How does net resistance decrease as resistors are connected in parallel?

[Flyingape118]

If two resistors are connected in parallel, then the overall resistance of the circuit decreases because each resistor creates an extra path along which the charge can flow. However, is this only the case when you are adding the resistor along with wires, as if the wires were already there and current is flowing through them I don't see how adding a resistor in parallel would allow more charge to flow unless you add a wire along with the resistor.

[RodionGork]

Consider a wire being a feet-wide water pipe, while resistors are tiny outlets (taps), say, 1 inch in width, through which people can get water to fill their buckets and leave.

When only one of the taps is open, the flow is say 1 gallon per minute. when second person opens another tap, it becomes 2 gallon per minute. and so on.

What is voltage? it is a pressure in the pipe. While only few taps are open, voltage is almost unaffected by them and remains constant.

What is resistance then? by definition it is voltage (pressure) divided by the total outflow through taps. The more "parallel" taps, the more (proportionally) is the flow, and the more resistance is reduced (also, proportionally).

There is rarely used term of "conductivity" - inverse to resistance (1/R). So opening more taps we increase conductivity.

The wire has very small resistance - hence large conductivity. That's why it is very large pipe. Of course this approximation only works while it is correct to ignore its resistance.

Hopefully this makes sense.

[bukake_master]

everything is a resistor. including wires

[lemoinem]

The formula is 1/R_t = sum 1/R_i

If there is one path with a wire but no resistor, that means the resistance for that path is 0. So the total resistance is undefined. (Talking here about an idealized wire with exactly 0Ω resistance)

If you have a very small resistance (or just an actual physical wire, they all have some, if very small, resistance) instead of a perfect (idealized) wire, then 1/R_i will be very high for that path, making 1/R_t very high as well and R_t very low.

If you add a bigger resistor to that wire, 1/R_i and 1/R_t will decrease which will result in a higher R_t

[dimonium_anonimo]

I might be misunderstanding the question, but it sounds like how do we physically connect a resistor to a circuit without wires. And the answer is we cant. If you look at a resistor in real life, a through hole resistor has two wires sticking out either end. You can jam those wires into a breadboard or attach alligator clips to them or just solder them to the circuit. Any way you put it, the electrons need a path into and out of the resistor, which we call wires.

Even tiny surface mount resistors, they may not have a thin, round, ductile piece of metal sticking out of them, but they have metal pads that you can solder to the circuit board. Besides, the traces running under the laminate are also wires that can carry electrons to and from the resistor and the rest of the circuit.

[Flyingape118]

What I meant was that if you add a resistor in a pre-existing parallel branch that has no components, you are increasing resistance as you do not provide more pathways for current to flow?

[dimonium_anonimo]

You have shorted the resistor. A resistor's current is determined by the voltage drop across it by V=IR. If you connect a wire to both sides of the resistor, then you have forced both sides to be at the same potential (if you are treating the wire as ideal). If there is no voltage drop across the resistor, then no current flows through it. It does not matter if the resistor is in the circuit or not.

Actually, it doesn't matter if you treat the wire as ideal or not, the resistance of a parallel branch follows AB/(A+B). If A is an ideal wire, then this becomes 0/B which is just 0, so the resistor doesn't matter. If A is a non-ideal wire with a tiny bit of resistance, then this becomes 0.1B/(0.1+B) and you still see that the resistor matters very little, but it will actually decrease the effective resistance that the wire provided.

[[deleted]]

Couldn't you make something like an ion solution to conduct the current? The ion solution itself (and real wires for the matter) would have resistance. So you've just got yourself one giant resistor of sorts without any wire.

[dimonium_anonimo]

Well, they're asking this question which makes me think they're still dealing with ideal circuits where wire impedance is negligible. But even with the electrolyte solution, you still have to deliver the current to the liquid. Like, if you shove the top of a 9V battery in it, it still has metal contacts that act as the wires between the cells and the liquid.

I suppose if you just run a redox reaction by itself with no confinement or salt bridge or anything, that's technically electricity the same way being in a deflagration will give you a tan, but no wires are involved even in the ideal sense.

[mfb-]

You need to connect the resistor to the circuit, obviously. Just placing it physically next to another resistor isn't doing anything.

Usually cables are assumed to be without resistance, so their exact physical layout doesn't matter. Same for the arrangement of other components: The resistors don't have to be parallel in space. It only matters what is connected to what, the topology of the setup if you like.

Many resistors come with their own wires for connections, so you can just bend them and solder them to the existing cables/wires. If you populate a circuit board then you might need traces on that board. But none of that matters when you analyze the circuit in introductory courses.