Ohm's Law confusion -- How to handle an incandescent bulb?

[hendrix7]

Homework Statement

A filament lamp is connected to a 15 V power supply. At this potential difference, the lamp has a resistance of 60 Ohms. Calculate the current passing through the lamp.

Relevant Equations

V = IR

I know that Ohm's Law gives me the answer of 0.25 A but what I don't understand is how is it ok to use Ohm's Law when I know that the lamp doesn't obey the law? I know that as the current is increased through the lamp, the resistance increases due to temperature rise which, I am told, makes it non-Ohmic. So how is it still valid to use the law for the question?

 

[berkeman]

I think since they give you the lamp's resistance at that voltage, they are saying that it's in thermal equilibrium at that point (hot, so higher resistance than when measured cold). If they had said that the lamp had a resistance of 60 Ohms before it was connected in the circuit, they you are right that you would not have enough information to solve the problem.

 

[hendrix7]

I think since they give you the lamp's resistance at that voltage, they are saying that it's in thermal equilibrium at that point (hot, so higher resistance than when measured cold). If they had said that the lamp had a resistance of 60 Ohms before it was connected in the circuit, they you are right that you would not have enough information to solve the problem.

Thanks for your reply. In an earlier part of the question, it said that the voltage was varied to investigate the effect on the current and gave some readings: 3 V gave a current reading of 0.1 A and 5 V gave a reading of 0.14 A. Would I still be able to use Ohm's Law for this data to find the resistance at these points, i.e. 30 Ohms and 35.7 Ohms respectively?

 

[DaveE]

Homework Statement: A filament lamp is connected to a 15 V power supply. At this potential difference, the lamp has a resistance of 60 Ohms. Calculate the current passing through the lamp.
Relevant Equations: V = IR

I know that Ohm's Law gives me the answer of 0.25 A but what I don't understand is how is it ok to use Ohm's Law when I know that the lamp doesn't obey the law? I know that as the current is increased through the lamp, the resistance increases due to temperature rise which, I am told, makes it non-Ohmic. So how is it still valid to use the law for the question?

Yep. You're on it. I guess you must assume that the incandescent bulb is acting as a simple resistor as @berkeman said, since that's the implication in this poorly constructed question.

I think there are a lot of people that just don't know about the V-I characteristics of incandescent bulbs so they think it's a simple and relatable component to put in their problems. It's not. Also, electrician types tend to only consider widely spaced discrete voltages and steady state conditions. They don't need to know how complex it can be.

 

[DaveE]

Would I still be able to use Ohm's Law for this data to find the resistance at these points, i.e. 30 Ohms and 35.7 Ohms respectively?

Um... no, not really. You can use the local resistance $r = \frac {\Delta v}{\Delta i}$ for calculations near the (V,I) DC operating point. This is distinct from Ohm's Law which is global $R = \frac{V}{I}$. Because the dominant factor in the resistance change is thermal, for fast events you can use the global resistance $R$, but this opens a can of worms about how fast is fast, since $R$ is changed as it heats/cools. It is a resistor, but it's value changes with temperature (thus operating point).

(Pretend two of those are straight lines, also that (V, I) is really (I, V), LOL)

 

[DaveE]

Sorry that explanation kind of sucks, but I have to go.

 

[hendrix7]

Sorry that explanation kind of sucks, but I have to go.

Thanks, Dave, for replying.
I now think I know why I was confused. I've read a little more on this and this is what I found out:
Ohm's Law is only V = IR if R is constant, i.e. is not affected by a change in temperature or any other external factors. Ohm's Law only states that, in this case, V is proportional to I. However, V = IR can be used for ALL cases, whether V is proportional to I or not. But, of course, if we are using V = IR for cases where resistance is not constant, then this does not follow Ohm's Law. I think I've got this right now. Can you confirm this for me?

 

[kuruman]

If I wrote the expression $~A=b~h~$ and asked whether $A$ proportional to $h$ there might be people who would see a linear relationship between the two quantities and answer "yes". If I then said that $A$ stands for "area", $b$ for "base" and $h$ for "height" of a rectangle, it is clear that the area is proportional to the height only if the base is kept constant: if I multiply the base by a factor $f$ and keep the base constant, the area is multiplied by the same factor $f.$ Nevertheless, it is true that if you know the area $A$ and the height $h$, you can always find the base $b$ by using $b=\dfrac{A}{h}.$

With Ohm's law $V=IR~$ only the symbols differ but the ideas are the same. You can always find $R$ using $R=\dfrac{V}{I}$ but $V$ is not proportional to $I$ unless $R$ is constant.