Describe what happens to the impedance of an inductor at very high frequencies qualitatively.thanks in advance!
I'm only answering this because it's a very important concept that isn't always clearly stated in the books. DC conditions: Capacitors become open circuits. Inductors become short circuits. AC conditions: Capacitors become resistors that decrease in resistance aka impedance as the frequency increases. Inductors become resistors that increase in resistance aka impedance as the frequency increases. DC is like a frequency of 0Hz so you can see why the DC conditions are as they are.
For inductors the following relationship holds: v(t) L(di/dt) v(t): time varying voltage L: inductance i(t): time varying current consider the time varying current to be given by: i(t) I?sin(ωt) now v(t) L(di/dt) LI?ωcos(ωt) Impedance is given by: Z v(t) / i(t) LI?ωcos(ωt) / I?sin(ωt) Lωsin(ωt + ?π) / sin(ωt) Thus |Z| ωL ω 2πf So the impedance of an inductor at a high frequency f is given by: (2πf)L Thus the impedance is proportional to the frequency with a proportionality constant equal to 2πL. It should also be mentioned that the phase angle φ by which the voltage leads the current is unaffected by the frequency. φ 90°. The phase angle differs when you look at the combined impedance of an RLC circuit for example. So increasing the frequency results in an increased impedence! Impedence is another form of resistance in a circuit and thus with a frequency approaching infinity the amount of current passing through the inductor is negligible! I hope this helped :D
