# Why is the capacitor short-circuited in this example?

Please forgive me if anything wrong in my question. I am a school student.

I have following circuit diagram:

simulate this circuit – Schematic created using CircuitLab

There are three capacitors with equal capacitance $C$. As they are in series combination, the total capacitance should be $C/3$.

But my teacher said that the second capacitor (C2) is short circuited, so the output will be $C/2$.

My question is: why is C2 termed as short-circuited in this diagram?

I asked him again on why C2 is short circuited, but he said me that I should search about it. Unfortunately, I couldn't find anything helpful on google.

• It's short circuited because there is a wire between its terminals. But your initial analysis is wrong too - capacitance in series does not add, instead it reduces (capacitance and resistance behave oppositely in serial and parallel combination). – Chris Stratton Sep 30 '14 at 15:45
• @ChrisStratton "It's short circuited because there is a wire between its terminals" but why?...I couldn't understand it. please explain in answer. – krupal Sep 30 '14 at 15:48
• Those lines in your drawing are effectively wires. – Chris Stratton Sep 30 '14 at 15:48
• @krupal because that is the definition of "short circuit". A component "is short-circuited"/"is shorted"/"has a short circuit across its terminals" if there is a wire that directly connects both terminals (and plain lines in the diagram represent wires, as you probably already know). – immibis Oct 1 '14 at 3:08
• "Short" is slang for "zero ohm resistance". Even if the wire is longer than the part it is bypassing, it's still called "a short". – gbarry Oct 1 '14 at 4:29

The short-circuit in red puts the 2 points A and B at the same voltage, bypassing the capacitor C2.

• A picture s worth than hundred lines of text..I usually feel bored to read the text...thanks for your time. – krupal Oct 1 '14 at 19:27

Any element for which terminals are connected by a conductor, as the capacitor in the figure, is said to be shorted.
By having their shorted terminals, the voltage thereof is zero (more precisely, the potential difference between them), so that this element is not operational in the circuit, and can be removed for analysis.

The other two capacitors are in series, hence that:

$$C_{eq} = \dfrac{C}{2}$$

provided that the capacitors are the same value.

Another correction:

As they are in series combination, the total capacitance should be 3C.

If they were in series, the capacity would be:

$$C_{eq} = \dfrac{C}{3}$$

Take a look a this previous answer.

• sorry, "another correction" was my mistake..corrected it – krupal Sep 30 '14 at 15:51
• The two remaining capacitors are connected in series, not parallel. The resulting capacitance is C/2. – Peter Bennett Sep 30 '14 at 15:53
• @PeterBennett I was editing at the time of your comment. Thank you very much for your contribution! – Martin Petrei Sep 30 '14 at 15:56
• How do you justify that the circuit isn't an open circuit due to the capacitors at the input? – Carlos - the Mongoose - Danger Oct 1 '14 at 0:44
• @IllegalImmigrant An open circuit, implies that not current can flow trough it. For this circuit, this is true, but for DC current only. For AC current, the capacitors can be a very low impedance... – Martin Petrei Oct 1 '14 at 0:50

A short circuit here means that there is no resistance (impedance) between the two terminals of the shorted capacitor. The vertical wire drawn next to the vertical capacitor shorts the two terminals of the capacitor. Any current flowing through this circuit segment will flow through the vertical wire and completely bypass the vertical capacitor due to the short. This means you can ignore the shorted capacitor -- it has no effect on the circuit.

The two remaining capacitors are in series because they have one terminal each connected directly to each other by a wire. If they were in parallel then both terminals would be connected directly to each other by wires (i.e. they would be in parallel if you connected the two vertical wires on the left).

Also, the equivalent capacitance $C_{eq}$ of $n$ capacitors $C_{1}$, $C_{2}$, $\ldots C_{n}$ in series is

$$\frac{1}{C_{eq}} = \sum_{i = 1}^{n}\frac{1}{C_{i}}$$

Since all the capacitors have capacitance $C$ and one is shorted here the equivalent capacitance is

$$C_{eq} = \frac{C}{2}$$

Capacitance adds when capacitors are in parallel.

From your question: "Why is C2 termed as short-circuited in this diagram?"

it seems you are asking "What in this diagram indicates that C2 is short-circuited?"

But in your comment: "... but why?...I couldn't understand it.", it seems you could be saying:

"I (now) understand that it is short-circuited, but why is it short-circuited?"

as in, it makes no sense to short-circuit the capacitor, so why is it drawn this way?

If you are actually asking why as in "why is it drawn this way (short-circuited)?", then the answer most likely is that it was drawn that way to provide an example of a shorted capacitor in a circuit for the purpose of introducing the concept.

In "real life", a circuit diagram would not normally include a permanent wire connecting both ends of a capacitor.

This is an old post but worth saying that the confusion here is that you are assuming that since you are learning something that it has some use, some method of problem solving but your teacher was simply pointing out what a short circuit would look like if it were drawn on a diagram. Absolutely worthless information regardless of how you look at it mostly because if something is short circuited it is essentially not there. Typical schooling to teach problems and not solutions. Sounds like your teacher doesn’t know the answer by telling you to search on google. Either way the concept is simpler than it seems: circuits may be corrupted by short circuits.