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I am confused about one of the core fundamentals of electrical components. How does a capacitor work under AC conditions? I know that a capacitor has two states (transient and steady.) This happens for DC circuits as well.

Let us assume that we have built an AC RC circuit with a sinusoidal source. Initially the capacitor will be in its transient state as it was completely chargeless beforehand. We already know that a capacitor tries to hinder the change in its potential difference so when a source tries to induce charge, it obstructs and gets charged gradually and hence potential difference across it changes slowly depending on its capacitance. At the initial stage the capacitor shows some weird behavior but eventually it gets stable which we call the steady state of the capacitor. During steady state, the capacitor has its potential difference changed sinusoidally. If the capacitor intends to obstruct the change in its potential difference then why is it able to change that so easily in steady state? It is understandable for lower frequency, but at higher frequency shouldn't it pose some problems as it has to change its potential difference almost abruptly?

Again if it can change so smoothly along a sinusoid then why is there a transient state to even start with? I mean in steady state there is also some point when \$V_C = 0\$ but still it is able to change its voltage so smoothly but at the initial stage it acts weirdly. enter image description here enter image description here

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    \$\begingroup\$ What is the weird behaviour you mention? Capacitors don't have states; they always follow the same simple rule with no weird behaviour. \$\endgroup\$
    – Andy aka
    Dec 29, 2023 at 13:56
  • \$\begingroup\$ I mentioned about the steady state analysis and transient state analysis of capacitor \$\endgroup\$
    – MSKB
    Dec 29, 2023 at 13:58
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    \$\begingroup\$ @MSKB There are no two states. There will be a transient only if you apply a transient yourself. Please show which kind of circuit you are simulating to know where the transient comes from. If you don't want a transient then you need to set up the simulation conditions to start from a scenario as if the simulation had already been running forever or at least long enough that the transient caused by non-steady initial conditions have settled below the level you want. \$\endgroup\$
    – Justme
    Dec 29, 2023 at 14:25
  • \$\begingroup\$ @Justme I know about the initial condition thing but I wanted to know how capacitor works at the initial phase \$\endgroup\$
    – MSKB
    Dec 29, 2023 at 14:28
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    \$\begingroup\$ @MSKB It does not work any differently at any phase, it's just a capacitor. Your circuit just does not start from a steady state. Same as applying a DC step to a capacitor, it takes time for the circuit to settle to new DC conditions. \$\endgroup\$
    – Justme
    Dec 29, 2023 at 14:34

3 Answers 3

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The transient state is there because the voltage source was started at phase zero. That's not where it would be in the steady state when the capacitor's instantaneous voltage was zero.

Look at the phase shift between the voltage source and the capacitor voltage in the steady state.

Since this is an RC circuit, the voltage source and capacitor voltage are two separate waveforms. It helps to plot them both at the same graph - you'll see how the phase shift stabilizes in the steady state.

Then all you have to do to avoid the startup transient is to start the voltage source at the correct phase - same it would have in the steady state instantaneously at the times when the capacitor voltage is zero.

When plotting, fit fewer cycles on the graph, so it's easier to see the relative phase of the two waveforms.

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A capacitor follows this rule: -

$$I = C\dfrac{dv}{dt}$$

And basically this means that it takes a higher current when the rate of change of voltage applied to its terminals is greater. This governs how a capacitor works all the time. So, a higher frequency voltage applied to its terminals produces a higher RMS current through the capacitor.

DC has zero rate of change of voltage hence, a capacitor takes no current but, if you turned-up the DC voltage, the capacitor "resists" that change by taking a lot of current until the voltage ceases changing. Then the capacitor takes zero current again. The faster the change in voltage, the higher the current taken by the capacitor.

No weirdness.

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A capacitor always does the same thing, it integrates the current to voltage, via a constant of proportionality called the capacitance.

'Steady state' and 'transient' are our human descriptions, designed to simplify describing this behaviour.

If we apply a step input, then it's simplest to describe that as a transient, and fit the response to decaying exponentials.

If we apply a single sinusoidal input, then it's simplest to generalise Ohm's Law to AC, define an impedance for the capacitor at that frequency, and write AC_voltage = AC_current * capacitor_impedance.

Of course depending on the instant of connection of the sinewave, what phase it is when we connect it, we may also have a step in voltage. The capacitor doesn't care, it just integrates current to voltage as usual. We find it mathematically simpler though to separate its response into a transient response to the step, and a steady state response to the sinewave.

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  • \$\begingroup\$ This is true, but it doesn't really solve the OP's problem. What OP essentially asks is why is there a transient state that differs from the steady state. There is a solution to that :) \$\endgroup\$ Dec 29, 2023 at 14:13

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