Calculating maximum output of Wall Socket

[mooeypoo]

Hey guys,

 

As I promised, I have another question in the series of "Help me be a good equipment-mastah" topic

 

I have a question that goes as follows:

 

At work, we have about 20 chargers for magnetometers (we have more devices, but they're not all charged at once) and about 30 for motorola 2-way radios. Up until a few days ago, almost all of the chargers were connected to about 2 electricity sockets. I changed that for now to have about a limit of approximately 8 devices per socket (out of - for now - limitations of room, sockets and chargers).

 

How do I figure out how many chargers can "sit" on the same socket without us waiting a few DAYS for the device to be charged? In other words, if I know the maximum time I am willing to give for charging, and I know the electrical charge of the socket (obviously), how can I know how many to hook up to the same socket?

 

We have a problem of laymen at work, too, so whatever changes I make, I'm going to have to make sure it's easy, non hassle, and most importantly - flexible enough to allow for the stupidities of people who take devices before they're fully charged or are just too damn impatient and declare a device malfunctioning when its only flaw is that its batteries are discharged.

 

In short, I'm supposed to make sure everything works in a workplace where people cry that things malfunction but don't give a crap enough to maintain them. Just throwing this to the equasion so the solutions take it into account. If I split chargers into separate sockets, I'm in need to seal SHUT the other sockets so that people won't connect anything to it. Yah. Talking about 'feeding with a spoon'.

 

Oh, another piece of information: during the night, we have about 10 hours that devices can be charged (though, our "smart" masses at work never shut off the radios, so they don't charge, but hey.. who asked for Geek-IQ when they hired, right?), but during the day we are using them almost continuously. We figured we would buy extra batteries, which is doable for the radios, but as for the magnetometers, the batteries are a hassle to change, so we're stuck with trying to figure out how much we can have while charging. We don't have a lot of charge-time during the day, but every night we should have about 10 hours regularly.

 

The manufacturer states that 16 hours of charge gives 40 hours of work. While I doubt we'll get that today, I assume it relates to a situation where the charger gets maximum charge from the socket.. hence my attempt to try and make the best out of the situation. We don't have 16 hours, and we don't have max output, but.. any ideas how I can make it best considering?

 

 

Anyhoos, thanks in advance

 

~moo

[Spyman]

How do I figure out how many chargers can "sit" on the same socket without us waiting a few DAYS for the device to be charged? In other words, if I know the maximum time I am willing to give for charging, and I know the electrical charge of the socket (obviously), how can I know how many to hook up to the same socket?

Normally the power supply to wall sockets is able to support more current than the socket and the electrical wiring can handle, so to prevent overloading and fires a fuse is used.

 

If the sockets are supplied differently, you need to tell us how.

 

Assuming normal conditions, you can hook up plenty of chargers on a single socket. More load on the socket should not increase charging time. If you load the fuse to much it will blow and all chargers will stop. If the fuse don't blow then all chargers will use their normal charging time, independent of how many that are connected.

 

The chargers should be marked with how much current they need and the fuse with it's maximum current.

 

Add all the loads on each fuse together and just make sure it's below the fuse rating.

 

Some easy Examples:

20 Amp fuse, chargers which use 1.0 Amp each -> 20 chargers max.

16 Amp fuse, chargers which use 1.0 Amp each -> 16 chargers max.

10 Amp fuse, chargers which use 1.0 Amp each -> 10 chargers max.

6 Amp fuse, chargers which use 1.0 Amp each -> 6 chargers max.

10 Amp fuse, chargers which use 0.5 Amp each -> 20 chargers max.

10 Amp fuse, chargers which use 0.1 Amp each -> 100 chargers max.

 

Note that there can be plenty of sockets in different rooms on the same fuse and it's also possible that two sockets close to each other, in the same room, can be connected to different fuses. Lightbulbs and other equipment installed can also be connected to a fuse used for sockets or vice versa.

 

How to find out which fuse is connected to which socket:

1) Read the electrical schematic diagrams, (eg. industrial areas)

2) Find individual markings on the fuses and sockets which match, (eg. hospital areas)

3) At least the fuses should be marked with a working area in the building, (eg. homes, appartments)

4) Follow the wirings from the fuse to all it's connections,

5) Turn on a load in a socket, remove a fuse and check if the power is gone.

 

In most cases I use first No.3 and then No.5 to verify removing correct fuse for a specific socket.

[mooeypoo]

I have a related question, though:

 

When you have a connected to a socket, and you hook up something that takes a bunch of "juice", the light dims. I always thought it's because both of these are using the same flow. So of course I know that it's not "split to 2" between them, but I was sure there's still a limit to what a socket can output, and then if you put 2 devices that take a bunch of juice, they get less each, or - if one just hogs on it - the other one (lamp, in this case) gets less.

 

Am I wrong?

 

If this is true, though, then if I have 10 chargers in the same socket and another 2 devices that seriously hog electricity, won't it limit the flow for the chargers?

 

Thanks for the thorough response, Spyman

 

~moo

[Rocket Man]

the technical term for that would be a "current shunt" you put too much load on a circuit, there's not enough current to go around. normally the fuse goes out before the circuit dips appreciably, but as for your workplace, it might be a little different.

 

a wall socket is one of the simplest items you might chance across. its three wires attached between three bigger wires and three holes in the wall.

draw too much current out, there's not much power left for the rest of the equipment.

 

a number of wall sockets in close proximity has the same problem. except that instead of looking like a power board, it looks like an unfeasable number of wall sockets.

 

a wall socket can generally give 10A. but if the wider circuit has too much load, you have a bigger problem

[Spyman]

I have a related question, though:

 

When you have a connected to a socket, and you hook up something that takes a bunch of "juice", the light dims. I always thought it's because both of these are using the same flow. So of course I know that it's not "split to 2" between them, but I was sure there's still a limit to what a socket can output, and then if you put 2 devices that take a bunch of juice, they get less each, or - if one just hogs on it - the other one (lamp, in this case) gets less.

 

Am I wrong?

 

If this is true, though, then if I have 10 chargers in the same socket and another 2 devices that seriously hog electricity, won't it limit the flow for the chargers?

 

Thanks for the thorough response, Spyman

 

~moo

All devices on the same fuse shares the flow, but if everything is normal and functioning they should not limit each others flow. The only limit should be the fuse, which blows when overloading.

 

The juice doesn't show up instantly, electricity has a speed too. If you turn on the light, it seems instant, but if you later on plug in a big consumer, like a powerful vacuum cleaner, which for a short moment during startup draws much, it will cause the voltage to drop temporarily, so the lights dims, but only for a short moment. If the lights gets more and more dim as you load the socket with more and more devices, I would advice to call an electrican, because something is seriously wrong.

 

Your chargers is only a drop in the ocean for the HUGE generators supplying the power.

(If not a small private portable generator.)

 

Lets take my home and neighborhood as an example. The city has large wires providing power for the city and our industries, several transformation stations shares that power and distributes it out to smaller transformators, from where it goes to small distributions cabinets and eventually to my house.

 

The cable to my home is fused with 63 Amps in the cabinet at the street corner, and it has plenty of such fuses for the other dozens houses in that area. The cabinet is able to distribute enough power for those houses to keep warm during winter, and it is backed up with more than that power. Inside my house the main fuses are of 16 Amps and then comes the smaller fuses, (10 Amps & 6 Amps), for different rooms, (light & sockets), and other devices like house heater, stove and refrigerator.

 

Theoretically, I have the full power from the main generator, (except losses), available in any socket in my house. The purpose of the fuse is to protect the cables from melting.

 

There are exceptions as always, (a few listed below):

- In some areas, typically farms far out in the country, the length of the cable itself can strangle the avaliable power so the fuse won't blow and the lights stays dimmed for the duration of a heavy load.

- Or like in my area, we have a heavy steel industry which can load the whole supply to our town, so much that the lights can dim for several minutes in my house.

- Also in an industry it is sometimes useful to limit the current by other means than a fuse, which in such case also would cause the lights to stay dim when applying a heavy load.

 

Thats way I asked if your sockets are supplied normally or specially.

 

 

Hmm, I don't know if I managed to explain that good enough...

 

There is normally nothing strangling the available power to your wall sockets !

(Except losses in the wires, which are brought down to minimum.)

 

The only thing limiting the power is:

* the generator, probably very powerful,

* the fuse, which only will blow or not, nothing else,

* the connections, to weak results in Fire or malfunction,

* the wires, to weak results in Fire or malfunction,

* the socket, to weak results in Fire or malfunction,

(* very long wires can cause a voltage limit, not current.)

 

The fuses are there because if you load the wire with to much power it will melt the insulation, glow like the wire in a light bulb and start a fire. The fuses should be dimensioned to protect the weakest point in the circuit.

 

But if nothing is wrong, then you have the current the fuse is marked with, at any one socket it is supplying until the fuse blow and the voltage should only drop very temporarily when a heavy load is connected and then return to normal.

 

If you have a load that needs 2 Amps connected in a socket, it drains 2 Amps through the fuse. When you connect a second load of 4 Amps, it drains 4 Amps through the fuse, the first load still drains 2 Amps and the fuse now have a current of 2+4=6 Amps flowing through it. If the fuse is marked 6 Amps, you are very close to the blow limit, it might blow after several hours or last days, depending of manufacturing quality. Adding a third load of 1 Amps will cause the fuse to get hot and blow after some time depending of the speed of the fuse, normally it is "slow", (marked with a snail), which allows some small overloads for a short duration of time, (like when a vacuum cleaner starts). But until the fuse blows you will have 2+4+1=7 Amps without restrictions, (except for some very small losses). If nothing is wrong, (or special), then the voltage stays normal during all the connections, (except for small dips).

 

The generators supplying the power to your company is probably supplying power to a large part of your town, unless your company is so big it needs it's own generator, in which case it's still supplying more than your small working area, (building), so it's not likely it will be noticeable affected if you connect one, ten or hundreds of small battery chargers. It is capable of supplying the rated voltage and enough current to blow your tiny fuse without noticing.

 

Hopefully that is clear enough...

 

 

If you don't trust me, you can always test it, yourself...

 

- Drain a set of batteries to a certain known level.

- Measure the charging time for one charger only connected.

- Drain several sets of batteries to the same level.

- Measure the charging time with several chargers connected.

 

If I am right the time would be the same, except for small differences in the batteries/chargers. If you are correct 5 chargers will take 5 times longer to charge the batteries, (or at least much longer).

 

Another way could be to measure the voltage/current at/through the socket with one and several chargers connected. If I am correct the voltage won't change and the current will rise to what the chargers use together.

 

Please note that the power in the socket is harmful and deadly !

(Don't mess with it, if you don't know what you are doing.)

[Rocket Man]

ooh, i just rememered, vacuums etc have very high start up currents and fuses have minimum burnout times.

as the fuse heats, the resistance increases reducing the overall power to the circuit. then motor manages to spin up to a low current level before the fuse actually blows.

as the current in the circuit decreases agian, the fuse can cool reducing resistance restoring the power to the light. it's still a current shunt but it's sitting on the fuses burn limit.

you can't sustain a current shunt if you have a fuse as spyman said so the only limit is where everything goes out. it will still work perfectly till then.

[Spyman]

ooh, i just rememered, vacuums etc have very high start up currents and fuses have minimum burnout times.

as the fuse heats, the resistance increases reducing the overall power to the circuit. then motor manages to spin up to a low current level before the fuse actually blows.

as the current in the circuit decreases agian, the fuse can cool reducing resistance restoring the power to the light. it's still a current shunt but it's sitting on the fuses burn limit.

you can't sustain a current shunt if you have a fuse as spyman said so the only limit is where everything goes out. it will still work perfectly till then.

AFAIK Fuses are generally not thought to change resistance or act like "current shunts".

[richard]

The dip is caused by a high current device being connected. This makes a surge current. Its explained from here http://en.wikipedia.org/wiki/Surge_current

[Rocket Man]

the motor acts as the current shunt, while the fuse become resistive at high temperatures

[Spyman]

the motor acts as the current shunt, while the fuse become resistive at high temperatures

No, the motor acts like a dead-short and causes a surge current, the fuse becomes highly resistive after it blowed.

[Rocket Man]

a dead short is a particularly effective current shunt.

 

have you seen one of those magnetic ring levitators? it's a coil with an extended core, the ring sits just above the coil due to eddie currents. if you put a red hot ring over it, it doesn't get enough induced current to lift due to resistance, put a chilled ring on it, it can barely sit over the core.

a fuse gets to red heat just before it blows and becomes very resistive without actually cutting the supply. if the short is disconnected in time(befrore the fuse melts), the fuse will cool down and conduct normally.

i'll bet that if you put a fuse on burnout on a current plot, you'll see a substantial dip before it cuts off completely.

 

though i doubt mooeypoo is going to hold her fuses at red heat with just a bunch of battery chargers.

[Spyman]

Yes, thats right, most standard fuses has a positive temperature coefficient.

 

But what I have been arguing is that the main point of a fuse is to have as low resistance as possible until it blows, due to limit losses. A device acting to limit the current by increasing resistance is NOT what I would call a normal fuse or circuit breaker.

 

This is normal electrical fuses: http://en.wikipedia.org/wiki/Fuse_%28electrical%29

 

This is normal circuit breakers: http://en.wikipedia.org/wiki/Circuit_breaker

 

Lets make an example:

 

A 110 VAC supply with a 10 Amp slow-blow fuse of 0.01 Ohms, (cold), drained by a 11 Ohms power resistor.

 

When the power is switched on the fuse is cold:

- Current in the circuit is:110/(11+0.01)=9.99 Amps.

- Voltage drop over the fuse is: 9.99*0.01=0.0999 Volts.

- Energy generated as heat by the fuse is: 9.99*0.0999=0.998 Watts

 

After a short while the temperature inside the fuse is raised and causes an increase in electrical resistance. How much? Well if it's going to limit a dead-short to 10 Amps in needs to be 11 Ohms, but lets say it only doubles to 0.02 Ohms. (You can try any increase of resistance but the outcome will be the same.)

 

With increased resistance when the fuse is hot:

- Current in the circuit is: 110/(11+0.02)=9.98 Amps.

- Voltage drop over the fuse is: 9.98*0.02=0.1996 Volts.

- Energy generated as heat by the fuse is: 9.98*0.1996=1.99 Watts

 

As you can see the fuse gets hotter with increasing resistance, which will cause it to blow even faster.

 

When the fuse is no longer able to dissipate the excess heat, the internal temperature will raise fast, causing it to melt and cut the circuit long before its increase in resistance is able to significantly limit the current.

 

I hope I made my point clear this time - the resistance in a normal fuse is insignificant.

[Pleiades]

lol, people still use fuses? Most homes are built with circuit breakers these days.

[Spyman]

Circuit breakers or fuses won't make any difference to the question asked in the OP.