# Dumb Question

#### nitrusoxyde

Joined Nov 6, 2005
2
I feel extremely dumb for asking this, but hopefully someone can explain something that's always confused me.

My parents taught me the basics of DC electricity when I was extemely small, about 4 years old..... now I manage a RadioShack. What a coincidence. I got into the concepts of digital electronics when I was in highschool and college, yet I still have a lingering issue.

I've long been perplexed by some of the simpler characteristics of DC electricity. In particular, its characteristic of following the "path of least resistance". I get really hung up on this. It seems that if I connect two LEDs in parallel, one with a 10 ohm resistor and one with a 10000000 ohm resistor, the 1ohm should be brighter - AND IT IS! ... but the LED with the high resistance value still lights.

Ok, so lets compare this with an LED with its inherant resistance by adding another branch to the parallel circuit - still the same.

Yet now, when I finally add a simple short to ground, the circuit is dead - I think I understand why to a small degree. Basically both sides of the LEDs and resistors are becoming electrically common. The difference is that there's no resistor or component in the shorted circuit. But what about the small amount of resistance in a copper wire? Is this not enough to keep it from being a "short"?

As I said, I feel really stupid. What logic do I follow here? Part of my problem is that it is difficult to explain my confusion!

#### n9352527

Joined Oct 14, 2005
1,198
Originally posted by nitrusoxyde+Nov 9 2005, 07:59 AM--><div class='quotetop'>QUOTE(nitrusoxyde @ Nov 9 2005, 07:59 AM)</div><div class='quotemain'>I've long been perplexed by some of the simpler characteristics of DC electricity. In particular, its characteristic of following the "path of least resistance". I get really hung up on this. It seems that if I connect two LEDs in parallel, one with a 10 ohm resistor and one with a 10000000 ohm resistor, the 1ohm should be brighter - AND IT IS! ... but the LED with the high resistance value still lights.
[post=11548]Quoted post[/post]​
[/b]

It is more correct to say that the path that has lower resistance will have greater current flowing through it than the parallel path with higher resistance. So it is not the case of the path with higher resistance will not have any current flowing through it.

Ohm's law says that the current is proportional to the voltage across the path and inversely proportional to the path resistance. i.e. I=V/R. From this formula we can see that current for both path depends on the resistance, because the voltage is equal.

<!--QuoteBegin-nitrusoxyde
@Nov 9 2005, 07:59 AM
Ok, so lets compare this with an LED with its inherant resistance by adding another branch to the parallel circuit - still the same.

Yet now, when I finally add a simple short to ground, the circuit is dead - I think I understand why to a small degree. Basically both sides of the LEDs and resistors are becoming electrically common. The difference is that there's no resistor or component in the shorted circuit. But what about the small amount of resistance in a copper wire? Is this not enough to keep it from being a "short"?
[post=11548]Quoted post[/post]​
[/quote]

There is a resistance in the short or copper wire as you said, but the value is very very small compared to the normal path resistance. Therefore, to maintain constant supply voltage across it the current would have to be much much greater. Let say the supply voltage across the path is 5V and the short circuit resistance of the copper is 0.001 ohm. Back to Ohm's law the current needed to maintain the 5V across the short would be 5000A! And because most supplies are limited in capability (internal resistance or current limiter etc.) and can not supply the 5000A, the supply voltage across the shorted path drops to a lower value.

We can calculate the voltage across the short copper wire. Let say that the maximum current of the supply when shorted is 5A , the voltage across the wire would only be 0.005V, which is very small and would not light up an LED. If you have sensitive and accurate volt meter you could measure it though.

#### nitrusoxyde

Joined Nov 6, 2005
2
Originally posted by n9352527@Nov 9 2005, 05:06 AM
It is more correct to say that the path that has lower resistance will have greater current flowing through it than the parallel path with higher resistance. So it is not the case of the path with higher resistance will not have any current flowing through it.

Ohm's law says that the current is proportional to the voltage across the path and inversely proportional to the path resistance. i.e. I=V/R. From this formula we can see that current for both path depends on the resistance, because the voltage is equal.
There is a resistance in the short or copper wire as you said, but the value is very very small compared to the normal path resistance. Therefore, to maintain constant supply voltage across it the current would have to be much much greater. Let say the supply voltage across the path is 5V and the short circuit resistance of the copper is 0.001 ohm. Back to Ohm's law the current needed to maintain the 5V across the short would be 5000A! And because most supplies are limited in capability (internal resistance or current limiter etc.) and can not supply the 5000A, the supply voltage across the shorted path drops to a lower value.

We can calculate the voltage across the short copper wire. Let say that the maximum current of the supply when shorted is 5A , the voltage across the wire would only be 0.005V, which is very small and would not light up an LED. If you have sensitive and accurate volt meter you could measure it though.
[post=11550]Quoted post[/post]​
Awesome, thank you for the detailed explanation. I feel pretty stupid when I ask a question that basic.

#### Firestorm

Joined Jan 24, 2005
353

Awesome, thank you for the detailed explanation. I feel pretty stupid when I ask a question that basic.
No questions are stupid therefore the person asking a question isn't stupid.

The electricity will go to the least resistive path, but not ALL of it will.