# What happen if a wire is connected in parallel to a capacitor?

simulate this circuit – Schematic created using CircuitLab

Does this have the same effect as shorting out a resistor. With a resistor, it makes sense, because in parallel, current wants to flow through the branch that has the less resistance. But I am not too sure about how much resistance a capacitor has? Or if it has one at all.

• Note that with parallel paths, more current will flow through the path with the lesser resistance but some will flow through the other path depending on the ratio of the resistances. For a perfect short (not realizable in practice), with 0 resistance, then all of the current will flow through the short. Commented Sep 28, 2017 at 22:59

Ideally all the current will flow through the short and that is what you should consider absolutely true at this stage in your learning.

In reality wires have resistance and inductance (and capacitance) and capacitors have inductance and resistance as well as capacitance. So if you have a (real) long-ish wire in parallel with a (real) well-made capacitor some significant part of the current may flow through the capacitor at higher frequencies. This will not show up in a simulation unless the imperfections are modelled.

A capacitor is defined by

$$Q = CV$$

from which we can also say

$$I = C\frac{dV}{dt}$$

If you put a short across the capacitor, what does that tell you about $V$?

And what does that tell you about the current that will flow through the capacitor?

• V=0? Like shorting out the battery? But how is battery shorted battery shorted by placing a wire in parallel with it? Commented Sep 28, 2017 at 23:19
• My equations are equations describing a capacitor. The V is the voltage across the capacitor, the I is the current through the capacitor. What made you think any of this answer has to do with a battery? Commented Sep 29, 2017 at 4:12

Short-circuiting a capacitor means that there can be no voltage across it. It is effectively removed from the circuit.

Does this have the same effect as shorting out a resistor.

The effect will be different in that the components were performing different functions when they were in circuit. The effect will be the same in that a short-circuit is a short-circuit.

With a resistor, it makes sense, because in parallel, current wants to flow through the branch that has the less resistance. But I am not too sure about how much resistance a capacitor has? Or if it has one at all.

An ideal capacitor has no resistance. A real one has some but it's low. It should not be considered a factor in your understanding of this problem.

• But why does current prefer to go through the branch with no cacpacitor it has no resistance? At first, both the capacitor and the empty branch has zero resistance, but the resistance of the capacitor as the charge accumulates. Commented Sep 28, 2017 at 23:23
• No charge can accumulate on the capacitor. It's two plates are short-circuited. They are connected together. Commented Sep 28, 2017 at 23:25
• @user132522 To reinforce what Transistor said: the two plates of the capacitor, in the hypothesis of perfect conductors (as it is implied by your basic circuit theory question), has its plates shorted by a perfect conductor, so it is no longer a capacitor, but just a funny looking piece of conductor. And the dielectric inside is, electrically, not different from the dielectric put on the outside to protect the plates from the environment. In other words, a capacitor works as it does just because its plates are insulated from each other. Commented Sep 29, 2017 at 7:16
• @LorenzoDonati: "... a capacitor works as it does just because its plates are insulated from each other." Exactly. I was fumbling around trying to come up with as good a wording. Commented Sep 29, 2017 at 7:30
• @Transistor thanks :-) (teaching to teenagers helps you in learning how to "boil down" problems to their roots before they lose interest! :-D) Commented Sep 29, 2017 at 7:36