Incandescent Bulb - suitable approximation for voltage-current curve

Click For Summary
When a 2 kW electric kettle and a 60 W incandescent bulb are plugged in series, the kettle's nearly constant resistance aligns with Ohm's law, while the bulb's resistance varies significantly with temperature. The bulb's cold resistance is about ten times lower than its hot resistance, complicating calculations. Despite this, for educational purposes, the exercise remains valid as the kettle's resistance is negligible, and the bulb's voltage remains relatively stable. The resistance of a filament lamp is proportional to its temperature, with significant differences between cold and hot states. Understanding these principles is essential for accurate electrical calculations and teaching.
RodionGork
Messages
6
Reaction score
2
Thread moved from the technical forums to the schoolwork forums
Hi Friends!

Regard the 2 kW electric kettle and 60 W incandescent bulb - what happens if we plug them in series? It is the problem I often ask my students somewhere at the end of the term to check if they remember basic "laws" about electricity. Recently I started collecting some exercises on a web-site, and added this one too - here is the full text - but it is pretty same, the main difference is that now there is precise question to find the power on the kettle in this situation.

This annoys me because calculations are made in assumption both devices behave in concordance with Ohm's law. It is quite close for the kettle (their heating elements made of alloys have almost constant resistance with temperature) - but it is definitely not so with the bulb (whoever tried measuring its resistance in the cold state knows it is about 10 times lower).

It still seems less or more OK for the school exercise, since the kettle will give negligible additional resistance and voltage on a bulb doesn't change much - hence its resistance too. But still I wonder if there exist some popular resistance-temperature approximation for pure metals used in school courses. When I tried to look up the topic in the wiki, I quickly faded out among formulas which seemed quite alien to me - but I understand they aim to be as precise as possible over the whole range of temperatures from zero to thousands.
 
Physics news on Phys.org
The resistance of a filament lamp is proportional to the absolute temperature of the filament. Cold is about 300 K. Hot is about 3000 K. Hence the factor of ten in resistance, and the importance of colour temperature.
The simplest approximation is with two straight lines. R is initially constant until the filament generates IR heat. Once the filament starts to glow, it behaves more like a current regulator. The transmission spectrum of the glass has much to do with the transition from one line to the other.
 
From
https://hypertextbook.com/facts/2004/DeannaStewart.shtml


Screen Shot 2024-08-08 at 2.14.54 PM.png
 
  • Like
Likes Lord Jestocost and Tazerfish
If have close pipe system with water inside pressurized at P1= 200 000Pa absolute, density 1000kg/m3, wider pipe diameter=2cm, contraction pipe diameter=1.49cm, that is contraction area ratio A1/A2=1.8 a) If water is stationary(pump OFF) and if I drill a hole anywhere at pipe, water will leak out, because pressure(200kPa) inside is higher than atmospheric pressure (101 325Pa). b)If I turn on pump and water start flowing with with v1=10m/s in A1 wider section, from Bernoulli equation I...

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 53 ·
2
Replies
53
Views
7K
Replies
2
Views
1K
Replies
9
Views
2K
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 17 ·
Replies
17
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
16
Views
2K