Electrical wiring: the "last inch"

[Doug Miller]

Adding resistance to a circuit changes voltage, not current.

[Metspitzer]

I can't believe what I am hearing.

[David Nebenzahl]

On 8/30/2009 1:58 PM Doug Miller spake thus:

Changes current, too. E.g., a current-limiting resistor in series with a LED, sized so that the LED won't draw excess current and burn itself out.

[Doug Miller]

Oh? What's the problem?

[Doug Miller]

Wrong.

If you want to limit the current passing through an LED, you put a resistor in

*parallel* with it, not series.

Total current in the circuit remains the same.

[Tony Hwang]

Hi, If you are not joking trying to be funny. Your basic knowledge is LACKing. Please be quiet if you don't know about some thing.

[Tony Hwang]

Hmm, You are very funny. He was right.

[Josh]

If you add resistance in series to the destination load (which is what higher wire resistance does), the *total* R seen by the source is higher. Assuming a constant V at the source, this means I (current) through the wire will be lower.

At the *load*, V will be lower also (often called the "IR drop" of the wire) -- the load is now part of a "voltage divider".

So yes, adding resistance changes both voltage (away from the source) and current.

Another way to think of it -- if the V across the final load (constant R) is lower, the current (I) must also be lower. Circuit analysis can be kind of fun -- you can often approach it from multiple perspectives and get the same answer.

This assumes a constant resistance load, which a light bulb isn't completely, but is sufficient for this purpose.

Josh

[Doug Miller]

No, it doesn't. It means *voltage* on the other side of the resistance will be lower. Current is the same at all points in a series circuit.

^^^^ You misspelled "only".

No, it doesn't.

Wrong again. Current is the same at all points in a series circuit.

Evidently you've found multiple ways to get the same wrong answer.

Back to Circuit Analysis 101 for you, and this time pay attention when the instructor discusses Kirchoff's Current Law.

You're assuming a *lot* of things; unfortunately, almost none of them are correct.

[Tony Hwang]

Hi, R is lower and I is lower? Where does it come from?

[bud--]

Agreed. That was probably the nuttiest comment in this thread.

As others have said, length is a consideration only to limit voltage drop. Length has no relation to the current carrying capacity of the wire.

The code is rather pragmatic. #18 fixture wires on a 20A circuit are not a problem because the lamp socket can not have a bulb over the wire rating. (Someone could screw in an adapter-to-plug and run a space heater - the rules of natural selection would then apply.)

Apparently there are not major problems with overloading #18 extension cords on 20A circuits - dead bodies are very effective at promoting code changes, probably also effective at UL. If I remember right, a consideration was that the available fault (short circuit) current at the end of the #18 cord is high enough so a short will give "instantaneous" operation of the circuit breaker/fuse (not time delay). (That is not likely true for Christmas tree lights.) AFCI breakers may also help (but the selling point was cords that had been abused, like walking on them).

You might be upset with motor circuits, that can have a circuit breaker significantly larger than the wire ampacity. Welder circuits even bigger difference.

[Mark Lloyd]

I guess you didn't try that.

The resistor in parallel wastes power, and will have no effect on current through the LED (other than if it creates excessive voltage drop, a really inefficient way to dim the LED).

With no resistor in series, current (from a constant voltage source greater than about 2V) will approach infinity, until the LED is destroyed. I've seen that happen.

Not when you add series resistance. LEDs regulate the voltage across them, unless the current is excessive. Since that voltage is around 2V, a 120V source would require a really big series resistor. To avoid that, 120V would usually be applied to a series string of multiple LEDs (in series with one smaller resistor).

A string of 50 LEDs (common in holiday lights) using 20mA from a 120V source would require a resistor of 1000 ohms (that's assuming 2V voltage drop on a LED). There would be 2V across each LED and 20V across the resistor. Use multiple strings in parallel for more light. Current is 20mA and power dissipated by the resistor is .4W.

Changing that resistor to 2000 ohms will change the current to 10mA.

Changing it to 500 ohms will change the current to 40mA. The resistor will need to handle .8W. You may need a bigger resistor.

[Mark Lloyd]

I had one do that once. It was a surprisingly loud noise for something that little. Half the plastic case disappeared.

BTW, one time I saw a working LED with no apparent series resistor. It was in a little flashlight powered by 2 button cells (CR2032 IIRC). This seeming impossibility seemed to make use of the series resistance of the battery, which couldn't supply too much current to this LED.

[Gary H]

This is true if the circuit is powered by a constant-current source, something unusual.

There may be some confusion here between actual current and current CAPACITY (of a conductor).

[Josh]

Yes, I agree. I could have worded that better -- I meant "lower than it would have been without the extra resistance in series", not "lower at this point in the circuit than at some other point".

The earlier post I was responding to claimed that the current would not change if the resistance of the wire changed. That's what I was trying to refute. You and I both agree that the current at any point in one circuit will be the same as any other point in that (series) circuit.

I suspect a lot of the next comments stem from that unclear wording; here goes...

Yes -- again, I meant "The V across the load will be lower than it would have been without the extra series resistance, just as the I is lower than it would have been without the extra series resistance"

Adding resistance to the wire changes the voltage across the load, and thus the current through the load (and the rest of the circuit). It doesn't change the voltage across the source (until you consider battery internal resistance, regulator quality, and other factors).

I agree. But it's not the same as it would be in a *different* circuit, one with more or less resistance in series.

I've never been accused of getting things right the first time :-)

Hopefully I've reworded my statements in a way that Kirchoff would agree with -- it's been 18 years since my Circuit Analysis classes; I have an ECE degree but do logic design/architecture for a living, so rarely get to break out the Ohms law these days; the Fourier transforms scared me away from analog design for good :-)

Josh

[clare]

It may not affect current, but it sure as heck affects power. The current stays the same, being a series circuit, but the voltage is shared between the wire and the device. - hense power to the device is reduced.

[clare]

I just rethought my answer and I was wrong. It not only drops the voltage but it DOES drop the current, because the total resistance of the circuit is higher. With a fixed supply voltage the current WILL drop. Combined with the voltage drop it is a double whammy on the power.

[clare]

It sure as heck does.

Yes, it is, but a higher resistance draws less current - so the current through the entire circuit is reduced.

You need to look at ohms law first. Then Kirchoff's applies.

And you are also wrong.

If a load has an effective resistance of 20 ohms on a 115 volt supply the circuit draws 5.75 amps.

If you add a .5 ohm resistance in the wiring to the device, the effective resistance is now 20.5 ohms, and the current is 5.609 amps. The voltage drop across the .5 ohm resistance is(.5X5.609=) 2.8 volts, so the device sees only (115-2.8=) 112.2 volts.

Power without resistance in the line is (5.75X115=) 661.25va (or watts if DC or straight resistance load) while it drops to (5.609X112.2=)629.33 va.

31.92 watts of power is wasted in the resistance of the wire.

That's true any way you slice it.

[clare]

Nope. And an LED is a "spacial case" too, because it is NOT a resistance. It is a fixed voltage drop.