Basic Circuit Theory

[danago]

Hey all. Ive never really done anything in the way of circuit theory up until now, where i have just started a unit on introduction to electrical/electronic engineering.

Lets say i have a 60W light globe. Does this mean that the maximum power it can dissipate is 60W, after which it will blow? Will it still be able to output less energy than stated?

For example, lets say i have a very basic circuit with a 60W globe and a 10V source voltage. According to KVL, the voltage across the globe will also be 10V, and then according to the relation P=VI, the current will be 6A? What if i now take a 0.01V source. Then by the same reasoning, a current 6000A would flow?

That definitely doesnt look right to me; it seems so very unrealisting for a 0.01V source to produce a current of 6000A, which would then tell me that the globe isnt actually outputting 60W anymore. If so, how can i find what power it is actually dissipating?

Am i on the right track, or have i completely missed something?

Thanks in advance,
Dan.

[Redbelly98]

The power rating of a light bulb is for a specific voltage. Change to a different voltage, you'll get a different power.

For example, if the 60W rating is for a voltage of 120V across the lamp, then 10V would result in a lower power and less than 6A.

[danago]

Oh ok, makes sense. What if i apply a much higher voltage than it was designed for? Is this what will cause it to blow?

[berkeman]

Oh ok, makes sense. What if i apply a much higher voltage than it was designed for? Is this what will cause it to blow?

Yes. I = V/R, so applying a higher RMS voltage will cause a higher current to flow through the resistance of the filimant, which will result in a higher power dissipation P = VI, which heats the filament beyond its operating temperature, and causes it to melt and fail open.

[danago]

Alright thanks for the replies :smile:

I was just doing some of the questions in my textbook and came across this one:

"A lightbulb rated at 60W will dissipate 60W as heat and light when connected with a 100V ideal voltage source. A lightbulb rated at 100W will dissipate 100W as heat and light when connected with the same voltage source.

If the bulbs are connected in series with the same 100V source, what is the power dissipated by each globe?"

To produce 60W and 100W respectively with a 100V source, the currents through each individual circuit must be 0.6A and 1A respectively. Then, using ohm's law, i can calculate the resistance caused by each globe (166.67 Ohms and 100 Ohms respectively). When wired up in series, the effective resistance of the circuit will be 266.67 Ohms, and so the current flowing will be 0.375A. Using P=I^2 R, i can then calculate the power dissipated by each element as 23.4W and 14.06W.

Have i reasoned through that correctly?

[Defennder]

Yeah it should be ok, since I don't think the resistance of the bulbs are non-linear.

[danago]

Yeah it should be ok, since I don't think the resistance of the bulbs are non-linear.

Alright cool, thanks :smile:

[Averagesupernova]

The resistance of incandescent bulbs ARE nonlinear. Measure the resistance of a cold filament and you will see this is true.

[Defennder]

They can be considered ohmic within a given temperature range. Since the question doesn't give you any other information, how else would you do it?

[Redbelly98]

While filament resistance changes drastically with temperature*, that fact is typically ignored in solving this type of introductory circuit homework problems.

* R is about 15 times higher at operating temperature (2800-3000 K) than at room temperature (290-300 K)