Author Topic: Noob Q's - Current Load, and Voltage Drop  (Read 810 times)

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Offline bryancostanich

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Noob Q's - Current Load, and Voltage Drop
« on: July 18, 2017, 08:10:02 pm »
Hi folks!

I’m trying to understand some of the fundamentals of electricity and I’m struggling with the concept of voltage drop, load, especially as it relates to Ohm’s law.


1) How does a load draw amperage? More specifically, how does it draw _just_ the amperage it needs? Let’s say for instance, we’re talking bout residential mains, in which we have a 12A (15A nominal) circuit with a wall outlet. At that wall outlet is ~117V (depending on the length of the wire that feeds) it, and _up to_ 12A of available current. Now, I have two loads that I can attach, let’s say, a high powered hair dryer that wants to suck 10A, and some sort of charger that maybe wants 1A.

What limits the charger to only use 1A, is it just the resistance built into the circuit? What about the hair dryer? Is it the total resistance of the device? Is this just a simple application of Ohm’s law?

2) Do all loads cause a voltage drop? How do we know what the voltage drop for any given component is? Is it just published with the component? Does Ohm’s law have anything to do with voltage drop? Is it somehow predictable/computable/calculatable?

2a) I understand Ohm’s law states that resistance will drop the amperage through a circuit by a proportionate amount, but a voltage divider works because there’s a voltage drop at the end of each resistor. How is that calculated? How does that relate to Ohm’s law?

Thanks!!!
 

Offline Johncanfield

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Re: Noob Q's - Current Load, and Voltage Drop
« Reply #1 on: July 18, 2017, 09:00:04 pm »
Time to review Kirchhoff’s Circuit Laws, Thevenin’s Theorem, etc.
 

Offline bryancostanich

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Re: Noob Q's - Current Load, and Voltage Drop
« Reply #2 on: July 18, 2017, 09:08:02 pm »
thanks, i guess. but i came here to ask this question because i'm struggling with some of the concepts that underly those things. so, back to the questions above...
 

Offline rstofer

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Re: Noob Q's - Current Load, and Voltage Drop
« Reply #3 on: July 18, 2017, 09:13:20 pm »
Hi folks!

I’m trying to understand some of the fundamentals of electricity and I’m struggling with the concept of voltage drop, load, especially as it relates to Ohm’s law.


1) How does a load draw amperage? More specifically, how does it draw _just_ the amperage it needs? Let’s say for instance, we’re talking bout residential mains, in which we have a 12A (15A nominal) circuit with a wall outlet. At that wall outlet is ~117V (depending on the length of the wire that feeds) it, and _up to_ 12A of available current. Now, I have two loads that I can attach, let’s say, a high powered hair dryer that wants to suck 10A, and some sort of charger that maybe wants 1A.

What limits the charger to only use 1A, is it just the resistance built into the circuit? What about the hair dryer? Is it the total resistance of the device? Is this just a simple application of Ohm’s law?


We would be dealing with impedance rather than simple resistance since this is an AC circuit.  But, in the end, it's the voltage and impedance that determine how much current the device will draw.  You're solving I=V/Z because both V and Z are constants (for this discussion).

Quote

2) Do all loads cause a voltage drop? How do we know what the voltage drop for any given component is? Is it just published with the component? Does Ohm’s law have anything to do with voltage drop? Is it somehow predictable/computable/calculatable?

Appliances often specify the current draw.  This includes all the complications of impedance which brings in complex numbers or vectors.  Just take their word for it, they measured it with an AC ammeter.  Of course Ohm's law applies!  It's a LAW, not a suggestion!  The voltage drop is calculated from V = I / Z.  If you have the parameters for Z, yes, you can compute the answer.  Not necessarily an easy calculation but definitely doable in the first semester of electric circuits.

Sometimes, appliances just give wattage.  The get amps, divide wattage by voltage.  A = P / V then Z = V / A.  At some point in the algebra, P = I2 * R.
Quote

2a) I understand Ohm’s law states that resistance will drop the amperage through a circuit by a proportionate amount, but a voltage divider works because there’s a voltage drop at the end of each resistor. How is that calculated? How does that relate to Ohm’s law?

Thanks!!!

Resistance doesn't 'drop' amperage, it drops a voltage based on the current passing through.  It will, of course, limit the current.

For the voltage divider, add the two resistors (R1 + R2) and call that RT.  Figure the total current I = V / RT then calculate the output voltage (across the bottom resistor, R2) as V = I * R2

ETA:  If you were to add a parallel load resistor to R2 to simulate some other circuitry, the value of R2 is the parallel combination of these resistors.  It would necessarily reduce the output voltage.  That's why voltage dividers are not an ideal way to generate a specific voltage unless you have some kind of voltage follower (op amp) to drive the load.

ETA:  Stay away from AC circuits until you have nailed Kirchoff, Thevenin and Norton for resistive circuits.  The math for AC circuits get truly ugly.
« Last Edit: July 18, 2017, 09:20:47 pm by rstofer »
 
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Offline Johncanfield

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Re: Noob Q's - Current Load, and Voltage Drop
« Reply #4 on: July 18, 2017, 09:16:45 pm »
https://youtu.be/m4jzgqZu-4s  He has some outstanding animated videos explaining physics, circuits, etc.
 

Offline bryancostanich

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Re: Noob Q's - Current Load, and Voltage Drop
« Reply #5 on: July 19, 2017, 04:58:06 am »
@RStofer, thanks!

i'd like to dive deeper on this: "Resistance doesn't 'drop' amperage, it drops a voltage based on the current passing through.  It will, of course, limit the current."

ok, so maybe "drop" amperage is the wrong way to say it, but to "limit the current" means to reduce the quantity of charge carriers/amperage, yes?

I think i understand what you mean by drop voltage. thank you @johncanfield for the youtube link. It drove home Kirchoff's voltage law.
 

Offline rstofer

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Re: Noob Q's - Current Load, and Voltage Drop
« Reply #6 on: July 19, 2017, 06:52:51 am »
@RStofer, thanks!

i'd like to dive deeper on this: "Resistance doesn't 'drop' amperage, it drops a voltage based on the current passing through.  It will, of course, limit the current."

ok, so maybe "drop" amperage is the wrong way to say it, but to "limit the current" means to reduce the quantity of charge carriers/amperage, yes?

I think i understand what you mean by drop voltage. thank you @johncanfield for the youtube link. It drove home Kirchoff's voltage law.

I don't prefer to think in terms of dropping amperage.  Amperage is a result of other variables like applied voltage and total resistance.  I can't go to the store and buy an amp but I can buy voltage (battery) and resistance.

Dropping amperage sounds a lot like having two resistors in series and the first one dropping the amperage of the second.  This is not correct.  The amperage is the same through both resistors and is simply the voltage divided by the sum of the resistances.  But the current is totally determined by the applied voltage and the sum of the two resistors.  It is a result, not a cause.

There are current sources that do whatever is necessary to the voltage to push a certain current through a load.  They aren't ideal because at some point they simply run out of voltage.  You can't push 1A through a 1 MOhm resistor with a 1 volt source.

I think Ohm's Law is only a first tool for circuit analysis.  When you really understand Kirchoff's Laws this whole thing of voltage, resistance and current will become a lot more clear.  Somehow the loop and node equations tie the whole thing together.
 


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