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Do capacitors act as an open circuits or closed circuits at time t=0? Why? What about inductors?

I tried it out, and what I got was this: Initially when I opened the switch, the capacitor acted like a short circuit. That should not be happening, right? A capacitor should block DC. I tried with a couple different caps. I am very confused.

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    \$\begingroup\$ What about WHAT inductor? It would probably be best to provide details for the network in question. Also, if you have access to a lab I suggest trying it out. Seeing it really helps you grasp what's going on. \$\endgroup\$ – Lou Dec 3 '09 at 14:57
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    \$\begingroup\$ A capacitor looks like an open circuit to a steady voltage but like a closed (or short) circuit to a change in voltage. And inductor looks like a closed circuit to a steady current, but like an open circuit to a change in current. \$\endgroup\$ – Chris Stratton Nov 30 '11 at 21:36
  • \$\begingroup\$ You probably should put this as an answer, as I believe that is what the OP is looking for. Perhaps with a brief explantion as to why (cap charging and magnetic fields and such). \$\endgroup\$ – Tevo D Dec 8 '11 at 18:49
  • \$\begingroup\$ @Tuva - Thanks, though I can't take all the credit - it was an improvement on a suggested edit. \$\endgroup\$ – Kevin Vermeer Dec 8 '11 at 21:15
  • \$\begingroup\$ @ChrisStratton I think it would be much easier for the OP to understand if you speak of the characteristics of these circuit elements in terms of their impedance in different applications instead of memorizing what they 'should' be. Although, this post is old so he most likely got it. \$\endgroup\$ – sherrellbc Aug 2 '13 at 12:20

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Short Answer:

Inductor: at t=0 is like an open circuit at 't=infinite' is like an closed circuit (act as a conductor)

Capacitor: at t=0 is like a closed circuit (short circuit) at 't=infinite' is like open circuit (no current through the capacitor)


Long Answer:

A capacitors charge is given by \$Vt=V(1-e^{(-t/RC)})\$ where V is the applied voltage to the circuit, R is the series resistance and C is the parallel capacitance.

At the exact instant power is applied, the capacitor has 0v of stored voltage and so consumes a theoretically infinite current limited by the series resistance. (A short circuit) As time continues and the charge accumulates, the capacitors voltage rises and it's current consumption drops until the capacitor voltage and the applied voltage are equal and no current flows into the capacitor (open circuit). This effect may not be immediately recognizable with smaller capacitors.

A nice page with graphs and some math explaining this is http://webphysics.davidson.edu/physlet_resources/bu_semester2/c11_rc.html

For an inductor, the opposite is true, at the moment of power-on, when voltage is first applied, it has a very high resistance to the changed voltage and carries little current (open circuit), as time continues, it will have a low resistance to the steady voltage and carry lots of current (short circuit).

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  • \$\begingroup\$ In the inductor, where does the back EMF come from at t=0? It seems like at this moment, you need some current o flow to create the change in the magnetic field, but if resistance is infinite at that moment then no current? \$\endgroup\$ – bigjosh Nov 7 '17 at 18:01
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Inductance and capacitance are effects that limit rate of change. Once things have settled out, there is no more change, and they have no further effect. So in the long-term, steady-state, capacitors and inductors look like what they are; they act like you'd expect them to act if you knew how they were constructed, but didn't know capacitance or inductance even existed.

An inductor is a wire. After it saturates the core, it behaves like a short circuit.

A capacitor is a gap between two conductors. After it charges, it behaves like an open circuit.

Their instantaneous behavior is the opposite. Until they charge, a cap acts like a short circuit, and an inductor acts like an open circuit.

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When you turn on an ideal switch from an ideal voltage source, to an ideal capacitor you get some odd solutions, in this case infinite current for an infinitesimal time. So it looks like a short for no time.

More realistic solutions include more ideal element to model the real world, the first might be a series resistance.

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For an uncharged capacitor connected to ground the other pin (the side of the switch) is also at ground potential. At the instant you close the switch the current goes to ground, that's what it sees. And the current is the same as when you would connect to ground without the capacitor: a short-circuit is a short-circuit.

That short-circuit current quickly drops when this big charge has to find it's way through the capacitor's series resistance to charge it.

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For capacitor:

$$V(t)=V(1-e^{(-t/RC)})$$

At \$t=0\$, \$V=0\$ so the capacitor behaves as an short circuit.

$$i(t)=\frac{V}{R} \cdot e^{(-t/RC)}$$

At \$t=\infty\$, \$i=0\$ so the capacitor behaves as an open circuit.


For inductor:

$$i(t)=\frac{V}{R} \cdot(1-e^{(-Rt/L)})$$

At \$t=0\$, \$i=0\$ so the inductor behaves as an open circuit.

$$V(t)=V_e^{(-Rt/L)}$$

At \$t=\infty\$, \$V=0\$ so the inductor behaves as an short circuit.

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Because capacitors store energy in the form of an electric field, they tend to act like small secondary-cell batteries, being able to store and release electrical energy. A fully discharged capacitor maintains zero volts across its terminals, and a charged capacitor maintains a steady quantity of voltage across its terminals, just like a battery. When capacitors are placed in a circuit with other sources of voltage, they will absorb energy from those sources, just as a secondary-cell battery will become charged as a result of being connected to a generator. A fully discharged capacitor, having a terminal voltage of zero, will initially act as a short-circuit when attached to a source of voltage, drawing maximum current as it begins to build a charge. Over time, the capacitor’s terminal voltage rises to meet the applied voltage from the source, and the current through the capacitor decreases correspondingly. Once the capacitor has reached the full voltage of the source, it will stop drawing current from it, and behave essentially as an open-circuit.

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I like to think of these in terms of their differential equations. Essentially the instantaneous equations for each are:

\$V = L \cdot \dfrac{dI}{dt} \$ for inductors

\$I = C \cdot \dfrac{dV}{dt} \$ for capacitors

From their differentials you can see for rate of current change \$\Big(\dfrac{dI}{dt}\Big)\$ you can get unlimited instant voltage changes across an inductor. The induced voltage across the inductor is the derivative of the current through the inductor: that is, proportional to the current's rate-of-change with respect to time.

Likewise for capacitors you can get large current changes based on the rate of change for voltage \$\Big(\dfrac{dV}{dt}\Big)\$. In your experiment the voltage was changed almost instantly say from 0V to 1V in 1us. This makes \$I = C \cdot \dfrac{1}{0.000001} \$ (which you can see could be quite large).

It is the differential terms for these components that make them interesting. Thus the higher the rate of change the bigger the V spike on inductors, or I spike in capacitors. Whereas the current for inductors and voltage for capacitors are limited to what is applied.

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The capacitor acts as open circuit when it is in its steady state like when the switch is closed or opened for long time.

As soon as the switch status is changed, the capacitor will act as short circuit for an infinitesimally short time depending upon time constant and after being in that state for some time it'll again continue to behave as open circuit. And for the inductor it'll behave as a short circuit in its steady state and open circuit when there's a change in the current.

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Capacitor acts like short circuit at t=0, the reason that capacitor have leading current in it. The inductor acts like an open circuit initially so the voltage leads in the inductor as voltage appears instantly across open terminals of inductor at t=0 and hence leads.

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You can check out my video that talks about this (step response) here:

https://www.youtube.com/watch?v=heufatGyL1s

Basically, a capacitor resists a change in voltage, and an inductor resists a change in current. So, at t=0 a capacitor acts as a short circuit and an inductor acts as an open circuit.

These two short videos might also be helpful, they look at the 3 effects of capacitors and inductors:

https://www.youtube.com/watch?v=m_P1rvhEeiI&index=7&list=PLzHyxysSubUlqBguuVZCeNn47GSK8rcso&t=101s

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simply remember capacitor rises voltage from 0 to high,so at intitally at ov capacitor acts as short ckt and for high voltage cap acts as open ckt, reverse in case of inductor

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  • \$\begingroup\$ This is not a correct definition. The current flow in capacitors depends on the rate of change of voltage, not the absolute voltage. The current in an inductor depends on the integral of voltage, not the absolute voltage. \$\endgroup\$ – Joe Hass Oct 17 '13 at 10:02
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    \$\begingroup\$ @JoeHass: The answer is badly worded, but it is not fundamentally incorrect. \$\endgroup\$ – Dave Tweed Sep 19 '14 at 13:51

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