Alternating Current

With the speed of light, yes it will be different. If you have long enough wire, you could measure it...get 150,000KM of wire, then it should be inverted from end to end ;)

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MVH,
Vidar

www.bitsex.net

At 50Hz the wavelength is 300000km/50 = 6000km. So half of it is only

3000km of wire ;) (Or maybe I did a mistake?)

I see.

Anyway. In Alternating Current. Isn't it the voltage is the one alternating from 110 volts to -110 volts. How come it's not called Alternating Voltage?

Suppose the voltage is going down from 110 volts to 0. What happens to the value of the current? How about the one where it goes down from 0 to -110. What happens to the value of the current in between the transition.

Tnx

Lorr

Because, in a purely resistive load, the current will follow the voltage exactly. However, in a inductive or capacitive load, the current won't follow the voltage, but both will alternate. In a inductive load, the current will be 90° before the voltage, in a capacitive load, 90° after the voltage.

The current decrease (again, in a purely resitive load) as voltage decrease. Current increase as voltage increase. Remember that 0V and

-110V still has a potential of 110V, you only redefine 0V. You could as well swap the terminals and say 110V and 0V. 0V is a question of how you define it;) Negative voltage means the current flows opposite direction of a positive voltage... The potential is still the same.

--
MVH,
Vidar

www.bitsex.net

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Thanks. Last.

Suppose in a DC (Direct Current) circuit with 10 volts source and 1 ampere. Is there a difference in the current in the wires if you use a resistor of 10 ohm vs a 10 ohm load (such as a light bulb)? Does the value of the resistor stand for the amount of current the load needs?? Does this mean heavier load such as dc motors has lesser resistance and more current?

Lorr

You made a mistake :-) The delay has nothing to do with the frequency, but with the resistance and capacitance of the used wire. Impossible to calculate without knowing material and diameter of it.

Regards,

Boris Mohar

Got Knock? - see: Viatrack Printed Circuit Designs (among other things)

U=R·I All electronics fall back to this. If you have a given voltage, and a given load, there is a given current. The other way too: If you have a given load, and a given current, you can easily find the voltage! Place 10V in U (Usually called V in english I realise, I'm probably using the norwegian symbols for this...),

10Ohm in R, and calculate I. (Twisting this: R=U/I I=U/R P=U·I)

--
MVH,
Vidar

www.bitsex.net

voltage

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Theoretically, yes. A resistance of 10 Ohms is a resistance of 10 Ohms.

Practically, no. The light bulb (mentioned above) has some built in inductance. This will cause the bulb to draw more current (more than 1 Ampere) when the switch is turned on. The current will then drop exponetially to the steady-state value of 1 Amp.

Knowing the resistance enables you calculate the current for any voltage. By applying a voltage of 10V to a resistance of 10 Ohms, the current will be

1 Amp.

Does this mean heavier load such

In theory, for the load current to increase the voltage must be increased or the resistance must drop.

Dwayne

No, the drop is due to the fact that resistance increase as the filament is heated. The inductance is so close to zero that'd you'd require equipment worth severel 1000$ to measure it...

In practice too....

--
MVH,
Vidar

www.bitsex.net

But considering an ideal situation (no capacitance, no resistance...)? Shouldn't the frequency have a lot to do with it? What am I missing?

At (say) 1Hz the wavelength is 300000 km, right? Using a 300000 km wire at the wire ends the phase is the same? No?

There would be very little difference (10 Ohms is 10 Ohms). However light bulbs aren't perfect resistors. The resistance varies with temperature so as the bulb heats up the current falls.

Essentially yes.

I would forget motors for the moment. Just think about resistors... What you say is a very good way to think of a load/resistor. I like to think of the load "sucking" the current out of the supply. The lower the resistance the more current it sucks out of the supply. The equation is I = V/R so the smaller R is the bigger I gets. It's interesting to work out what happens if you replace a resistor with a big fat wire which has nearly zero resistance.

On the other hand I like to think of voltage sources (batteries, power supplies etc) "pushing" current into the load. eg the higher the voltage the more current is pushed through the load. The same equation I = V/R .... Now think what happens if V is made large, very large.

DC Motors are much more complicated than they look and don't make good examples for beginners. A motor converts electrical energy into mechanical energy. The current/energy drawn (sucked) from the power supply depends on the mechanical load connected to the motor. Think about a food mixer. Put water in it and very little energy is needed. Put bread dough in it and it tries hard to go at the same speed but draw more current. A perfect motor would convert ALL the electrical energy into mechanical energy but real motors aren't perfect. The copper windings have some resistance and this causes energy to be lost as heat.

Oops, sorry, you're right... it seems i misplaced my decimal point (while calculating it in function of time, instead of wavelength). Resulted to 30000km instead of 3000km. So actually it's even less than

3000km, because electrons travel only at 2/3 of the lightspeed through a wire :-)

Btw: by saying that delay had nothing to do with frequency, i meant that the delay within a wire will always be the same, no matter the frequency. Of course, when using wires to shift signals, it should be taken into account eventually. :-)

In response to what Mark VB posted in news:04rke.99400 $ snipped-for-privacy@phobos.telenet-ops.be:

Electrons move through a wire extrememly slowly. It's the wavefront that moves quickly. Noise in air travels a mile in ~5sec, but the air moleculse don't.

--
Joe Soap.
JUNK is stuff that you keep for 20 years,
then throw away a week before you need it.

Guys,

Suppose you short the wire in the battery. The current will be maximum right? Suppose it takes 24 hours for the battery to be used up in a

10 ohm load. If you short it, would the battery be used up in say 1 hour. Is the operation of the battery in the shorting mode the same as in normal only accelerated?

How about AC lines. If you short the wires, it sparks and explodes. So I assume current is maximum. However, if you use wires that can never melt or break. What would happen to the other end in the power utility. Would it destroy their circuit?

Suppose I use a regulated dc supply and I short the + and - outputs. Would the circuit be damaged or would the wires just heat up without any damage to the source? I'm asking this because I plan to use a DC regulated power supply and make the signal vibrate at 1 Mhz and direct it to the antennae to experiment with em radiation. I don't plan to use any resistor because I want the current to be maximum so the em produced would be maximum. I wonder if this would damage the source. What sort of protection (diode?) that can be used in the source to detect and prevent shorting damage (in case this happens).

Thx

Lorr

You are not allowed to do that.

You are allowed to read this.

Read the whole thing. Twice.

Regards,

Boris Mohar

Got Knock? - see: Viatrack Printed Circuit Designs (among other things)

The laws of physics don't change, the load however goes to the minum (being the internal resistance of your battery plus the presumably negligable short-circuit).

See above. In this case the utility company relaises there are idiots around who short out their power lines so they put in circuit breakers.

If it's a good quality supply it will be itnernally protected, either shutting down or blowing a fuse. Maybe it'll break - how much are you going to pay for it?

Eh? Get a transmitter and appropriate licence or precautions.

Ken

The energy stored in a battery is measured in "Ampere-hours". Theoretically, if you have a 20 Ampere-hour battery, it can deliver 20 amperes for one hour, or 10 amperes for two hours, or one ampere for

20 hours, (or any combination of time and current where time X current = 20) before becoming fully discharged.

In practice, if you discharge a battery rapidly, you will get less total energy out of it than if you discharge it slowly. Lead-Acid storage batteries are normally rated at a "20 hour rate" - based on the current required to discharge them in 20 hours. So a 20 AH battery will deliver 1 amp for 20 hours, but may only deliver 15 amps for 1 hour.

There are normally several fuses or circuit breakers between you and the power company's generators - the lowest-rated fuse or circuit breaker will blow, and remove the power, when you short the circuit.

There is always some resistance in the wiring which will limit the maximum current you can draw.

Depends on the power supply - many regulated supplies are protected against excessive current drain - they will shut down, or reduce the output voltage to keep the current within limits.

I think you need to learn a LOT more electronics before you do this!

Consult some Amateur Radio manuals for information on radio transmitters - and be aware that transmitting a signal without an appropriate license is illegal.

--
Peter Bennett, VE7CEI  
peterbb4 (at) interchange.ubc.ca  
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