0
\$\begingroup\$

Why the transient current will die off in series RL circuit with AC source ?

Every time the ac voltage source reverses direction so does the current and the current has to pass through the same resistance and inductance every time. Voltage source, current, resistance and inductance all are same for every reversed direction and energy that electrons get in each reversed direction is also the same. Then why will the transient current die off with the passage of time ? I need intuition. Can anyone explain this to me please ?

enter image description here

\$\endgroup\$

4 Answers 4

1
\$\begingroup\$

The actual transient (upon applying a sine wave) is a DC level that exponentially decays. It decays because the source is not a DC source but was forced to produce a DC current transient due to the inductor. Given that the source doesn't naturally produce DC, the circulating DC current exponentially decays to zero due to energy dissipation in the series resistor.

If there wasn't a series resistor and you applied a sine wave to a pure inductor, the DC transient would remain without decaying for all time. Then, if you applied a cosine wave, you'd find there is no DC transient. Try it.

Here's my simulation of the circuit that clearly indicates that the peak magnitude of the current is higher in the first half cycle of applied voltage than at any other time: -

enter image description here

And if I lowered the resistor value from 180 Ω to 100 Ω, it's even clearer: -

enter image description here

If I reduced the resistor to zero ohms, the inductor current will never go negative: -

enter image description here

The transient is pure DC and remains pure DC for all time because although the supply source cannot naturally sustain DC (it's an AC source), there is no resistor in the loop to dissipate the acquired DC energy hence, it remains.

If I applied a cosine wave to the inductor we see this: -

enter image description here

There is no transient because an inductor will naturally produce no transient when driven from a "sine" waveform that begins at its peak value. Note that to make this work in the simulator I needed to set the initial current condition of the inductor to zero (.IC I(L1)=0)

\$\endgroup\$
12
  • \$\begingroup\$ "If there wasn't a series resistor and you applied a sine wave to a pure inductor, the DC transient would remain without decaying for all time. " @Andy aka I am aware of that but I do not understand that how the presence of resistor will decay the transients as every time the voltage source reverses we have same resistor and inductor in series. \$\endgroup\$
    – Alex
    Apr 5, 2021 at 21:54
  • 1
    \$\begingroup\$ The resistor burns the initial energy gradually to zero. It is not replenished by the source because the source is alternating. \$\endgroup\$
    – Andy aka
    Apr 5, 2021 at 22:05
  • 1
    \$\begingroup\$ The energy that the source supplies is alternating and not DC so, the source (AC) cannot sustain the DC transient demanded by the inductor during the first half cycle of the sinewave being applied. The inductor is attempting to integrate a full half sine cycle and that's were the anomaly comes in. Because the source is not naturally supplying DC current it can't sustain that initial transient and it gets gradually eaten by the resistor; turning it to heat. \$\endgroup\$
    – Andy aka
    Apr 6, 2021 at 10:10
  • 1
    \$\begingroup\$ Your figure is misleading you. The current into the inductor peaks in the first half cycle higher in magnitude than it peaks in the first negative half cycle. Your figure, as it stands is meaningless because we don't know what the waveforms are and there is no scale to them. I can fully assure you this. \$\endgroup\$
    – Andy aka
    Apr 6, 2021 at 10:26
  • 1
    \$\begingroup\$ In a pure inductor, with a sinewave applied, the current never falls to be a negative value. it always remains positive. So, when others have indicated that the resistor needs to be big to see the transient, they have got this wrong. The transient is pure constant DC with a pure inductor driven from a sine wave. \$\endgroup\$
    – Andy aka
    Apr 6, 2021 at 10:46
2
\$\begingroup\$

The "transient" response dies off simply because that is the definition of transient in this context. The transient response is the result of an event at a specific point in time, such as opening/closing a switch or turning on a source. This is in contrast to the steady-state response, which assumes that the stimulus is a periodic (usually sinusoidal) waveform that has existed forever and will continue forever.

You can have both a transient and a steady-state response if, as you indicate, you suddenly connect a source of any kind.

\$\endgroup\$
1
\$\begingroup\$

In the steady-state-situation there is a constant phase shift between current and voltage (in your example this phase shift is about 54 degrees measured at the voltage source). At the moment you switch on the voltage source the current is zero. At this moment the voltage can have any phase or what's basically the same any value between \$-V_{peak}\$ and \$+V_{peak}\$. Hence the phase between voltage and current when switching on the voltage source will not have the value required by the steady state situation. The transient will reduce the phase mismatch over time.

If you switch on the voltage source at the right moment there will be no transient.

\$\endgroup\$
1
\$\begingroup\$

When you solve the Laplace equation of that linear circuit, for example for the current I of the inductor, you get 2 terms:

The first term is: A * exp(-t * R/L)

The second term is: B * sin((2 * pi * f * t) + phi)

A, B and phi can be determined by the initial conditions.


The first term decay rapidly after some time and that's why it is called the transient term. It disappears after some time.

The second term is called the steady state term.


Build that circuit on a breadboard and put a probe of a scope on a node.

Turn the voltage source on, you won't see the transient term unless R is pretty big.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.