3
\$\begingroup\$

I was looking at an explanation for why parallel resistors have less resistance:

Resistors in Parallel

One of the answers involved simulating parallel resistors as a black box and it made sense.

If you have a "black box" with two wires connected and are told that there is a resistor inside you could measure voltage applied and current drawn to determine the internal resistance.

Now consider that there are TWO resistors inside and that they are in parallel. Again apply 10V and you will see that 2 mA (not 1 mA as before) is drawn. 1mA will flow through 1 resistor and 1 mA will flow through the other resistor.

The rest of the answer uses Ohm's law to explain why the parallel resistors must therefore have a certain equivalent resistance.

This seems to imply that resistors have a "pulling" force to draw electrons, which they definitely do not (since the whole point of a resistor is to resist the flow of electrons, right?).

But this answer does make sense. So why does adding more resistors result in more current being pulled?

\$\endgroup\$
15
  • 7
    \$\begingroup\$ the resistor does not pull current ... the power supply voltage is able to push only a specific current through the resistor ... the resistor is like a garden hose ... the pressure at the tap is able to push only a certain current flow through the hose, dependent on the hose diameter and length (resistance to water current flow) \$\endgroup\$
    – jsotola
    Nov 1 at 21:50
  • 3
    \$\begingroup\$ The resistance of a conductor is inverse-proportional to it's cross section. The more resistors are in parallel, the larger is the total cross-section. \$\endgroup\$
    – Eugene Sh.
    Nov 1 at 21:50
  • 2
    \$\begingroup\$ By the way, the same setup on a constant current source would not result in more current flow. This only applies (current increases) to a voltage source \$\endgroup\$
    – marco-a
    Nov 1 at 21:52
  • 1
    \$\begingroup\$ @Roymunson Cross-section is not the only factor. Resistance is proportional to the length of the conductor. When you put two resistors in series, you are effectively increasing the length. It is not "my" model, it is a physical model. physicsclassroom.com/class/circuits/Lesson-3/Resistance \$\endgroup\$
    – Eugene Sh.
    Nov 1 at 23:58
  • 3
    \$\begingroup\$ Do larger highways pull more cars? \$\endgroup\$ Nov 2 at 13:11
8
\$\begingroup\$

Resistors only "pull" current in the context of providing a path for electrons to flow from an area of higher energy (voltage) to an area of lower energy, dissipating the difference as heat. To use the fluid analogy, think of your voltage source as a tank of water at elevation. A resistor could be modeled as a pipe with a known flow rate for a given pressure difference across it. Your original scenario has one of these pipes stuck out of the bottom of the tank. Adding a second pipe next to the first would double the flow rate.

\$\endgroup\$
3
  • 4
    \$\begingroup\$ Resistors pull nothing - emf pushes current through them, as @jsotola correctly pointed out on the question. \$\endgroup\$
    – TonyM
    Nov 1 at 22:46
  • 1
    \$\begingroup\$ ... and if you're going to discuss electrons don't they flow from lower potential to higher? Conventional current flows from higher to lower. \$\endgroup\$
    – Transistor
    Nov 2 at 14:11
  • \$\begingroup\$ In other words, any resistor with resistance R can also be viewed as a conductor with conductivity 1/R, depending on their function in the circuit. Viewing them in this way is what made me really grasp the formula for parallel resistors. \$\endgroup\$ Nov 24 at 12:07
9
\$\begingroup\$

Resistors "draw" current the same way you "draw" beer out of a keg. Indeed there is nothing in the faucet or the mug which attracts the beer: it's the pressure inside the keg which pushes it out. With resistors, it's the voltage of the power supply which does the job, resistors just provide a path for the current.

\$\endgroup\$
5
\$\begingroup\$

‘Drawing’ or ‘pulling’ current or power are figures of speech. ‘Pull’ is synonymous with ‘draw’ in this context, and they both mean ‘load the circuit’, even though we know that electric charge is both ‘pushed’ and ‘pulled’ through a load (like a resistor) by potential difference, that is, electromotive force.

Draw/pull goes with the ‘water analogy’ of electricity. One ‘draws’ water from a well, so by that analogy one ‘draws’ power from a power source. You have this resource - power - and you’re taking some of it from a source and using it to do something.

So we might say a resistor will ‘draw/pull 10A on this branch’ meaning it will ‘load this branch 10A’. And the more parallel loads (resistances) you add, the more the load, or ‘draw’, from the supply.

This draw/pull stuff is more electrician-speak than engineering, though plenty of engineers use these terms too. For their part, electricians understand electromotive force and current perfectly well. After all, their (and our) lives and livelihoods literally depend on it.

With a resistor or resistive load, the energy ‘drawn’ gets converted into something else: light, heat, radio waves, mechanical motion, magnetic field, chemical bonds, etc.

\$\endgroup\$
4
\$\begingroup\$

Forget about "electrons" for regular circuit analysis. Just deal with current which flows from + to -.

Parallel resistors provide alternate paths for current to flow in the same way that parallel pipes provide alternate paths for fluid to flow. The more parallel paths, the lower the resistance.

This seems to imply that resistors have a "pulling" force to draw electrons, ...

No, resistors "resist" the electro-motive force (the voltage) pushing current through them.

But this answer does make sense. So why does adding more resistors result in more current being pulled?

It doesn't. Adding resistors allows more current to flow for a given voltage.

\$\endgroup\$
1
\$\begingroup\$

I'll give it shot with a water analogy to explain this:

Your voltage source is like a barrel full of water.

Now you stick a pipe (resistor) at the bottom of the barrel.

What happens? Water flows (current) from the barrel through the pipe out of the barrel.

Ok, now what happens if you add a second pipe? The flow is doubled, and the amount of water that is drained depends on the diameter of the pipe (bigger pipe = less resistance, tighter pipe = more resistance).

Of course that analogy is a bit flawed because current doesn't just leak out (like water does here) of circuits (it flows in loops instead).

However, I think the concept is applicable to electrical circuits.

It's only to be expected that more current (water) flows if you add more resistors (pipes) in parallel.

The resistor just provides mean for current to flow, just like any resistive load would (lamps for example).

If your voltage source isn't connected to anything, no current is flowing at all*, because it doesn't have the means to do so. But as soon as you hook a load (resistor for example) on it, current will flow because of the difference in potential.

So to summarize: it's the difference in potential that is causing the current to flow, it has nothing to do with the resistor itself (ever observed a resistor conducting current on its own?), the resistor is just creating a path for current to flow. This has been mentioned in the comments: add more paths and more current will flow.

*Unless your voltage source is in the kV range, then it could overcome the air resistance and arc over.

\$\endgroup\$
3
  • 1
    \$\begingroup\$ In a parallel arrangement, the resistors are completely independent of each other: the other resistor doesn't know nor care if you remove its neighboring resistor friend. \$\endgroup\$
    – marco-a
    Nov 2 at 2:58
  • 1
    \$\begingroup\$ excessively pedantic nitpick of your asterisk: it's not really connected to nothing, it's connected to air! \$\endgroup\$
    – Hearth
    Nov 2 at 3:00
  • 1
    \$\begingroup\$ Chemical engineer here: The analogy between resisters and pipes is more than just an analogy. The formulas used in calculating flow (water or current), potential (voltage or "head" (the vertical distance between the source of the flow and the outlet) and the resistance (in flow, this is generally measure in pipe-length (but is dependent on the type of pipe, the fittings, etc) are exactly the same in electrical land as in pipe land. Add parallel pipes, you double the flow. Make the pipe "longer" (without changing the vertical drop) and you increase the resistance. \$\endgroup\$
    – Flydog57
    Nov 2 at 16:58
1
\$\begingroup\$

I disliked that term, "draw", when learning electronics as a kid; it is quite misleading. Electrical resistance merely allows current to flow--that's it. (Proportional to the EMF/voltage, of course.)

That said, decades later, I now work in an electrical field. The perspective is different in that we need to know how much current/power a particular resistive load will allow to flow/consume from the total of limited supplied power, so the English term "draw" makes more sense there (like drawing water from a well). Drawing too much current from this fixed amount of supply could cause blown fuses, tripped circuit breakers, malfunction or even fires.

\$\endgroup\$

Not the answer you're looking for? Browse other questions tagged or ask your own question.