How does the capacitor block the flow of electricity to the negative without using the electricity?

First, the lamp doesn’t “stop” the electricity. It continues along the circuit until it reaches the other battery terminal. What the lamp does is act like a resistance impeding the flow. Across the battery terminals is a voltage “drop” that equals the battery voltage. The voltage drops but there still is a current in the return wire (to the ground terminal). This concept of a voltage drop is important.

Now. A capacitor doesn’t directly connect the two wires inside. Rather, it is two large plates separated by an insulator. When a voltage is applied, the plates build up opposite charges. Their charge increases until the charges are strong enough so the capacitor cannot charge any further. The two plates repel any further charge. This situation results in the capacitor having the same voltage as the battery. The battery cannot “push” any further charge into the capacitor. Thus it will block any further current into the capacitor, making it appear as if is blocking the voltage from the battery.

 

Are these circuits now valid?

No. The battery shouldn't be in series with both the capacitor and the coil when the Fire switch closes.
Use Danko's circuit (post #22).

 

I see. The battery does not short with a light bulb because the light bulb stops the electricity reaching the negative battery terminal by consuming the electricity.

This is probably only a half truth. I imagine protons return to the battery negative terminal while electrons are used for creating light, thus converting into photons?

How does the capacitor block the flow of electricity to the negative without using the electricity?

You need to change your ideas about what is electricity. Stop thinking about particles (electrons, protons) and start thinking about electrical energy. Electricity doesn't get consumed, electricity (the pre-existing charge that is moved around) is used to transfer electrical energy. You really need to take some time to understand some basic concepts of what you want to build like understanding electricity beyond lay uses of the term.

 

are talking about EMPs that are so strong that the they are virtually always generated by explosive devices (including nuclear warheads).

Not necessarily.

Find the Popular Mechanics issues from about 20-25yrs ago, it shows how to produce EMP from cutting a high amp coil very quickly. There's no need for any big explosion to make the EMP.

 

Not necessarily.

Find the Popular Mechanics issues from about 20-25yrs ago, it shows how to produce EMP from cutting a high amp coil very quickly. There's no need for any big explosion to make the EMP.

Notice that I did not say "always".

Cutting a high amp coil that is energized is one of the faster ways to end up dead if you aren't extremely careful. It will generate whatever voltage is required to maintain continuity of current through whatever path is available.

 

Notice that I did not say "always".

Cutting a high amp coil that is energized is one of the faster ways to end up dead if you aren't extremely careful. It will generate whatever voltage is required to maintain continuity of current through whatever path is available.

Not cutting it by hand (not fast enough), hence the article, or other knowledge.

But in general, a fast mechanical "scissor" that cuts all the coils very quickly. And it scales very nicely.

 

Of course not cutting it by hand -- but you don't want to be anywhere near it when it is open-circuited. Maybe you are satisfied that someone with no knowledge of electrical circuits or devices, who doesn't want to bother with anything beyond a lay description of anything, and who has no apparent regard for safety considerations can read an article in Popular Mechanics and do this safely. I'm far from convinced of that.

 

djsfantasi

Thanks for explaining the voltage drop. That was very interesting.

Also, thanks for explaining how a capacitor does not connect two wires. Does this mean a capacitor is an open in a circuit, or does the capacitor become a power source itself from that point in the circuit?


Alec

Danko's diagram is not making any sense to me. It looks like a short will occur if both switches are closed, as the first path in the series which runs through the fire switch, goes straight to the battery negative. What am I missing here?

 

Okay, funny, I had a look at Danko's hand drawn diagram and I instantly got that both switches shouldn't be closed at the same time to prevent a short.

I still don't get the diagram though. It appears to me that if the charge switch is closed, the capacitor will charge and release its energy through the coil, as a closed circuit is created. And it also appears to me that the fire switch doesn't serve a purpose.

 

Wouldn't a capacitor never charge if its by itself in a circuit because it instantly passes through the electeicity? Or does it wait until its full before releasing it?
This is has been on my mind for some time.

 

Wouldn't a capacitor never charge if its by itself in a circuit because it instantly passes through the electeicity? Or does it wait until its full before releasing it?
This is has been on my mind for some time.

This is ridiculous. You are trying to bake bread without knowing what flour, water, salt and yeast are. Study some basic tutorials and come back with questions about what you don’t understand. All of your misconceptions would be addressed in the first lecture of an introductory class in electronics.

 

Of course not cutting it by hand -- but you don't want to be anywhere near it when it is open-circuited. Maybe you are satisfied that someone with no knowledge of electrical circuits or devices, who doesn't want to bother with anything beyond a lay description of anything, and who has no apparent regard for safety considerations can read an article in Popular Mechanics and do this safely. I'm far from convinced of that.

The concept of, not a practical method. I thought I made that clear?

 

Wouldn't a capacitor never charge if its by itself in a circuit because it instantly passes through the electeicity? Or does it wait until its full before releasing it?
This is has been on my mind for some time.

The cap is like a sponge for charge. It will store charge up to the volts you apply to it. Once that sponge is full it sits there until there's a path for that charge to move out. That charging and uncharging of the cap can happen really slowly, or very fast depending on ckt design, with limitations. Example: 1F cap 25v vs 1uF cap 25v. The 1st one has a boat load more energy in it.

Really no different than a dry sponge under the faucet. Drip water on it and the sponge fills up slowly, or run water fast and the sponge fills quickly, but at some point the sponge will be full. Then you can squeeze the sponge slowly or fast to get the water out, but at some point the sponge becomes empty. There's also all the places in between full and empty.

But like others have stated, you need to study the basic components 1st, which can be done right here on AAC.

 

OK, but capacitors store energy, not charge as in electrons or protons. Existing charge is separated by some 'force' to produce an electric field that stores electrical energy between the plates of the capacitor.

https://www.khanacademy.org/science...arallel-plate-capacitors/a/capacitors-article

Do capacitors store charge?
Capacitors do not store charge. Capacitors actually store an imbalance of charge. If one plate of a capacitor has 111 coulomb of charge stored on it, the other plate will have −1−1minus, 1 coulomb, making the total charge (added up across both plates) zero. If you short circuit the capacitor by connecting the two plates with a wire of negligible resistance, you’ll see a sudden rush of current (depending on the size of the capacitor, this can result in sparks) as the electrons on the −1−1minus, 1 coulomb plate rush onto the +1+1plus, 1 coulomb plate. This sudden rush of current releases all the energy that’s stored in the capacitor.

 

For many reasons, I do not read stuff on "Khan Academy"

Same site also says:

A capacitor with a large capacitance will store a lot of charge, and a capacitor with a small capacitance will only store a little charge. Capacitance equals the charge stored on a capacitor, divided by the voltage across that capacitor.

.

Charge always needs a differential for amps. 20C/V in Cap1, and 20C/V in Cap2, there's zero diff between those two specific plates, but connect those two plates together and you now have 40C/V (parallel caps).

The farad (symbol: F) is the unit of electrical capacitance, the ability of a body to store an electrical charge, in the International System of Units (SI), equivalent to 1 coulomb per volt (C/V).[1] It is named after the English physicist Michael Faraday (1791–1867). In SI base units 1 F = 1 kg−1⋅m−2⋅s4⋅A2.

There might be confusion as to what "stored charge" means for a cap. It simply means a potential exists.

An interesting experiment for novices, two caps of 1000uF each (rated 35v), charge each to 20v, then observe a wire between + of Cap1 to the - of Cap2.

 

For many reasons, I do not read stuff on "Khan Academy"

Same site also says:
.

Charge always needs a differential for amps. 20C/V in Cap1, and 20C/V in Cap2, there's zero diff between those two specific plates, but connect those two plates together and you now have 40C/V (parallel caps).



There might be confusion as to what "stored charge" means for a cap. It simply means a potential exists.

An interesting experiment for novices, two caps of 1000uF each (rated 35v), charge each to 20v, then observe a wire between + of Cap1 to the - of Cap2.

The "Khan Academy" is correct.
I was just clarifying for the OP not to think about particles but to think energy as the kinetic energy in particles is usually wasted in the circuit.. There is no physics confusion on what charging and discharging means in a capacitor. It doesn't mean stuffing more physical charge into it. It's always energy.

 

If the cap has ability to "create" a charge diff between two plates, as noted by Khan, that cap is said to be charged.
That 'charged' cap then also has UE (electrical potential energy).

The process of moving charge from one side to the other to create differential between the plates, is "charging" the cap. Work is done to move it.

I guess it's the viewpoint. Look at each plate, vs look at both plates at same time.

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng2.html

 

If the cap has ability to "create" a charge diff between two plates, as noted by Khan, that cap is said to be charged.
That 'charged' cap then also has UE (electrical potential energy).

The process of moving charge from one side to the other to create differential between the plates, is "charging" the cap. Work is done to move it.

I guess it's the viewpoint. Look at each plate, vs look at both plates at same time.

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng2.html

Sure, charging the process and 'Charge' q, the physical property (electricity) are two separate things can lead to a misunderstanding of how electrical energy is manipulated, transformed and transported. I think we can all agree it's charged with electrical energy.

http://physics.bu.edu/~duffy/py106/Charge.html

 

The cap is like a sponge for charge. It will store charge up to the volts you apply to it. Once that sponge is full it sits there until there's a path for that charge to move out.

Isn't that path back to the battery negative terminal that path though?

Battery(+)___
(-) x x x x x x | Capactor
|____________|

(Ignore the x's, this website doesn't honour consecutive white space input)

If you replaced the capacitor in the above diagram with a light bulb, the light bulb would 'work'. With a capactor instead, will it 'work' by charging up?

 

Isn't that path back to the battery negative terminal that path though?

Battery(+)___
(-) x x x x x x | Capactor
|____________|

(Ignore the x's, this website doesn't honour consecutive white space input)

If you replaced the capacitor in the above diagram with a light bulb, the light bulb would 'work'. With a capactor instead, will it 'work' by charging up?

If you want electrical analogies use a better one.
http://amasci.com/emotor/cap1.html

My favorite capacitor analogy is a heavy hollow sphere which is completely full of water and is divided in half with a flexible rubber plate through its middle. Hoses are connected to the two halves of the thick irong sphere, and they act as connecting wires. The rubber plate is an analogy for the dielectric. The two regions of water symbolize the capacitor plates.

Now think: in this analogy, water corresponds to electric charge. How much water have I put into my iron sphere? None! The sphere started out full, and for every bit of water that I took out of one side, I put an equal amount into the other at the same time. Same as when running a current through a conductor. When the pump pushed water into one side, this extra water also forced some water out of the other side. No water passed through the rubber, instead there was some rubber-current in the divide. Even so, essentially I drove a water current through my hydraulic capacitor, and this current pushed on the rubber plate and bent it sideways. Where is the energy stored? Not in the water, but in the potential energy of the stretched rubber plate. The rubber plate is an analogy to the electrostatic field in the dielectric of a real capacitor.

I never really understood capacitors until I started trying to construct proper water-analogies for them. Then I discovered that my electronics and physics classes had sent me down a dead-end path with their garbage about "capacitors store electric charge." Since my discovery, I've gained significantly more expertise in circuit design, which leads me to a sad thought. Maybe the more skilled of electrical engineers and scientists gain their extreme expertise not through classroom learning. Instead they gain expertise in spite of our K-12 classroom learning. Maybe the experts are experts only because they have fought free of the wrong parts of grade school science, while the rest of us are still living under the yoke of the many physics misconceptions we were carefully taught in early grades.