
Re: PARALLEL CURRENT
genebren Mar 6, 2018 2:53 PM (in response to salesm21)7 of 7 people found this helpfulAll current comes from the source that is providing the potential (assuming that it is an ideal source, with no series resistance).
In the initial steps of understanding electronics the supply is ideal (no limitations). As you move forward in your understanding of electronics you will view sources as having real word limitations (some source resistance and some maximum current available).
When you have resistors in parallel, they act as a single resistor, i.e your 5 resistors of 5 ohms each. The equivalent resistance is 1 ohm. So for your example the total current flowing through the load would be 10A.
Gene

Re: PARALLEL CURRENT
jw0752 Mar 6, 2018 4:53 PM (in response to salesm21)3 of 3 people found this helpfulHi Mitchell,
Many years ago when I was just starting in electronics I first encountered Ohms law. Using the formula I determined that I should be able to get an awesome flash of current by putting a dead short across a 6 volt lantern battery. I got everything set up for the experiment and using a piece of 12 Gauge wire I made the connection. Needless to say there was no spectacular current. At the time there was no internet and I had no one around me that knew anything about electricity so the reason why my experiment failed remained a mystery to me for some time. I remember the aha moment when I eventually came across the idea of internal resistance in all power sources. Under normal operating conditions the internal resistance is low enough compared to the external resistance so that it is irrelevant. However as the external resistance gets lower and approaches the internal resistance it becomes a greater and greater factor. You can calculate the internal resistance by seeing how much the voltage of the battery drops when put it under an external load. Calculate the current in the circuit by measuring the voltage drop across the load and divide it by the resistance of the load. Next divide the voltage drop of the battery by the current and you will get an approximation of the internal resistance of your source at this current level. Keep in mind that the resistance of a source may not be linear with respect to external current as there are several factors that produce the internal resistance.
John

Re: PARALLEL CURRENT
fmilburn Mar 6, 2018 7:41 PM (in response to salesm21)3 of 3 people found this helpfulMechanical Engineer here :)
I don't think of it this way but the water analogy might help. Imagine a large elevated tank with water. Now imagine 5 small diameter pipes of fixed length L hooked together in series connected to it  they will restrict flow (current). Now imagine the 5 pipes in parallel connected to the same tank  they will provide less restriction and there will be more flow (current). The 5 pipes in parallel could serve as a replacement for one larger diameter pipe of fixed length L while the 5 in series could serve as a replacement for one smaller diameter pipe of fixed length L.
If the pipe is large enough the diameter and size of the tank small enough then pressure drop and change in water depth may impact the flow  kind of like the nonideal nature of a real battery.
Having said this, the sooner you get to thinking about Ohm's Law and electricity as electricity instead of water the better :)

Re: PARALLEL CURRENT
mcb1 Mar 6, 2018 10:35 PM (in response to salesm21)4 of 4 people found this helpfulMy question is where does all the extra current come from?
Ohms law assumes the supply voltage and current is whatever your figure ends up.
5 x 5ohm resistors in series across 10v will be I=E/R = 10/25 = 0.4A as you calculated.
There is only one path for current to flow which is throught the first resistor, then the second and so on.
If you measured the voltage across each resistor it would be 2 volts as it gets 'spread' across each resistor.
BUT 5 x 5 ohm resistors in parallel means there are 5 seperate circuits that are across the 10v source.
The current in each path is I=E/R = 10/5 = 2A, but there are 5 paths so the total current is 2 x 5 = 10A
The equivalent resistance is 1/Rt = 1/R1 + 1/R2, etc = 1/5 + 1/5 + 1/5 + 1/5 + 1/5 = 1 so 1/Rt = 1 which equates to 1.
I=E/R 10/1 = 10A.
The picture and calculations here might help.
https://www.swtc.edu/Ag_Power/electrical/lecture/parallel_circuits.htm
So these calculations show what the expected current will be flowing in the circuit using these values.
If it was implemented, you would require a supply capable of supplying 10v at 10 Amps.
Mark
Recently been thinking about the ways current flows through a parallel circuit and it confuses me. So all connections on a parallel circuit are at equal potentials with the source. So no two components are getting anymore or less energy than the others. However based on the resistive nature of that branch will determine the current that is pushed through that part. So id like to simplify this by taking two circuits with equivalent parts. Both circuits have a DC source of 10V and each has 5 resistors equaling 5 ohms each. In series each resistor would draw away not current but the energy that is propelling electrons down the conductor. The circuit has .4A of current. In a parallel circuit all five have 10volts dropped on them and therefore each has its own current (In this case they would all be identical but each is drawing a separate current.) My question is where does all the extra current come from? Is it front the extra paths or was it always in the source?