Electronics > Beginners
Some noob questions
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PA4TIM:

--- Quote from: Terry01 on July 16, 2018, 09:03:32 pm ---I think you've got it the paddy way round buddy. Volts=power current=flow of volts.
Current is the flow of volts and volts is the potential energy between 2 points.
Think about filling a sink with a hose(current) the water(volts) then filling a paddling pool with the same hose(current). You'd crank the tap up more when filling the pool compared to the sink but the hose/"current" stays the same but the water(volts) is more powerful.

It'll click soon  :)

--- End quote ---


Current is the stuff that makes the lamp burn. It needs a difference in voltage between two points and something between those points to flow through. If the voltage is high enough and the (resistance not to high) current can flow.
Mr D:

--- Quote from: PA4TIM on July 16, 2018, 09:11:04 pm ---
If you use a voltage source and you set the voltage at 50V, the current through the 10 ohm lamp will be 5A and through the 20 ohm amp it will be 2,5A


--- End quote ---

I don't understand this. The more power-hungry the lamp, the less amps it needs? Seems like the wrong way round?!

Or is it the case that the amps are what's left over after work has been extracted by the resistance?
rstofer:
In the usual workings of electronics, current is a result, not a cause.  Yes, there are current sources but they are rare unless working with op amps built from discrete components.  I can't recall ever playing with them but I do know they exist.

Usually, I have a voltage source (battery or power supply) and I have a load (as simple as a resistor or as complex as a PC) and the result of operating the load on the source results in a certain current flow.  The current is the result, not the cause.

It doesn't matter much because Ohm's Law relates all 3 variables and we can swap them back and forth at will.  When you get to Kirchhoff's Laws, you will be writing equations in terms of current flow or voltage drop.  This is where it will all make sense.  In fact, it isn't even possible to analyze a simple op amp circuit without writing Kirchhoff's Current Law at the inputs (explicitly or just winging it).  It's simple to write (in the case of linear feedback, less so if the device is used as an integrator) but key to understanding the closed loop gain is calculated.

Get the idea that water pressure is kind of like voltage, pipe friction resistance is kind of like electrical resistance and water flow rate is like current and then put the analogy away.  It was just created to give an intuitive feeling toward the 3 fundamental units.
PA4TIM:
Current sources are not very rare, but voltage sources are much more common.  Some loads need a current source, like a led, because they will draw more current then they can handle. Other parts will need a voltage. The voltage divided by the resistance will be the current draw. And the voltage times the current gives the power the load dissipates and that should be not more as the load can handle or the smoke escapes.

rstofer:

--- Quote from: Mr D on July 16, 2018, 09:28:20 pm ---
--- Quote from: PA4TIM on July 16, 2018, 09:11:04 pm ---
If you use a voltage source and you set the voltage at 50V, the current through the 10 ohm lamp will be 5A and through the 20 ohm amp it will be 2,5A


--- End quote ---

I don't understand this. The more power-hungry the lamp, the less amps it needs? Seems like the wrong way round?!

Or is it the case that the amps are what's left over after work has been extracted by the resistance?

--- End quote ---

No, this is bass ackwards!

Current is a result in most cases.  A lamp will have a rated voltage and a rated power.  Let's say 100W at 120V...  So from P=E^2 / R (E squared over R), we can calculate that the hot resistance is 144 Ohms.  We can calculate the hot current from P = I^2 * R or I = sqrt(P / R) or, in this case, 0.8333 Amps.  We can also use P = I * E to get watts so 100 = 0.8333 * 120 (which it does).  All of the manipulations of Ohm's Law should be consistent,  P, I, E and R are all tied together.

So, what is really happening?  We apply a voltage to a lamp.  The internal resistance (assume hot) connected to a certain voltage will produce a certain current flow.  That current flow times the voltage will result in a certain amount of power being dissipated.  Mostly, it will come off as heat in the case of incandescent lamps.

What I have avoided saying is that the incandescent lamp changes resistance as the filament heats  up.  The value is much lower when cold.  You can just grab a 100W lamp and measure the cold resistance if you are interested.  I suspect it will be quite low and nowhere near 144 Ohms.  But I don't want to go there...

If you really want to chase the topic down a rathole, calculate the instantaneous current when you apply 120V to that cold filament resistance.  There are a lot of reasons why the current may not go quite that high but it does explain why lamps burn out when they are switched on.
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