Why does a capacitor act like a short-circuit during a current impulse?

Question

While trying to solve questions involving impulses and step functions, we are supposed to assume that an uncharged capacitor or an uncharged inductor acts as a short circuit and open-circuit respectively. But, I don't see the theoretical reasoning behind it. Furthermore, can an impulse show up against a capacitor or inductor with only a step source?

Answer 1

The basic equation for a capacitor is charge, Q = voltage, V x capacitance, C. If this is differentiated we get: -

\$\dfrac{dQ}{dt} = C\cdot\dfrac{dV}{dt}\$

And rate of change of charge is current, therefore: -

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

A voltage impulse has very high dV/dt therefore I is also very high.

This to me, represents a short when dV/dt is infinite.

I'll leave you to use the formulas for an inductor to satisfy your curiosity and my laziness.

Answer 2

Why does a capacitor act like a short-circuit during a current impulse?

It doesn't act like a short circuit for a current impulse. Here's the equation that defines the ideal capacitor:

$$i_C(t) = C\cdot \frac{d}{dt}v_C(t) $$

Applying the Laplace transform to this equation (assuming zero initial conditions) yields

$$I_C(s) = sC\cdot V_C(s)$$

The Laplace transform for the unit impulse is

$$\delta(t) \Leftrightarrow 1$$

Thus, if the capacitor current is the unit impulse, the Laplace voltage is

$$V_C(s) = \frac{1A}{sC}$$

Going back to the time domain, we have

$$v_C(t) = \frac{1}{C}u(t)\, V$$

Thus, the voltage across the capacitor is a scaled step.

If we instead approximate the unit current impulse with a current pulse

$$i_C(t) = \frac{1}{T}[u(t) - u(t - T)] $$

then the capacitor voltage is

$$v_C(t) = \begin{cases} 0 & \text{if } t \le 0\\ \frac{t}{CT} & 0 \lt t \le T\\ \frac{1}{C} & t \gt T \end{cases} $$

In any case, this is different behavior than that of an ideal short circuit where the voltage across is zero for any current through.

Answer 3

Andy gives a great answer addressing the motivation for such statements. This answer will be the theoretical answer requested in the post.

Given an input voltage function \$V(t)\$ then the current through the capacitor is \$I(t) = C \frac{dV(t)}{dt}\$.

Strictly speaking if \$V(t)\$ is the unit step function then \$I(t)\$ is not defined at \$0\$. The impulse "function" is even more problematic in that it is not even a function.

We basically have two options:

1) Realize that while \$C \frac{d}{dt} \$ is not strictly defined on this space it is defined on the space of differentiable functions. However, this operator is linear on this dense set and extends uniquely to a linear operator on the entire space of functions. In layman's terms, we can estimate the step/impulse functions by differentiable functions \$f_i(t)\$ and then \$C\frac{df_i(t)}{dt}\$ will estimate \$C\frac{dV(t)}{dt}\$.

2) Use that the space of voltage functions, with appropriate mathematical adjectives, is self-dual to figure out how one can write this in the language of distributions.

Option 1) is pretty clear. Any reasonable estimates of the step or impulse functions will clearly have a derivatives approaching infinity at 0.

Option 2) obviously takes more study but the answer is interesting to think about. As a distribution the impulse functional \$\delta\$ is defined as \$\delta(f)=f(0)\$ for all voltage input functions \$f\$ and the resulting current functional is (uniquely defined by) \$(C\frac{d \delta}{dt})(f) = C\frac{df}{dt}|_{t=0} \$ for all (smooth) voltage input functions \$f\$. This is why you will see an "instantaneous" infinite short current when you feed in the impulse function. You can do a similar computation for the unit step function.

Answer 4

First, you must understand what you are dealing with in terms of voltage and current. A capacitor is an electrical component across which the voltage can only change in a continuous manner; i.e. there can be no 'instant' jumps from one voltage to another. This is always true whether the capacitor is charged or not. This happens because the capacitor is designed to store voltages on its plates: as a external voltage is applied across a capacitor, it starts charging or discharging until it matches the voltage.

Similarly, an inductor forces the current going through it to always be continuous, regardless of whether it is charged or not because it is storing the charge in its magnetic fields.

The question specified uncharged capacitors and inductors as a means of setting both to voltage on the capacitors and the current on the inductors to zero before the impulse or steps occur.

So what happens when a step function first hits a capacitor? The voltage across the capacitor, which had been zero, cannot change instantly, so it stays at zero, while the current through it changes instantly to match the step function. For that instant, this is exactly the same behavior any wire or short-circuit would have.

A step function hitting a induction results in an instant change in voltage while the current flowing through remains at zero. This is exactly the same behavior as an open circuit.

Now, both of these components start changing over time. Given enough time, the capacitor starts acting as an open circuit and the inductor as a short-circuit. But you aren't dealing with that right now. You are just dealing with the instantaneous responses.

As to whether an impulse can show up against a capacitor or inductor with only a step source, the answer is it depends entirely on what part of the impulse you are looking for. If you are looking for the voltage across an inductor, for example, it will most definately show up. If you were looking for a current through the inductor however, no, an impulse will be invisible.

Answer 5

I don't know if this helps, but its the way I think about it: in short, the reason is that an impulse is imagined to be an infinite current in an infinitely short time. (the total charge is finite)

So when dealing with infinities, anything not infinite is effectively zero. I.e. the capacitor can only pick up a finite charge (Q = CV) therefore it can only develop a finite voltage, which is treated as approximately zero in the heuristic impulse analysis.

Resistors will get infinite voltage across them, so they conduct according to V=IR (V is infinite).

However, inductors are open circuits for rapidly changing current, so do not conduct.

As for the step analysis, I think Brian Drozd above had a good answer.

Answer 6

At t=0, Vc(0)(=0) can not jump to step input voltage. So, Vc(0)=0, which means that the capacitor is short circuited at that moment. Thus, the step input voltage is transferred to the resistor.