# Explanation of the current-voltage relationship of a capacitor

I came across the following statement in a textbook:

The voltage on a capacitor cannot change abruptly. According to

$$i(t)=C\frac{dV}{dt}$$

a discontinuous change in voltage requires an infinite current, which is physically impossible.

How does this relationship prove that a discontinuous change in voltage requires an infinite current?

• What would dV/dt be for a discontinuous step in voltage ? – jp314 Jan 4 '16 at 2:36
• This site will never cease to amaze me. I never thought that anyone could downvote such a question. Wow! – Adam Jan 4 '16 at 2:41
• @Adam you may have got the downvote because this question, at it's root is a mathematical one and your lack of understanding appears to be one of you not realizing that the derivative of a step change is infinity. On this basis alone I am voting to close the question but that's just a personal thing and no down-vote necessary. – Andy aka Jan 4 '16 at 9:38
• I'm voting to close this question as off-topic because it's about math and not EE. – Andy aka Jan 4 '16 at 9:38
• To @Andyaka and other potential people voting to close this question: I beg to differ. The question is on-topic as stated in the help page as it verses about the theory and simulation of electromagnetic forces. The second Adam mixed his equation, its discontinuity and its tendency to infinity to voltage and current, the question became one about theory of electromagnetic forces. – Ricardo Jan 12 '16 at 16:31

How does this relationship proves that a discontinuous change in voltage requires an infinite current?

First, note that the equation given in your question defines an ideal (non-physical) capacitor and so this is the context of my answer.

Second, note that if the capacitor voltage is discontinuous at some instant(s) of time, the time derivative of the voltage does not exist there.

However, one can approximate a discontinuity by, e.g., letting the voltage change linearly with time over some short interval. For example, let the capacitor voltage change linearly from $0\mathrm V$ to $1\mathrm V$ in $\Delta t$ seconds.

Then, according to the ideal capacitor equation, the capacitor current during the transition is

$$i(t) = C\frac{1 \mathrm V}{\Delta t}$$

In the limit as $\Delta t \rightarrow 0$, the capacitor voltage becomes discontinuous (finite change in zero time) and the capacitor current goes to an infinity large, infinitesimally short pulse; a current impulse.

But this is academic since physical capacitors obey the ideal capacitor equation only approximately and over a relatively narrow region of operation.

You can do a pretty simple rational explanation without bringing calculus into it by going to another (related) capacitor equation:

$$Q=C \times V$$

or

$$V = \frac{Q}{C}$$

The capacitor voltage is a result of the charge on it, not the cause of it. In order to change the voltage on the capacitor, you would need to add or remove charge from it... which is physically a current.

Infinite current might be imagined as charge popping into existence on the capacitor -- but any real current would manifest as charge carriers traveling to the capacitor. The charge carriers will begin to affect the voltage on the capacitor as they approach, and this will cause a non-instantaneous change in voltage, no matter how fast (below infinite speed) the charge carriers are traveling.

Paraphrasing Feynman, 'Nature is the thing that we are studying. The math just describes the way nature already works.'

A lot of good answers, but they don't really answer the question. $$i(t) = C\frac{dv}{dt}(t) = C v'(t),$$ or by using the dot notation $$i = C \dot{v}.$$

"The voltage on a capacitor cannot change abruptly. According to .. a discontinuous change in voltage requires an infinite current, which is physically impossible."

The voltage rate-of-change (i.e. Volts per second) is directly proportional to the current; $$\dot{v} = \frac{1}{C} \cdot i,$$ so if the current jumps, then the rate-of-change jumps.

How does this relationship prove that a discontinuous change in voltage requires an infinite current?

Mathematically, this is actually a deep and very involved subject involving the Newton–Leibniz axiom following the Fundamental Theorem of Calculus. Essentially it tells us that as long as $v'(t)$ is Riemann integrable, then $$\int_a^b v'(t) dt = v(b) - v(a),$$ and also that the function $$v(t) = \int_a^t v'(\tau)\,d\tau$$ is continuous.

From $$v(t) = \int_0^t v'(\tau)\,d\tau = \frac{1}{C} \int_0^t i(\tau)\,d\tau,$$ we see that the voltage must be continuous if the current is Riemann integrable. Physical (real) signals are certainly Riemann integrable, from which we conclude that $v(t)$ must be continuous. Note: If $i(t)$ is assumed to be piece-wise continuous, then there's an even stronger conclusion; that $v(t)$ is absolutely continuous.

A discontinuous change in voltage requires the voltage to change in no time. (Draw a jump in voltage - it'll look like there's a slope of infinity between the two points.) That means that $\frac{dV}{dt}$ at that time is infinite, so your equation tells you that infinite current is required.

You'll only get an infinite current if you have a perfect voltage source and the wires have zero inductance. two other things which are also physcally impossible.

The relationship Q=CV (charge in the capacitor equals capacitance times voltage), leads to the reasoning that a step change in voltage would cause a step change in charge, thus an infinite current.

Real world devices only approximate the ideal described by that relation, typically also having internal resistance and inductance which reduces the current to something finite.

Still, an "approximately infinite" current can do real damage in carelessly designed circuits. Earlier today someone was asking about failed rectifiers in a circuit that had much more capacitance than was reasonable.