Help with Capacitors: Short Circuits & Connecting Terminals

[TsAmE]

Homework Statement

I don't get how this circuit is a short-circuit as there is a resistor connected in this circuit, which is opposing the flow of current.

How can both terminals of the capacitor be connected to earth, shouldn't one be connected to the positive and the other to the negative (in order for a current to flow)?

Homework Equations

None.

The Attempt at a Solution

n/a.

 

[n.karthick]

hey, here the capacitor is going to discharge its charge (if there is any).
It is like when you release a compressed spring. The spring will tend to come back to its original shape isn't it? The charged capacitor can be thought of as a compressed spring. Can you tell the significance of resistance value in discharging of capacitor, like what will be the discharging time if you have large resistance? will the discharging time change?

 

[poor mystic]

Hi TsAmE
it looks like they want you to think of a previously charged capacitor and how it discharges when connected to a short circuit.

Capacitors work by storing electric charge. If you were to connect a capacitor to a battery for a moment, electrons would migrate from the battery onto one 'plate' of the capacitor.
Then you could connect the capacitor to a load (usually symbolised by a resistor) and for a moment there would be some electric current flowing in the load resistor.

A charge can be kept on a capacitor for quite long periods. Any imbalance of charges on the capacitor would soon be removed by this circuit, with the current in the resistor at any time being exactly proportional to the charge left on the capacitor.

 

[TsAmE]

hey, here the capacitor is going to discharge its charge (if there is any).
It is like when you release a compressed spring. The spring will tend to come back to its original shape isn't it? The charged capacitor can be thought of as a compressed spring. Can you tell the significance of resistance value in discharging of capacitor, like what will be the discharging time if you have large resistance? will the discharging time change?

I know that the larger the resistance, the larger the discharging time (Time constant RC).

Sorry but I still don't get why it is a short-circuit if a resistor is contained in the circuit.

Since a capacitor is like a battery which stores charge, why isn't it connected with the positive terminal as the supply and the negative terminal as earth? (Refer to new attachment).

 

[Integral]

The short circuit is just referring to the red section. Perhaps in a different problem they but some sort of load, eg a resistance, in that place. Both terminals of the cap are not connected to directly to ground, the positive side (as shown in your pic) is connected to the resister. Current must flow through it to reach ground.

 

[TsAmE]

Both terminals of the cap are not connected to directly to ground, the positive side (as shown in your pic) is connected to the resister. Current must flow through it to reach ground.

If that is the case why is it that the discharging current starts and ends at the terminals of the short-circuit

I don't see how the conventional current flows from the positive to negative in this circuit.

 

[Integral]

If that is the case why is it that the discharging current starts and ends at the terminals of the short-circuit

I don't see how the conventional current flows from the positive to negative in this circuit.

I don't understand what you are saying. Why do you think that the current "starts and ends", what ever that means, with the short?

 

[poor mystic]

I remember similar levels of confusion in early classes. The thing to remember is that the whole mechanism is just as simple as a bucket of water which suddenly develops a hole in the bottom. If it seems more complicated than that, you're still confused.

The 'water' is the electric charge imbalance on the capacitor. This charge will move through whatever conducting medium is available to get to the other side of the capacitor and balance the charge.

The smaller the hole in the bucket, the higher the resistance the hole has to water flowing through it, and the slower the discharge. For the discharging capacitor a similar property of the resistor in the circuit slows charge movement.

By the way, don't read too much into the term 'short circuit'. It is confusing, but in this case it doesn't mean a 'dead short' or blowing fuses.

 

[kntsy]

Homework Statement

I don't get how this circuit is a short-circuit as there is a resistor connected in this circuit, which is opposing the flow of current.

How can both terminals of the capacitor be connected to earth, shouldn't one be connected to the positive and the other to the negative (in order for a current to flow)?

Homework Equations

None.

The Attempt at a Solution

n/a.

1.short circuit refers to the colored area only.
2.There is no need to connect the charged capacitor to a battery to discharge it; Connecting both plates by non-insulators(wires/Earth) enables the flow of electrons and discharge occurs.

 

[TsAmE]

I remember similar levels of confusion in early classes. The thing to remember is that the whole mechanism is just as simple as a bucket of water which suddenly develops a hole in the bottom. If it seems more complicated than that, you're still confused.

The 'water' is the electric charge imbalance on the capacitor. This charge will move through whatever conducting medium is available to get to the other side of the capacitor and balance the charge.

The smaller the hole in the bucket, the higher the resistance the hole has to water flowing through it, and the slower the discharge. For the discharging capacitor a similar property of the resistor in the circuit slows charge movement.

By the way, don't read too much into the term 'short circuit'. It is confusing, but in this case it doesn't mean a 'dead short' or blowing fuses.

Yeah I understand the way a capacitor operates, but it is just the way in which the current flows which is confusing me.

1.short circuit refers to the colored area only.
2.There is no need to connect the charged capacitor to a battery to discharge it; Connecting both plates by non-insulators(wires/Earth) enables the flow of electrons and discharge occurs.

I understand this, but if the flow of electrons are enabled and discharge occurs, where would the electrons start (i.e. negative terminal?) and end (i.e. positive terminal)?

 

[RajChakrabrty]

TsAme,

I am considering,the 1st picture you have attached.
one has rightly said that short circuit area is the red part only,so concentrate on remaining part.
But, there is another confusion ,anyone could have mentioned, that switch is not closed.
if switch will remain open then Discharging of Capacitor is not possible.

here we can make one anticipation , based on the given ckt...

** the short ckt part contains an Inductor----
as, when the switch will be closed the inductor will be considered as short ckt for t=0+
0+(zero +) means the time is just after closing of switch.
this may be the reason for showing the short ckt there.

[you may know,
the inductor acts as ahort ckt and capacitor as open ckt at t =0+
and
inductor as open ckt and capacitor as short ckt for t=infinite]

this is the answer of short ckt issue.



now, the current flowing in the ckt and the GND problem.
I have seen that some one has discussed beautifully with water flowing concept.
so,
if you have still any doubt,
ask agin.Hope your confusion will be solved soon.
Best of Luck.

 

[TsAmE]

TsAme,

I am considering,the 1st picture you have attached.
one has rightly said that short circuit area is the red part only,so concentrate on remaining part.
But, there is another confusion ,anyone could have mentioned, that switch is not closed.
if switch will remain open then Discharging of Capacitor is not possible.

here we can make one anticipation , based on the given ckt...

** the short ckt part contains an Inductor----
as, when the switch will be closed the inductor will be considered as short ckt for t=0+
0+(zero +) means the time is just after closing of switch.
this may be the reason for showing the short ckt there.

[you may know,
the inductor acts as ahort ckt and capacitor as open ckt at t =0+
and
inductor as open ckt and capacitor as short ckt for t=infinite]

this is the answer of short ckt issue.



now, the current flowing in the ckt and the GND problem.
I have seen that some one has discussed beautifully with water flowing concept.
so,
if you have still any doubt,
ask agin.Hope your confusion will be solved soon.
Best of Luck.

Thanks. I understand the water flowing concept, but what I think is confusing me is that I am thinking of this circuit in terms of any circuit, where one wire must be connected to + and the other to - (thus producing a conventional current), but here both wires are connected to Earth (where the dot is). This is why I can't see how a current can flow.

 

[poor mystic]

Ah well. In this case, the + and - charges have been forced to take up residence on the capacitor. Any conduction path will do, that charge must be balanced.
But notice! the capacitor works like a battery, dry cell or whatever... it has a negative and a positive terminal, and current certainly flows in a loop from the device by one wire and back to the device by the other wire.

 

[poor mystic]

Take a dry cell and a small lamp. Connect the lamp to the cell, check that the lamp lights.
Connect one side of the dry cell to earth, the other side to the lamp and through the lamp to earth. The lamp lights.
The Earth can be a conductor as well as a terminal.

 

[RajChakrabrty]

discussion by Poor mystic is wonderful.

 

[poor mystic]

poor poor mystic suspects irony

 

[TsAmE]

Thanks! I understand now :). Why is it that in the diagram the discharging current doesn't point from the + terminal of the capacitor to the - ?

 

[Zryn]

The blue loop indicating the polarity of the current is a poor illustration of where the electrons move from and to. They do not start at ground, and move through the capacitor to the switch but rather, will travel from the top (+) of the capacitor to the bottom (-) along the path of least resistance, through whatever medium necessary, providing they have sufficient force (voltage) to do so.

 

[TsAmE]

I see. Another doubt which I have had is how the capacitor charges up to the supply rail, even if a resistor is present. Say in the diagram you used a +13V supply rail and the capacitor is charging. A certain amount of voltage will be dropped across the resistor (say 5V is across R). Now what I want to know is why is it that the capacitor can still charge up to +13V and not Vc (13V - 5V = 6V)? As the 5V was lost when the current traveled through R.

 

[Zryn]

What do you know about capacitors charging? Click on the 'capacitor' link in your post and scroll down to 'Inverse exponential rate of charging'.

Capacitors are non-linear (exponential) when they are charging / discharging (over time). When they reach equillibrium (always 5 RC time constants worth of charging / discharging; though like an exponential they can never truly reach 100% charge or discharge) you can use linear equations such as Ohms Law.

* If the switch in your diagram opens faster than 5 RC time constants, the capacitor will always be charging or discharging and so you will not be able to use Ohms Law to determine the current in the circuit.

 

[TsAmE]

What do you know about capacitors charging? Click on the 'capacitor' link in your post and scroll down to 'Inverse exponential rate of charging'.

Capacitors are non-linear (exponential) when they are charging / discharging (over time). When they reach equillibrium (always 5 RC time constants worth of charging / discharging; though like an exponential they can never truly reach 100% charge or discharge) you can use linear equations such as Ohms Law.

* If the switch in your diagram opens faster than 5 RC time constants, the capacitor will always be charging or discharging and so you will not be able to use Ohms Law to determine the current in the circuit.

Yeah I understand this, but still don't get why it (almost) charges up to the supply voltage, despite there being a resistor in the way.

 

[Zryn]

As long as the capacitor's charge is not equal to the source potential it will continue to charge.

As long as the capacitor is charging, the voltage dissipated across the resistance will be decreasing over time, the same as the capacitor voltage will be increasing over time, and the value of the circuit current will change over time, as per the non-linear exponential charging equations.

Eventually, when time reaches infinity, the capacitor will hold the same voltage as the source, and the current will be 0A, and therefore by Ohms law, the voltage drop across the resistor will be 0A*R = 0V.

This all happens because when time has not yet reached infinity, the circuit has not yet reached a stable equilibrium yet, much like a spring that is trying to uncoil, but takes infinity time to do so, it will continue trying to uncoil until it reaches its stable equillibrium.

 

[vk6kro]

Yeah I understand this, but still don't get why it (almost) charges up to the supply voltage, despite there being a resistor in the way.

The resistor does not have a fixed voltage across it.

The capacitor starts off with zero voltage on it. So all the supply voltage must be across the resistor. (so you can calculate the current. I = E / R).
With a current flowing into it, the capacitor starts to charge up and get a voltage across it.
This means the resistor has less voltage across it, so the current decreases.

If the supply is 12 volts, and the capacitor has charged to 6 volts, there must be 6 volts across the resistor.

As the capacitor charges more and more, the current keeps dropping. If you wait long enough, the capacitor voltage will get very close to the supply voltage. Depending on the component values, this may only need a millisecond or so, or it could take hours with big capacitors and resistors.

 

[TsAmE]

As long as the capacitor's charge is not equal to the source potential it will continue to charge.

As long as the capacitor is charging, the voltage dissipated across the resistance will be decreasing over time, the same as the capacitor voltage will be increasing over time

I see, so voltage division actually occurs between R and the capacitor while it is busy charging? (as there is a certain resistance across the capacitor?)

Also when the capacitor has fully charged and there is no voltage drop across the resistor, could it be said that the capacitor has an infinite resistance? (as all of the supply voltage is dropped across it?)

 

[Zryn]

A capacitor is comprised of two conductive plates separated by a dielectric, (air, ceramic polymer, could be almost anything non conductive) as the capacitance of a capacitor is related to the distance between the plates, the stronger the dielectric strength, the larger a voltage it can separate without failing and creating a short circuit.

This means when fully charged the capacitor looks like an open circuit, i.e. infinite resistance because of this dielectric separating the two conducting plates.

When the capacitor is charging, current doesn't flow >>through<< it, like current flows through a resistor (creating a voltage drop), since that would mean the dielectric has failed and the capacitor would be a short circuit. Rather, charged particles flow onto the positive plate, and can then flow out of the positive plate when the capacitor is discharging.

Ideal capacitors have no resistance across them like you imply (in a DC circuit), but yes, there is a voltage division between the resistor and the capacitor (since kirchhoffs voltage law - based on conservation of energy - still has to be adhered to, and there is only two components to divide the voltage between).

 

[vk6kro]

I see, so voltage division actually occurs between R and the capacitor while it is busy charging? (as there is a certain resistance across the capacitor?)

Also when the capacitor has fully charged and there is no voltage drop across the resistor, could it be said that the capacitor has an infinite resistance? (as all of the supply voltage is dropped across it?)

The capacitor should always have infinite resistance, however you can send current into it if you have a supply that is greater than the voltage already on the capacitor.
This is what capacitors do. They store charge and develop a voltage across themselves when they do this.

It is probably better to look on the capacitor as a voltage source for analysis purposes. It combines with the main voltage source to reduce the voltage across the resistor.
Ohms Law still applies for the resistor, so you can work out the charging current at any level of charge on the capacitor.

 

[flufffrost]

I cannot figure out exactly how a capacitor stores electrical charge on one plate? Why is a capacitor any different from two wires connected to a battery, their other ends separated only by paper?

 

[flufffrost]

sorry, I also meant to ask why can capacitors instantly discharge while batteries can't?
thanks!

 

[Zryn]

Capacitance is proportional to the area of the conductive plates, and is only different in the design.

A capacitor is designed to store maximum charge and so the plates area is purposely made large. A wire is not intended to store charge and since metal is reasonably expensive, a wire is only made large enough to carry the current desired.

That being said, the two ends of wire separated only by paper will exhibit capacitance, but probably on a scale that is insignificant and maybe not even measureable without sensitive instruments.

Capacitance is also inversely proportional to the distance between the plates, so the larger the distance, the less capacitance.

When designing circuits, the capacitance between two wires running next to each other can be a significant problem, as voltages can appear that are unintended and have serious effects for the operation of the circuit.

 

[vk6kro]

"Instantly" is a dangerous term.

Capacitors cannot discharge instantly because there is always some resistance in the plates themselves and the wires leading to them. This limits the discharge time to a finite value.

Capacitors could be built to give very high discharge rates, but they would have to be built like car batteries with huge terminals and connecting wires and very thick plates. They could end up being as big as a truck and cost plenty.
Normal combinations of store-bought capacitors are nowhere near as good as this, but they can still be lethal if they are charged to high voltages.

Batteries are similar. They have internal resistance which is partly due to internal connections and partly due to the chemical processes. However if you ever see a short circuited car battery with hundreds of amps flowing in an external short circuit, you might consider this as being close to instantaneous.
Like they say, "don't try this at home".