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I was doing a problem where I was asked to find the current in an R-C circuit 'just after' the switch had been closed i.e at t = 0. All of the capacitors are initially uncharged. Could someone explain why the capacitors behave as short circuits and why the resistors in parallel with the capacitors can be 'eliminated' from the circuit when solving the problem?

Thanks!

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In our ideal capacitor model, an uncharged capacitor is assumed to contain nothing that would slow down the flow of current. The device has an instantaneous resistance of zero, corresponding to a short circuit at that particular time (which is also a reasonable approximation for adjacent time points much less than the charging time constant of the circuit).

As we know from Ohm's Law, current flows through a resistor only when there is a potential voltage applied across it. No potential voltage can appear across a short circuit; therefore, no current flows in that parallel link. A resistor with a current of zero acts like an open circuit.

Putting this all together, as we begin to pass current through an uncharged capacitor, it acts like a short circuit, and resistors connected in parallel act like they don't exist. Does this make sense?

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  • $\begingroup$ So as a capacitor charges, its resistance increases? Also would it be correct to use Kirchhoff's loop rule to say that in the loop formed by the parallel assembly, if the voltage across the capacitor is zero i.e q/c is zero therefore iR across the resistor must also be zero hence there is no current in that branch of the circuit. is this a correct way of going about this? $\endgroup$
    – Thomas
    May 21, 2021 at 20:45
  • $\begingroup$ Yes (referring to its effective instantaneous resistance; the capacitor is not Ohmic) and yes. $\endgroup$ May 21, 2021 at 20:50
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    $\begingroup$ I would be very careful in saying a capacitor's resistance increases. That's not quite the case though the voltage it develops across it will try to impede current flow through it in a direction that charges the cap in the same direction. $\endgroup$
    – DKNguyen
    May 21, 2021 at 20:55
  • $\begingroup$ Agreed. Thus my parenthetical comment that I'm referring to V(t)/I(t) and not an inherent device characteristic. $\endgroup$ May 21, 2021 at 21:03
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A uncharged ideal capacitor is a short in that very first instant when current starts to flow through it. Why? In the simple terms: because the voltage across a capacitor cannot change instantly and being uncharged means it starts out with 0V across. The fact it can't change instantly means that the very first instant where current flows through it, it will still have 0V across it just like an ideal wire.

A capacitor's behaviour is described by $I=C\frac{dV}{dt}$ and one implication is the presence of the $\frac{dV}{dt}$ means the capacitor's voltage has to be smooth and continuous from one instant to the next.

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Could someone explain why the capacitors behave as short circuits and why the resistors in parallel with the capacitors can be 'eliminated' from the circuit when solving the problem?

Whether or not capacitors behave as short circuits after a switching event depends on the details of the circuit, i.e., the location of the switch, capacitor(s) and resistors, which you haven't provided.

But the basic rule of thumb is that you cannot change the voltage across an ideal capacitor instantaneously, i.e., in zero time. This means the voltage across an ideal capacitor just prior to switching is the same as the voltage across the capacitor the instant after the switching occurs.

So if there is a voltage across the capacitor before switching there will be a voltage immediately after switching, i.e., the capacitor will not behave as a short circuit.

On the other hand, if the voltage across the capacitor is zero before switching, it will be zero after switching. Meaning that if there are resistor in parallel with the capacitor, they will be shorted ("eliminated" as you say") the instant after switching.

All this behavior of capacitors is based on the relationship between voltage and current in an ideal capacitor, which is

$$i_{C}(t)=C\frac{dV_{C}(t)}{dt}$$

Or,

$$v_{C}(t)=\frac{1}{C}\int i_{C}(t)dt$$

Hope this helps.

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