and whyalso does that work when you open circiut a capacitor.
An ideal inductor appears as a short to DC. Real inductors have some resistance due to the wire they are wound with. Remember the voltage across an inductor V L di/dt if I is constant V 0. An ideal capacitor will appear open to DC. It will charge up to the DC voltage that would appear across it if it were not in the circuit (thevenin voltage). Ic CdV/dt. if V is constant Ic is zero. Electolytic capacitors have some leakage current.
Remember that when a component is shorted, the situation is effectively the same thing as taking a wire from one terminal of the component to the other. That is, it's like having a wire in parallel with the component itself. Current flows through the short (or wire) instead of through the component. This is as true in AC circuits as it is in DC circuits. So if an inductor (or any other component) is shorted, then any component in parallel with it will also be shorted because it will also not have any current flowing through it. Components in series will not be shorted provided the short starts at one of the inductor's terminals and ends the other.
