# Behavior of inductors and capacitors after switch to new source

#### Javri

Joined Jan 14, 2016
5
What happens to the behavior of fully-charged inductor and capacitor elements when they are switched into a circuit with a new source?

The circuit I am curious about involves two resistors in series with a 2A current source and an inductor. There is also a capacitor in parallel with one of the resistors. After a long time, a switch connecting the 2A source changes to a 9V source, effectively replacing sources.

I know that in the original circuit with the current source, the inductor is charged to 2A and acts as a short, while the capacitor acts as open. Right after the switch is made, does the inductor continue acting as a short for that instant with a 2A current flowing through it? And does the capacitor in that moment still act as an open circuit, with an open circuit voltage of however many volts it had right before the switch? I'm inclined to believe so, but I wanted to verify.

#### crutschow

Joined Mar 14, 2008
29,785
If I understand you correctly, no.
The inductor will try to maintain the same current (so the voltage across it will instantly change accordingly) and the capacitor will try to maintain the same voltage (so its current will instantly change accordingly) for any sudden change in the circuit source voltage.
So for an instantaneous voltage change, the inductor acts like an open-circuit and the capacitor acts like a short circuit.

#### Javri

Joined Jan 14, 2016
5
No current will flow through the inductor during the instantaneous voltage change? What current is it trying to maintain then?

#### Jony130

Joined Feb 17, 2009
5,335
The right after the switch, the coil acting as a constant current source, and capacitor is acting as a voltage source.

#### MrAl

Joined Jun 17, 2014
8,984
What happens to the behavior of fully-charged inductor and capacitor elements when they are switched into a circuit with a new source?
The circuit I am curious about involves two resistors in series with a 2A current source and an inductor. There is also a capacitor in parallel with one of the resistors. After a long time, a switch connecting the 2A source changes to a 9V source, effectively replacing sources.

I know that in the original circuit with the current source, the inductor is charged to 2A and acts as a short, while the capacitor acts as open. Right after the switch is made, does the inductor continue acting as a short for that instant with a 2A current flowing through it? And does the capacitor in that moment still act as an open circuit, with an open circuit voltage of however many volts it had right before the switch? I'm inclined to believe so, but I wanted to verify.
Hi,

If you could post a schematic that would be nice as then we could look at it.

For two DC levels that change from one level to another after a long time:

A capacitor acts as an open circuit after long time period, and acts as a short (or voltage source) for short time periods, given a change in voltage level.
The inductor acts as a short circuit after a long time period, and acts as an open circuit (or a current source) for a short time period given a change in voltage level.

The actual current that flows however is based on the initial conditions, BOTH current AND voltage.

So if you have a current source of 2 amps in series with a 1 ohm resistor and 1mH inductor which are also in series, then the VOLTAGE is 2 volts and the current is 2 amps. If you replace the current source with a voltage source of 9v then the inductor will act as an open circuit momentarily, then start to draw more current.

If on the other hand the current is 2 amps and the resistor is 4.5 ohms with the same inductor of 1mH, then the voltage is 9v and so when the current source is replaced with the voltage source of 9v, nothing changes unless the polarity is reversed.

For the time after an instant however, nothing changes for either circuit.

You can figure out the capacitor situation in a similar manner.

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#### crutschow

Joined Mar 14, 2008
29,785
No current will flow through the inductor during the instantaneous voltage change? What current is it trying to maintain then?
Current is flowing.
Why do you think otherwise?
You stated that "the inductor is charged to 2A" when the source is switched.
That is the current the inductor tries to maintain.

#### WBahn

Joined Mar 31, 2012
26,398
I know that in the original circuit with the current source, the inductor is charged to 2A and acts as a short, while the capacitor acts as open. Right after the switch is made, does the inductor continue acting as a short for that instant with a 2A current flowing through it?
In general, these are two contradictory conditions -- the inductor either continues to act as a short (meaning that it will allow whatever current through it that is needed to maintain zero volts across it) or it will continue to have 2A of current flowing through it (meaning that it will act as a current source and produce whatever voltage is necessary across it to maintain 2A of current through it).

The latter is the case.

And does the capacitor in that moment still act as an open circuit, with an open circuit voltage of however many volts it had right before the switch?
Same thing -- two contradictory conditions. An open circuit will allow whatever voltage is necessary to appear across it that is needed to prevent any current flowing through it, while maintaining the same voltage means it acts like a voltage source and allows whatever current through it is necessary in order to keep the save voltage across it.

Again, the latter is the case.

The reasons are the, for an inductor, the energy stored in the inductor is a function of the current through it. Since energy cannot change instantaneously, neither can the current. The same with voltage across a capacitor and the energy stored in it.[/QUOTE]

#### RBR1317

Joined Nov 13, 2010
691 I believe this is the circuit drawn as described. At t=0 the switch shifts from the current source to the voltage source. The concept of a short/open-circuit component is useful at t= ∞, but less so at t=0+. A better concept is to identify which circuit values remain the same from t=0- to t=0+.

As long as there are no impulse functions applied to the circuit at t=0 (impulse functions can cause stored energy levels to change instantaneously), the circuit values which stay the same across a switching transient from t=0- to t=0+ are the inductor currents and the capacitor voltages. Other values in the circuit will adjust themselves according to Kirchhoff's laws to ensure that this is so.

What happens at t=0++ & beyond requires a transient analysis. The circuit shown here will require a second-order differential equation for the transient analysis. But at t= ∞ the capacitor will appear to be an open-circuit (no current flow) and the inductor will appear to be a short-circuit (zero voltage drop).

#### MrAl

Joined Jun 17, 2014
8,984
View attachment 98696
I believe this is the circuit drawn as described. At t=0 the switch shifts from the current source to the voltage source. The concept of a short/open-circuit component is useful at t= ∞, but less so at t=0+. A better concept is to identify which circuit values remain the same from t=0- to t=0+.

As long as there are no impulse functions applied to the circuit at t=0 (impulse functions can cause stored energy levels to change instantaneously), the circuit values which stay the same across a switching transient from t=0- to t=0+ are the inductor currents and the capacitor voltages. Other values in the circuit will adjust themselves according to Kirchhoff's laws to ensure that this is so.

What happens at t=0++ & beyond requires a transient analysis. The circuit shown here will require a second-order differential equation for the transient analysis. But at t= ∞ the capacitor will appear to be an open-circuit (no current flow) and the inductor will appear to be a short-circuit (zero voltage drop).

Hi,

Nice drawing there, very clear and easy to see and read and understand. I wish everybody did that when they talk about a circuit.

Yes, the short and open ideas are just for a circuit with zero energy. But to expand on that isnt too much harder although it does lead to some pretty complicated time domain solutions. The better idea is to use initial conditions generators, which are usually in the form of a battery or current source, for the cap and inductor respectively.

With initial conditions in place, the time domain solution comes out pretty complicated for both voltage across the cap and current through the inductor, so i'll only post one solution for the capacitor voltage here, then we can see how the cap becomes a voltage source at t=0.

Vc(t)=
(e^(-(t*(C*R1*R2+L))/(2*C*L*R2))*(((
((C*R1*R2+L)*((C*E-v*C)*L*R2^2-v*C*L*R1*R2))/(R2+R1)-
(2*C*L*R2*(((C*E-v*C)*R1-i*L)*R2^2+(-v*C*R1^2-i*L*R1+E*L)*R2))/(R2+R1))*
sin((t*sqrt((4*C*L-C^2*R1^2)*R2^2+2*C*L*R1*R2-L^2))/(2*C*L*R2)))/(
sqrt((4*C*L-C^2*R1^2)*R2^2+2*C*L*R1*R2-L^2))-(((C*E-v*C)*L*R2^2-v*C*L*R1*R2)*
cos((t*sqrt((4*C*L-C^2*R1^2)*R2^2+2*C*L*R1*R2-L^2))/(2*C*L*R2)))/(R2+R1)))/(C
*L*R2)+(E*R2)/(R2+R1)

[Note: 'v' is the initial cap voltage, 'i' is the initial inductor current]

That's more complicated than most people really want to get into unless they have to, and that's only one of three possible solutions for the cap voltage in the time domain. This solution assumes that there is some oscillations before damping out hence the sin and cos terms. it can be reduced somewhat, but still wont be very simple without lumping some constants, and then the constants will be complicated.

But anyway to make a long story short, if we set time 't'=0 we get:
Vc(t)=v

and since v is the initial cap voltage, we see that the cap looks like a voltage source for a tiny tiny fraction of time at the very least. It could stay that way however even at t toward infinity depending on the value of the current source and value of the voltage source as indicated in a previous post.

The simple answer is that the inductor maintains the current for a tiny tiny fraction of time, and the capacitor maintains it's voltage for at least a short period of time too, so we have a simpler circuit with the original voltage source and then one more voltage source and one more current source. The various values are then much easier to calculate.