I’m really confused about how current works in a parallel circuit. It seems to me that the current flowing between two nodes is completely independent from any other node.

In this circuit the current after each resistor is V_{i}/R1, V_{i}/R2, V_{i}/3, right? So the value of R1 does not affect the I after R2 correct? If this is true, then why would the circuit short if R3 was removed? I guess the problem is, I don’t understand why a circuit would short. Mathematically speaking, shouldn’t current flow through R1 and R2 regardless of whether or not the last set of nodes has a load? I’m sure there is some kind of mathematical explanation but I can’t figure it out. Thanks for the help.

By removing R3 and replacing it with a wire, you have created a ultra low resistance path for a large amount of current to flow, and thus a short circuit.

Total resistance in parallel is calculated 1 / R = 1 / R1 + 1 / R2 + 1 / R3 +..., different from series.

You can figure out the total equivalent resistance seen by the battery with the 3 parallel loads.
Resistance of 2 Rs in parallel is R1R2/(R1+R2)
So say all 3 were 1K, then
R(1||2) = (10001000)/(1000+1000) = 500,
and that in parallel with R3 =
R1||R2||R3 = 500*1000/(500+1000) = 333.3 ohm.

If the battery was 5V, and the same 1K resistor, then 5V/1000 = 0.005A would flow.
With 3 parallel resistors = 333.3 ohm, then 5V/333.3 ohm = 0.015A.
Makes sense, yes? Each path allows 5mA to flow, 3 paths in parallel = 15mA.

Now if R3 changes, you can do the math and figure out the total current flow.
If R3 is removed, you have two 5mA paths.
If R3 is decreased, the current flow in that path increases. If 5V is maintained across the battery, then the current flow calculations will remain true. If the battery cannot maintain that current level, the voltage will droop - such as if R3 changed to1 ohm, but the battery only had 500mA capacity:
5V/1ohm = 5A, but battery only has 500mA capacity -> voltage will drop as you are effectively shorting out the battery.
The battery will try, and may overheat and fail. Or the resistor may overheat: power dissipated = V^2/R, so 5V*5V/1ohm = 25W, so the resistor may overheat and fail. If it fails open, voltage remains at 5V, and current flow returns to that calculated for R1 & R2.

LarryD sorry if I didn't make this clear, I meant replace R3 with a wire. CrossRoads I followed most of what you said. Correct me if I'm wrong, but what you're essentially saying is that when you replace a resistor with a wire, it demands such a high current that the voltage can't keep up and that in this case V=IM would not apply?

So the old adage that electricity always follows the path of least resistance is a myth. In a parallel circuit current flows down all possibles paths and any one path does not effect the current in any other.

That is also true of a short as long as the battery holds up. Even if it doesn't the current in any leg of a paralle circuit is still what ohms law would give you if it were the only leg and you had the same input voltage as your now hobbled battery.

Grumpy_Mike:
So the old adage that electricity always follows the path of least resistance is a myth. In a parallel circuit current flows down all possibles paths and any one path does not effect the current in any other.

Many many years ago, I was taught that, that myth was a fact in school by my physics teacher. I can't even begin to explain how much confusion "that fact" has caused me when I started getting into electronics years later. I had to take the long path to learn the truth, because of one statement.

Electricity follows the path of least resistance applies to insulators breaking down, not conductors.

ie lightning striking something. The existence of any path with a lower resistance than the rest
of the insulator (we are talking 10^12 ohms rather than 10^16 ohms for instance), means pretty
much all the energy is concentrating into a path of weakness, then highly accelerated charge
carriers blast through it and cause more damage until the entire path is a plasma (which conducts
like a metal).

Of course in ordinary dc circuits more current flows in the paths of lower resistance, but that's not
really a process of finding a path.

Even so all possible paths have current in them determined by ohms law. It is just that more current flows in the lower resistance paths. Conductors / insulators are actually a continuum and is the second widest range of physical phenomena known.

For a bonus point name the phenomenon with the widest range of values.

No, the Gnd connection takes all available current, so the voltage at the input is 0V.

With the switch open, 5v/R1 is available to charge up the input capacitance of the gates of MOSFET transistors that make up the input pin.

If the resistor is the internal pullup resistor, spec'ed at 30-50K, than that's as low as 0.1mA, so it doesn't take much, and in fact input current for digital pins is spec'ed at 1uA max (0.000001A).

DRLDavis:
So I asked this question because I don't understand how a pull up resistor works.

Why would a closed button cause the input to read zero? Isn't the input pin still receiving the same amount of current as it did with an open button?

R1 is a high value. When the switch is open the input is "pulled up" to 5V. When the button is closed it is "shorted" to ground. The reset pin doesn't have to be ZERO to reset. It only has to be a low value. Check the data sheet if you really want to know the voltage.

If you have a meter you can check this pretty easy. Measure the voltage from ground to the pin and press the button.

You might want to check up on voltage dividers. You are setting one up here but the 'bottom' value is ZERO OHMS so you get no voltage across it. If you replaced the switch with a resistor of equal value to R1, lets call it R2 then you would divide the voltage in half. As the resistance of R2 is lowered the voltage across it becomes lower. When you get to ZERO resistance you have ZERO voltage across it and therefore the pin is at ZERO in relation to ground.

I almost understand!!! I think... But how does it work with a pull-down resistor?

My question on this is how does the input receive any current at all when the switch is open? Wouldn't the pin receive a floating signal? It is completely disconnected from the source of voltage. I know its connected to ground but how does this matter if there is no electricity flowing through it?

P.S. Sorry for being such a confused soul. I have searched the internet everywhere for answers to these questions but I can't find them. Thank you for your patience and help!

Current flows out of the input, trough the pulldown resistor, to ground.
This pulls the input low.
If the resistor is increased in value, the current through it will result in a higher voltage and may approach an in determinant logic level, neither high or low.
This is why there is a maximum effective value for a pulldown resistor.
For Arduino we use 10K to produce an effective logic low.

When the switch is closed 5V (vin) pulls the input high.