate any power. The voltage in an inductor will lead the current by 90°. Power V*I* cos 90 0 watts. The impedance of the inductor L is XL 2π*f*L 377*L. Real inductors have resistance too, this is where power is dissipated in an inductor. In a real inductor the voltage leads the current but not by 90°. The power dissipated in the winding resistance as (i^2)R where R is the winding resistance. Energy stored in an inductor (1/2)(L)(i^2)
If there is no losses in the inductor it is simple. This means no equivalent series or parallel R. Use 2pi x f x L to get the reactance, and then the current by ohms law. There is no power dissipated. Current is drawn but it recirculates without any heating. The inductor charges and discharges. If the losses are significant the calculation needs the impedance Z determined by the vector sum of the equivalent resistive and inductive components. This is often stated as a complex number. See the link to get started.
