Current Path - Im getting confused - Page 1
Forum section: electronics / projects

author: AngraMelo
author_rank: Regular Contributor
timestamp: December 10, 2018, 01:09:40 pm
content: Ok, first I would like to thank you. You have been tremendously helpful.Having said that, here goes one more question that is bothering me.Im struggling with understanding the phrase: current chooses the path of least resistance.Now imagine a voltage source and in parallel with a light. The current goes through the light and it lights up. Great.Now if I put a resistance in parallel with the light bulb, the lights shines brightly and the resistance heats up.Now imagine the resistance goes short circuit.My intuition says the bulb will not light up given the short.My question is: In the real world most short circuits would still have resistance. So basically the short is a very low resistance, given that the source can give the necessary amps, what/when/how is this threshold where the current stops going to the lightbulb and just goes through the short.I dont understand thatSorry if this is a stupid question

author: Ice-Tea
author_rank: Super Contributor
timestamp: December 10, 2018, 01:22:12 pm
content: As you say, a short circuit still has "some" resistance. In the most cases, it will present several orders of magnitude less resistance and can be considered "zero" for practical purposes.For other cases, you can calculate the current in both parallel sections you calculate for current by using the actual resistance in your calculations. It then becomes a "classic" case of two resistances in parallel.

author: MrW0lf
author_rank: Frequent Contributor
timestamp: December 10, 2018, 01:22:18 pm
content: Quote from: AngraMelo on December 10, 2018, 01:09:40 pmMore correct would be it will distribute proportionally preferring paths of least resistance, so current to bulb will never completely stop. Suggest to play with some (online) circuit simulator.

author: Berni
author_rank: Super Contributor
timestamp: December 10, 2018, 01:25:44 pm
content: Given an ideal battery the lamp would actually never go out, even if you shorted the battery directly at the terminals.Current does chose the path of least resistance but its not deciding what direction it goes into and puts all the current there. Its just a fancy way of saying that more current flows where there is less resistance. If something has half the resistance than the alternative path then twice as much current flows there as a result. Its all just Ohms law.The reason why the lamp goes out when you short the battery is that the battery has a limit of how much current it can provide. As the battery is overloaded the voltage on the terminals drops and if it drops to below the voltage the lamp needs then the lamp goes out. If you had an ideal battery that can supply an infinite amount of current, then the lamp would be unaffected during the short while the wire shorting the battery would quickly start to glow red hot and melt from the massive current flowing trough it.

author: ArthurDent
author_rank: Super Contributor
timestamp: December 10, 2018, 03:02:17 pm
content: Like most sayings, “current chooses the path of least resistance” tries to convey a concept but it isn’t a law and is not precise. Math is the precise language of science and the parallel resistor drawing below will explain what the saying actually means using ohms law.The current I1, flowing through R1 is E/R1. The current I2, flowing through R2 is E/R2. If R1 and R2 are equal then I1=I2 so the current is equal.If R1 is 1 ohm and R2 is 1,000,000 ohms then almost all the current will flow through R1, the 1 ohm resistor, and almost none will flow through the 1,000,000 ohm resistor, or as the saying goes, “current chooses the path of least resistance”. More precisely worded: “most of the current chooses the path of least resistance but the rest of the current will choose the other path”.The drawing below is also simplified but in real life the battery will have some internal resistance and the wiring will have some resistance as well. If the internal resistance of the battery is extremely low then the short circuit current can will be high and if the internal resistance is high then the short circuit current the battery can produce will be low, following ohms law. If you have a small 9 volt battery and you accidently short the terminals probably nothing will happen. If you have a 12 volt auto battery and accidently short the terminals you will probably get an explosion because the extremely low internal resistance means it can produce very high currents.Bottom line is that in all cases ohms law will determine what happens and you have to consider all sources of resistance in a circuit to be precise and predict the effects.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 10, 2018, 03:21:14 pm
content: Right, "chooses the path of least resistance" is an embarrassingly bad oversimplification that should die a gruesome and ignominious death.Current flows proportional to the resistances in the path. For the final say, follow the math.Tim

author: rstofer
author_rank: Super Contributor
timestamp: December 10, 2018, 03:35:35 pm
content: Quote from: T3sl4co1l on December 10, 2018, 03:21:14 pmExactly! The current follows ALL of the paths.Ohm's Law is a LAW, not a suggestion! You can't treat it like a speed limit.Given two resistors in parallel, both see the same voltage. The current through each resistor is simply the voltage divided by the resistance. Since we are dividing by resistance, a high value results in a low current and vice versa. Dividing by 0 results in a value approaching infinity. That's why the car battery is so dangerous - the internal resistance is close to 0 Ohms.As to what voltage makes the light visible, it depends on the filament and how 'visible' is being measured. An eyeball may not be the most effective measuring device for low levels.

author: AngraMelo
author_rank: Regular Contributor
timestamp: December 10, 2018, 03:53:25 pm
content: So let me see if I got this.If the source can supply infinite amps and I short the resistance in parallel with the lightbulb then the light bulb would keep its brightness and the short would cause infinite (proportional to the resistance of the cables that are shorted) amps to flow through them, is that right?Now, this is where this question came up. On the example attached is the most common currently limiting circuit I see around. As I understand it it is based on the idea that Q2 steals current from the base of Q1 when the voltage drop on the sense resistor goes above the diode drop voltage on the base of Q2. Now, I believe it is a fair assumption to say that most power supplies would be able to supply current to the base of Q1 and the collector of Q2 without problems. So why does the circuit works?

author: IanB
author_rank: Super Contributor
timestamp: December 10, 2018, 04:06:00 pm
content: Quote from: AngraMelo on December 10, 2018, 03:53:25 pmBecause of R1. The maximum current that can flow through R1 is Vcc/R1, and this current has to be shared between Q1 base and Q2 collector. If Q2 turns on then more current will go through Q2 and less into Q1.

author: AngraMelo
author_rank: Regular Contributor
timestamp: December 10, 2018, 04:18:28 pm
content: Oh my goodness, of course... my bad. I should've seen that!Thank you!

author: coppercone2
author_rank: Super Contributor
timestamp: December 10, 2018, 04:38:25 pm
content: Quote from: T3sl4co1l on December 10, 2018, 03:21:14 pmhow do you know if there is not a path thats so nice some electrons go back and tell the other ones to completely avoid all the other ones?