New electronics student here.

Can anyone explain what happens to the current when the capacitor opens in an RLC circuit?

Thanks!

 

There are several possible configurations for an RLC circuit. Did you have a specific one in mind?

Current won't flow through open components.

An "open capacitor" will be a capacitor with a broken lead or a bad connection. Or crushed to powder or otherwise missing completely.

 

New electronics student here.

Can anyone explain what happens to the current when the capacitor opens in an RLC circuit?

Thanks!

Welcome to All About Circuits!

Can you post a circuit diagram? As thingmaker3 suggest there are several RLC circuit configurations.

Dave

 

hi
thats link may give you some information about the RLC circuits and it equipment
http://en.wikipedia.org/wiki/RLC_circuit

 

in a series circuit, when any component opens, current wil stop flowing as continuity for the circuit is lost.

 

Thanks for the responses. I appreciate them.

I figured the answer was simply that current would stop flowing, but that seemed too simple of an answer. I was wondering if there was some other phenomenon that occured that I had overlooked.

 

It's kind of a trick question though...

If we're talking strictly DC, and the capacitor hadn't fully charged, but current was flowing through the inductor, the sudden interruption of the current flow would cause the magnetic field around the inductor to collapse, and a large voltage reversal of short duration of the opposite polarity would be observed across it.

 

Doesn't the confusion in the OP arise out of the statement "what happens to the current when the capacitor opens in an RLC circuit" - by definition/design the capacitor is open - there is no "when".

If the circuit was say a Thevenin source with LC parallel arrangement, we could interpret the question as meaning, "what happens to the current when the capacitor branch opens in an RLC circuit"

Hence why the ambiguity could be cleared if there were a circuit diagram so we could ascertain the circuit arrangement.

Dave

 

Hi cljohnson72,

You really have not formulated the question well-enough.

It is likely that the current will not "just stop", since there is an inductor involved, and maybe a capacitor with one end disconnected, or maybe even a portion of a capacitor.

If you want to know exactly HOW the current stops, i.e. obtain an equation or graph of the current versus time, you could do that. But you would need to have a specific schematic, with defined source(s), and also define exactly what "the capacitor opens" means, and then define the location for which you want to characterize the current, and define the time that the capacitor opens.

The solution would involve writing and solving the simple differential equation for the circuit, with the capacitor intact, and then using that solution to provide initial conditions for the modified equation that corresponded to the capacitor being open.

If your circuit were not limited to being DC, it might be interesting to try to get a three-dimensional plot, with current plotted versus both the time since the capacitor opened and the time that the capacitor opened, or maybe at least look for the extrema plots (i.e. depending on the time the capacitor opened). For that type of analysis, it would probably be easiest to simply convert the differential equations to discrete-time difference equations, which are literally trivial to convert to, and trivial to express iteratively in software, and then let a digital computer simulate them and make whatever plots you desire, etc. (Or, you could probably use an already-available circuit simulator, such as the excellent, free LTspice software from linear.com, and possibly also plot the output data with third-party software.) [Sorry to have blathered-on about all of that.]

- Tom Gootee

http://www.fullnet.com/~tomg/index.html

 

It is likely that the current will not "just stop", since there is an inductor involved

would you explain to me how current will continue in an open series circuit. based on the fact that he is a new electronic student, i think he was looking a beginners answer. granted a magnetic field will collapse and induce a voltage potential on the field windings, etc., in a dampened manner. but will current flow in the circuit???

 

New electronics student here.

Can anyone explain what happens to the current when the capacitor opens in an RLC circuit?

Thanks!

Are you sure you mean RLC circuit or RCL? An RCL circuit contains resistance, capacitance, and inductance whereas anRL circuit is only resistance and inductance.

 

Are you sure you mean RLC circuit or RCL? An RCL circuit contains resistance, capacitance, and inductance whereas anRL circuit is only resistance and inductance.

Why would the OP ask what happens to the current when the capacitor opens in an RL circuit? There is no capacitor.

RLC or RCL - doesn't matter, its the arrangement that will affect the behaviour.

Dave

 

would you explain to me how current will continue in an open series circuit. based on the fact that he is a new electronic student, i think he was looking a beginners answer. granted a magnetic field will collapse and induce a voltage potential on the field windings, etc., in a dampened manner. but will current flow in the circuit???

Hi techroomt,

Thanks. With only an ideal inductor and ideal resistor in series, opening the circuit should immediately stop the current. And you are also correct that my post was probably inappropriate for most beginning electronics students.

- Tom