Regarding first order circuit basic question

[berry1991]

With the statement of:
"The circuit is at steady state before the switch closes"

Does this means when t<0, the inductor is at steady state and it is short circuit. Am I correct?

Then when t>0, the switch is closed and the inductor is also in a steady state?

When t=∞, the inductor is definitely in a steady state. Thus the inductor can be represented with a short-circuit.

 

[gneill]

With the statement of:
"The circuit is at steady state before the switch closes"

Does this means when t<0, the inductor is at steady state and it is short circuit. Am I correct?

The circuit is at steady state, which means that nothing is changing (all currents and potentials are constant). The inductor is behaving as a short circuit and is carrying a constant current.

Then when t>0, the switch is closed and the inductor is also in a steady state?

At the instant the switch closes the circuit is not longer in steady state; it will move towards a NEW steady state from that instant onwards, eventually reaching it after some long time period (t --> ∞). This is called the transient period.

When t=∞, the inductor is definitely in a steady state. Thus the inductor can be represented with a short-circuit.

Yes.

 

[berry1991]

At the instant the switch closes the circuit is not longer in steady state; it will move towards a NEW steady state from that instant onwards, eventually reaching it after some long time period (t --> ∞). This is called the transient period.

So the current before and after the switch is closed:

I(0-) = I(0+)=I(0), but not equal to I(∞)

Is this correct?

Another question:

The following attachment is the circuit diagram:

The answer given for the inductor current[I(∞)] = 0A

From what that I understand, as time=∞,the switch will be closed. So all the current will be flowing through the closed-switch and there will not be any current flow into the 20Ω and the short-circuited capacitor. This explain why I(∞) for inductor is 0A.

Am I correct?

 

[gneill]

So the current before and after the switch is closed:

I(0-) = I(0+)=I(0), but not equal to I(∞)

Is this correct?

It is correct assuming that I(t) refers to the inductor current.

Another question:

The following attachment is the circuit diagram:

The answer given for the inductor current[I(∞)] = 0A

From what that I understand, as time=∞,the switch will be closed. So all the current will be flowing through the closed-switch and there will not be any current flow into the 20Ω and the short-circuited capacitor. This explain why I(∞) for inductor is 0A.

Am I correct?

Closing the switch cuts off the inductor and 20Ω resistor from the energy source (the voltage source). The inductor is left in a circuit consisting of itself and the 20Ω resistor.

The 20Ω resistor will dissipate the energy that was stored in the inductor as the current flows through it.

 

[berry1991]

Closing the switch cuts off the inductor and 20Ω resistor from the energy source (the voltage source). The inductor is left in a circuit consisting of itself and the 20Ω resistor.

The 20Ω resistor will dissipate the energy that was stored in the inductor as the current flows through it.

So in other words, after an amount of time(that is t=infinite) the current in the inductor will be zero due to the resistor dissipate the energy being stored in the inductor till 0?

 

[gneill]

So in other words, after an amount of time(that is t=infinite) the current in the inductor will be zero due to the resistor dissipate the energy being stored in the inductor till 0?

That's it, yes.

 

[berry1991]

That's it, yes.

Now everything is clear to me now. thanks for your cooperation.