# Christmas Lights Question

#### dporay

Joined Oct 10, 2015
18
A string of standard ( non-LED ) Christmas Lights (100 lights, Two 50 light strings connected in parallel ) has me confused. The total resistance of the string is 55 ohms. When using the formula I=E/R the calculated current is 2.2amps using 120 volts for E. Measuring the current with an AC ammeter provides a reading of 320 ma. What am I missing between the measured and the calculated values?

#### MrChips

Joined Oct 2, 2009
28,137
You cannot used the measured resistance of cold lamps.
When lamps are cold the resistance is low.
The resistance will increase when the lamps get hot.

#### MisterBill2

Joined Jan 23, 2018
14,260
Mr Chips called it!! Incandescent lights have a much lower cold resistance and that is what you measured. Use the hot current value to calculate the resistance.

#### dporay

Joined Oct 10, 2015
18
You cannot used the measured resistance of cold lamps.
When lamps are cold the resistance is low.
The resistance will increase when the lamps get hot.

#### dporay

Joined Oct 10, 2015
18
Mr Chips called it!! Incandescent lights have a much lower cold resistance and that is what you measured. Use the hot current value to calculate the resistance.

#### cmartinez

Joined Jan 17, 2007
7,914
The answer that you're looking for lies in the concept of resistivity. Just as MrChips has stated, the resistance of any material is heavily dependent on its temperature.

Have a look at this chart, it should give you a pretty good idea of what it all means.

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#### WBahn

Joined Mar 31, 2012
28,182
A string of standard ( non-LED ) Christmas Lights (100 lights, Two 50 light strings connected in parallel ) has me confused. The total resistance of the string is 55 ohms. When using the formula I=E/R the calculated current is 2.2amps using 120 volts for E. Measuring the current with an AC ammeter provides a reading of 320 ma. What am I missing between the measured and the calculated values?
Measure the resistance again as quickly as you can after you unplug the lights. They cool down pretty quick so you probably won't see the full hot resistance, but you should see a higher resistance the drops over a few seconds (maybe quicker).

#### dporay

Joined Oct 10, 2015
18
The answer that you're looking for lies in the concept of resistivity. Just as MrChips has stated, the resistance of any material is heavily dependent on its temperature.

Have a look at this chart, it should give you a pretty good idea of what it all means.

#### dporay

Joined Oct 10, 2015
18
Measure the resistance again as quickly as you can after you unplug the lights. They cool down pretty quick so you probably won't see the full hot resistance, but you should see a higher resistance the drops over a few seconds (maybe quicker).

#### MrSalts

Joined Apr 2, 2020
2,767
Just measure the voltage drop across one lamp when it is on (you could pierce the insulation with a needle and clip to the needle) - you've already measured the series current draw, use ohms law to get the hot resistance - if you're really set on measuring it.

alternatively, you can just calculate the resistance based on the series/parallel connection of the circuit, the 120vAC voltage drop across the strand and your measured current.

#### dporay

Joined Oct 10, 2015
18
Just measure the voltage drop across one lamp when it is on (you could pierce the insulation with a needle and clip to the needle) - you've already measured the series current draw, use ohms law to get the hot resistance - if you're really set on measuring it.

alternatively, you can just calculate the resistance based on the series/parallel connection of the circuit, the 120vAC voltage drop across the strand and your measured current.
Great Idea.

#### MrChips

Joined Oct 2, 2009
28,137
Of course, you can calculate the resistance of the filament from the voltage and current measured.
R = V / I = 120V / 0.32A = 375Ω

Your original intention was to compare theory and practice.
You cannot practically measure the hot resistance directly with an ohmmeter.

What is the temperature of the filament?
The filament glows because the temperature is high. Once the filament stops glowing the temperature is much lower.
The temperature of the glowing filament is above 2000°C.

#### dporay

Joined Oct 10, 2015
18
Of course, you can calculate the resistance of the filament from the voltage and current measured.
R = V / I = 120V / 0.32A = 375Ω

Your original intention was to compare theory and practice.
You cannot practically measure the hot resistance directly with an ohmmeter.

What is the temperature of the filament?
The filament glows because the temperature is high. Once the filament stops glowing the temperature is much lower.
The temperature of the glowing filament is above 2000°C.
I missed the point about the resistance of the filament being different when cold versus being hot.

#### MrChips

Joined Oct 2, 2009
28,137
Let's do a ballpark estimate.
The chart provided by @cmartinez in post #6 lists the resistivity temperature coefficient of tungsten as 0.0045/°C.

If we take your measurement of 55Ω at 25°C, we can extrapolate the temperature at 2000°C to be approximately
55 x 0.0045 x 1975 = 488Ω

That is higher than the calculated 375Ω but you get the idea.

What we have not taken into account is that Xmas lamps have bridge resistance in parallel with the filament. When a filament blows, the whole string still works. So it gets more complex trying to figure out the resistance of the total string of lights.

#### WBahn

Joined Mar 31, 2012
28,182
If the TS is going to be satisfied with a calculated result, why did he measure the current in the first place? He had a voltage and a measured resistance and a calculated current using Ohm's Law. But he measured the current and found that it disagreed with the calculated result. A proposed explanation has been offered, but why should he just accept that the resistance is actually given by Ohm's Law now when the current wasn't given by Ohm's Law before? Seems like a reasonable situation in which wanting to take a (at least seemingly) more direct measurement is completely reasonable.

IF the TS has the interest and the pieces parts needed, a good compromise would be to measure the current and voltage across the string as the current is increased. by putting resistors in series with the string. If they have two meters, measure the current with one and the voltage across the string with the other. Then calculate the resistance at a bunch of points, showing that the resistance at very small currents is close to the cold resistance measurement and that the calculated resistance decreases as the current increases.

#### MrChips

Joined Oct 2, 2009
28,137
I think there is a huge learning opportunity here.

1) On the onset, the TS was puzzled as to why his calculation did not agree with measurement. It is perfectly reasonable to be inquisitive at this stage.

2) TS was given the immediate answer, i.e. the cold resistance vs hot resistance. Most people would be happy with that and move on with life.

3) But it begs the question, what is the temperature of the glowing filament? This introduces knowledge about black body radiation and that is perfectly valid.

4) Now that we know the resistivity temperature coefficient of tungsten, can we estimate the resistance of the glowing light bulb. We came within 25% in agreement but why the discrepancy.

5) Now we need to learn how the filaments are actually wired taking in consideration that there is resistance wire loop in parallel with each filament. For the truly inquisitive who likes to get to the bottom of a mystery, one can disassemble an Xmas lamp and measure the resistance of the bridge. Then they can do a further analysis into the effect of the bridge on the total resistance at different temperatures.

#### MisterBill2

Joined Jan 23, 2018
14,260
Let's do a ballpark estimate.
The chart provided by @cmartinez in post #6 lists the resistivity temperature coefficient of tungsten as 0.0045/°C.

If we take your measurement of 55Ω at 25°C, we can extrapolate the temperature at 2000°C to be approximately
55 x 0.0045 x 1975 = 488Ω

That is higher than the calculated 375Ω but you get the idea.

What we have not taken into account is that Xmas lamps have bridge resistance in parallel with the filament. When a filament blows, the whole string still works. So it gets more complex trying to figure out the resistance of the total string of lights.
Generally that "bridge resistance" is an arranged shunt with a low insulation breakdown voltage. When the filament opens (fails), then the whole mains voltage is across the shunt and the insulation breaks down and the string usually conducts. That is how it is intended to work, and usually it does work that way. The concept is that the shunt heats and forms a permanent repair, and usually that happens. But as each light fails the voltage, and thus the current, in the string rises, increasing the stress on the remaining string. This is why it is common to see a string of lights with the black cloud on the glass of every bulb.

#### MrChips

Joined Oct 2, 2009
28,137
Generally that "bridge resistance" is an arranged shunt with a low insulation breakdown voltage. When the filament opens (fails), then the whole mains voltage is across the shunt and the insulation breaks down and the string usually conducts. That is how it is intended to work, and usually it does work that way. The concept is that the shunt heats and forms a permanent repair, and usually that happens. But as each light fails the voltage, and thus the current, in the string rises, increasing the stress on the remaining string. This is why it is common to see a string of lights with the black cloud on the glass of every bulb.
Thanks for that.
I was looking for the word "shunt" but it slipped my mind.

#### dporay

Joined Oct 10, 2015
18
Let's do a ballpark estimate.
The chart provided by @cmartinez in post #6 lists the resistivity temperature coefficient of tungsten as 0.0045/°C.

If we take your measurement of 55Ω at 25°C, we can extrapolate the temperature at 2000°C to be approximately
55 x 0.0045 x 1975 = 488Ω

That is higher than the calculated 375Ω but you get the idea.

What we have not taken into account is that Xmas lamps have bridge resistance in parallel with the filament. When a filament blows, the whole string still works. So it gets more complex trying to figure out the resistance of the total string of lights.
That is one point I overlooked. The bridge resistance of the lamp when the filament burns out.