Question

how to understand that an inductor behaves like short circuit and capacitor like an open circuit in steady state in a network

Answer 1

It depends on the steady-state frequency.

At DC (zero frequency) an inductor behaves like a short circuit and a capacitor behaves like an open circuit. As the frequency increases, the inductor acts more and more like an open circuit and a capacitor acts more and more like a short circuit.

The physics is pretty easy to see for a capacitor.Think of the capacitor as two metal plates separated by a dielectric. Apply a voltage across the capacitor. The plates build up a charge Q = CV, where C is the capacitance (in farads) and V is the applied voltage. (The capacitor is storing energy in an electric field.) Differentiating both sides with respect to time (and assuming the capacitance remains constant) we get:
dQ/dt = CdV/dt

But dQ/dt is just the change in charge with respect to time, which is defined as the current (I); so, dQ/dt = I = CdV/dt. So, suppose we connect a sine wave voltage source across the capacitor:
\[V=V _{0}\sin \omega t\]
The current is then:
\[I=\omega C V _{0}\cos \omega t\]

Look at the ratio of the amplitude of the voltage divided by the amplitude of the current:
\[V/I=1/\omega C\]
This is the "resistance" (actually called "impedance") of the capacitor. When \[\omega = 0\] the impedance becomes infinite (an open circuit). On the other hand, if the frequency gets very high, the impedance gets very low.

The inductor stores energy in a magnetic field, which is proportional to the current through the inductor times the value of inductance. For an inductor with an inductance of L henrys, the V-I relation is V = L dI/dt.

Suppose we have a circuit with a current source driving an inductor with inductance L, where the source current is given by:
\[I=I _{0}\sin \omega t\]
Then the voltage across the inductor is
\[V=\omega LI _{0}\cos \omega t\]

Now the ratio of the amplitudes, i.e., the effective impedance, is given by:
\[V/I=\omega L\]
So at zero frequency (DC), the inductance has zero impedance (a short circuit)

Note that in the second example it was specified that the inductance was driven by a current source to allow for the possibility of direct current. A good way to blow out a transformer, which is effectively two coupled inductors, would be to connect a DC voltage source across one of the windings.

So to recap, the answer to your question is that the steady-state behavior of capacitors and inductors depends on the steady-state frequency:
- at DC (zero frequency), an inductor acts like a short circuit and a capacitor acts like an open circuit
- at infinite frequency, an inductor acts like an open circuit and a capacitor acts like a short circuit

Answer 2

thanks a lot