As we know that in steady state capacitor acts as OC and inductor acts as SC. But in steady state AC we deal with term like jXL or -jXC and uses phasor and other stuffs to determine circuit parameters like current and voltage. How it can happen when already all the values impedance values are either zero(in inductor) and infinite(in capacitor) ?

• You should not confuse 'steady state' and 'DC'.
– RJR
Commented Sep 6, 2016 at 9:58
• @Rohit Ambasta: adding to RJR's comment: steady state means that there is a steady sinusoidal signal, i.e. a sinusoidal signal of constant amplitude and constant frequency; not DC. Capacitors act as OC for DC but not for steady state signals.
– Curd
Commented Sep 6, 2016 at 10:25
• Then why we r using DC batteries in case of steady state AC analysis or transient analysis ? Ex. We are using supply of 5V for solving circuit parameters. Commented Sep 6, 2016 at 11:08
• That means under steady state only there will be a fixed frequency and so capacitor and inductor have fixed reactance.Also they will behave like OC and SC respectively only under DC supply but not in steady state.Am I Right? Commented Sep 6, 2016 at 13:04
• @Rohit Ambasta: yes, you got it!
– Curd
Commented Sep 6, 2016 at 14:03

As we know that in steady state capacitor acts as OC and inductor acts as SC.

This is true only for DC analysis, where "steady state" refers to values of voltage and current that are not changing at all.

But in steady state AC we deal with term like jXL or -jXC and uses phasor and other stuffs to determine circuit parameters like current and voltage. How it can happen when already all the values impedance values are either zero(in inductor) and infinite(in capacitor) ?

"Steady state" in AC analysis is a completely different concept. It means that the parameters of the excitation such as frequency and amplitude are not varying. It does NOT mean, however, that the instantaneous values of voltage and current are not changing, or that you can continue to model inductors and capacitors as shorts and opens. You need to evaluate their actual complex impedances at the excitation frequency. Those values are finite and nonzero, and they remain constant as long as the frequency doesn't change.

• Nope: steady state does not refer to instantaneous voltage and current values, but to the assumption that phase and amplitude don't change (any more). Those quantities make sense only with AC.
– Curd
Commented Sep 6, 2016 at 12:37
• @Curd: Did you actually read the question? You seem to be addressing a point that the OP didn't raise. Commented Sep 6, 2016 at 13:31
• Of course I read the question and, yes, I (and RJR) addressed a point the OP didn't raise: the point that steady state does not mean that voltages and currents aren't changing (DC). It seems that the OP is subject to this misconception and that is the key to his question.
– Curd
Commented Sep 6, 2016 at 14:00

lets consider the impedance of an an inductor or capacitor with a steady state AC (sinusoidal) voltage applied across it.

We are considering $Z = \dfrac{V}{I}$

Now for a capacitor the general case for all waveforms is

$$I = C \cdot \dfrac{\text{d}}{\text{dt}}V$$

Considering a pure sine wave the current will be a pure cos function and the current will be proportional to the frequency leading the current by $90^o$

$$Z = \dfrac{-j}{2 \cdot \pi \cdot C} \Rightarrow I = \dfrac{V}{Z} = j \cdot V \cdot \pi \cdot f \cdot C$$

Now if we consider DC to be a special case of AC where the frequency is zero the current will be zero and a capacitor will look like an open circuit.

For an inductor

$$V = L \cdot \dfrac{\text{d}}{\text{dt}}I \Rightarrow I = \dfrac{1}{L} \cdot \int V \text{ } dt$$

Giving

$$Z = j \cdot 2 \cdot \pi \cdot f \cdot L$$

So the current will lag the voltage and taking DC as zero frequency the impedance will be a short circuit.