I have some confusion about capacitors.

There's mention that at DC T=0, a capacitor should be considered as short circuit.

So how does it work when we use it as a filter capacitor?

Won't it be shorted to ground as the sketch in a flash?

enter image description here

  • \$\begingroup\$ It's also important to consider where a capacitor is placed. It works as a low-pass filter here because (sufficiently) high-frequency components are shortcircuited to ground. \$\endgroup\$
    – MSalters
    Apr 29 '19 at 11:31

The capacitor is considered a short-circuit for sufficiently high frequency components relative to its capacitance. That's how it acts as a filter. The lower frequencies see it as an open circuit and ignore capacitor, but the high frequencies (i.e. noise frequencies) see it as a short-circuit and take the detour through the capacitor and are short-circuited, preventing them from travelling through the load.

It kind of acts as an almost infinite ohm resistor for "really low" frequencies, and an almost zero ohm resistor for "really high frequencies". Then there is an intermediate range where the capacitor kind acts in between. It's not a hard rule that a capacitor always acts as a short-circuit (or an open-circuit) for all frequencies.

You are missing some conditions when you say that that the capacitor is appear as a short-circuit at t = 0. This statement assumes that you apply a DC voltage to the capacitor at t = 0. You are basically applying an infinitely fast vertical voltage edge to the capacitor (turning on the power supply). That infinitely fast vertical needs infintely high frequency components. It's those super high frequency components that see the capacitor as a short-circuit. The lower and intermediate frequencies in this edge charge up the capacitor to varying degrees until the capacitor is full charged.

That's how AC "flows" through a capacitor (by AC I mean non-zero frequency components like sinusoids, anything that is not just a steady flat 0Hz voltage. I do not mean only the AC wall voltage). It's no different than DC, except that a AC/sinusoid voltage is always changing so the capacitor's voltage is never allowed to catch up and match the applied voltage so charge is always being added or removed from the capacitor in AC which is equivalent to current flow through the capacitor. If a frequency is higher, a given capacitance will provide less impedance to it and it appears more as a short-circuit and will flow through the capacitor more easily.


Consider the following schematic


simulate this circuit – Schematic created using CircuitLab

This represents the real world \$ R_{supply} \$ represents the output impedance of the supply and ESR is the equivalent series resistance of the capacitor. Lets assume the capacitor is fully discharged and we close the switch.

At time \$t = 0^+\$, the switch as just been closed, the ideal capacitor C can't change its voltage in zero time so looks like a short circuit. The current is limited by \$R_{supply} + ESR\$. \$R_{supply}\$ and ESR should be very small but will not be zero.

If we were to measure the current in the capacitor we would see it is: $$\dfrac{V_{DC}}{R_{supply} + ESR}\cdot \exp\left(\dfrac{-t}{C \cdot (R_{supply} + ESR)}\right)$$

The capacitor charges quickly and stops drawing any current.

In your application the capacitor is used to filter out high frequency noise.

$$Z_{cap} = \dfrac{-j}{2 \cdot \pi f \cdot C}$$

The capacitor looks like a low impedance to high frequency signals effectively 'shorting' them out. So the current does not flow in the supply.


At various moments when analysing a circuit, you can treat a capacitor as a short circuit.

When connecting a discharged capacitor across a DC voltage source, the capacitor wil act like a short circuit until it is charged. How long that is depends on the capacity and how much current the voltage source can deliver - the resistance of the voltage source, in other words.

When looking at AC circuits, the capacitor doesn't charge so it isn't useful to look at it that way.

In AC circuits, it is more useful to look at a capacitor as an impedance that varies with the frequency of the applied AC.

Take your example.

At 50 Hz, a 1uF has about 3kOhm of impedance. At 230V, about 76mA of current would flow.

If you used a 100 uF capacitor, then more like 7A would flow. That would look like a short circuit.

The point of a filter capacitor is to choose a value such the capacitor "looks like" a short circuit to the noise, but like a very large resistor to the line frequency.

Since the noise and line frequencies a (usually) very different, it is usually fairly easy to find such a value.

Assume a line frequency of 50 Hz, and a noise frequency of 100 kHz.

A 10 nF capacitor has an impedance of over 300kOhm at 50 Hz. Very little of the AC line power will go through the capacitor.

At 100kHz, the impedance is only 159 ohms. Noise at that frequency will pass much more easily through the capacitor. The noise is practically "short circuited" to ground through the capacitor.

The impedance of the capacitor drops further with increasing frequency, making it more effective at "short circuiting" the noise.

  • \$\begingroup\$ Just as a bolt-on to your answer: It is worth noting that there is a breakdown issue to be handled as well - the capacitor has to be able to handle both the expected amperage through it and the voltage across it. \$\endgroup\$ Apr 29 '19 at 8:03

Only for a little bit. Once the capacitor is charged up, it doesn't conduct any more. As it approaches fully charged, less and less current flows.

Wiring capacitors up in various ways and probing them in this simulator really helped me understand.


One useful view of a capacitor is how fast it will charge up

Q = C * V

which differentiated, with constant C, becomes

dQ/dT = C * dV/dT


I = C * dV/dT

For 50Hertz power, with 1 volt ripple, and 1 ampere current, the capacitor is 0.02 farads.


For T=0 this is true. But after some time T>0 the capacitor will be charged i.e. not being a short anymore (which is dependent to the capacity and the line resistances which forms a low pass filter with a time constant of R*C).


People talk about AC here but that's a distraction since AC behavior is stationary behavior while the "t=0" qualification defines transient behavior.

The basic statement is that a discharged capacitor (the typical assumption for initial conditions, namely "t=0") has a voltage of 0V across it regardless of the initial current. So its momentary resistance is 0 Ohms. The smaller its capacity, the faster its momentary resistance will rise if the current stays the same. But at t=0, it's still 0.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.