What prevents abrupt voltage change in a circuit with a capacitor?

Question

Let's say we have a simple circuit consisting of a power supply and a resistor, and currently the input voltage is 0V. We now apply a voltage of 5V to the circuit (like a step increase - instantaneously). The voltage across the resistor changes instantaneously to 5V.

If a capacitor is introduced into this circuit, it will gradually charge until the the voltage across it is also approximately 5V, and the current in this circuit will become zero.

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My question:
What is now preventing us from suddenly changing the voltage from 5V to let's say 10V (again like a step increase - instantaneously)? We could do it before the capacitor was introduced, but why not now?

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The answer I have thus far always gotten is that for that to happen, according to i=C*(dv/dt), the current flowing in the circuit must be infinite, and since that cannot happen, the voltage cannot be changed instantaneously.

The problem is that we ARE changing the voltage of the power source instantaneously, just like before the capacitor was introduced. The introduction of the capacitor has not somehow taken away our ability to change the voltage of the power source (instantaneously), has it???

So, in this circuit with the capacitor included:

1. Can we change the input voltage instantaneously or not? (theoretically)
2. If we can, what will happen to the current? Will it try to become infinity as i=C*(dv/dt) states? Or is there something else that I'm missing?
3. If we can't, is the capacitor somehow magically preventing this change? How can any component actually affect the input voltage of the circuit?

Answer 1

It's like the "paradox" of the immovable object meeting the irresistible force. In reality, neither can exist.

A real power source will have some impedance. A real capacitor will have some impedance. Real wires connecting them have resistance and inductance.

So in reality when you slap a fairly 'stiff' power source across a fairly good real capacitor there's a spark and the capacitor charges through those series resistances with some ringing and stuff due to the inductances. Ignoring the inductances, the voltage difference would simply divide in ratio to the internal resistance of the capacitor and the internal resistance of the power supply and the wire resistance.

To answer your specific question: If the capacitor and voltage source and wires are ideal, you have a mathematical problem, like division by zero. It's of no consequence in the real world- it just illustrates that the ideal models of the power source and the capacitor and wires are insufficiently accurate to describe their real-world behavior. Modelling any one of those as a real part with real resistance (and inductance) will make the mathematical problem go away, but it won't likely give you an accurate indication of what is actually happening.

For example, if the wire (or the capacitor or the power supply) had 10m\$\Omega\$ resistance, and there is a 10V difference, you could predict you'd see 1000A (which is very high, but not infinity) and the capacitor would charge very quickly. In reality that isn't likely going to happen because of other non-ideal factors. If the 10m\$\Omega\$ was modeled as in the power supply, the power supply voltage would drop. If the 10m\$\Omega\$ was modeled as in the capacitor, the voltage would suddenly appear across the capacitor terminals. If the 10m\$\Omega\$ was modeled as in the wire, the voltage would appear across the wire. But none of those is very realistic.

If you modeled as a circuit with no resistance at all and a tiny bit of inductance (even superconducting wires of any length have inductance) then a simple mathematical model would predict ringing that would persist forever, energy sloshing back and forth between the inductance and capacitance at an angular frequency of \$\omega_0 = {1 \over \sqrt{LC}}\$.

Answer 2

Can we change the input voltage instantaneously or not? (theoretically)

The answer is a qualified yes. Formally, the voltage across the capacitor can be of the form

$$v_C(t) = 5u(t)$$

where \$u(t)\$ is the unit step function. In that case, the capacitor current is

$$i_C(t) = C\cdot 5\delta(t)$$

where \$\delta (t)\$ is the unit impulse 'function' (distribution) which is roughly defined as being zero for all \$t\$ except when \$t=0\$ where the area is equal to one.

However, this solution is maximally non-physical since, in ideal circuit theory, we assume that voltage and current rates of change are small enough that electromagnetic radiation can be ignored.

When we get a solution that is not consistent with this assumption, the solution is non-physical and, thus, of 'academic' interest only.

Even if the voltage source internal impedance could be made arbitrarily small, one cannot avoid the inherent self-inductance of a circuit that encloses a non-zero area.

Thus, to find a physically relevant solution, we must model the self-inductance and radiation resistance of the circuit as is done here for the "two-capacitor paradox".

When this done, we find that the solution for the voltage across the capacitor does not instantaneously change and the circuit current is finite.

Answer 3

I like Spehro's answer, but I think he might be over-reading your question. In this 5V to 10V step is the resistor still in the circuit. If so then the step from 0V to 5V, will look pretty much like the step from 5V to 10. There will first be all the (extra) voltage across the resistor (5 Volts), and the C will start to charge up to the final 10V.

(Or maybe I'm making your question too simple?)