Lightbulb I/V curve question

In summary, the question is about hysteresis and its relation to the voltage-current graph of a light bulb. The lower curve represents the heating phase while the upper curve represents the cooling phase. Hysteresis is shown when there is a time delay between the effect and the cause, in this case, the change in voltage and current due to the heating and cooling of the filament. The 40Hz frequency may play a role in the hysteresis effect, but its exact impact is unclear.
  • #1
ninjaduck
10
0
The question

I've been given this question by a teacher and I'm clueless as to what b) and c) are. Could you please help? I would really appreciate it. The prize is a place to see a lecture on quantum simulation at our local University.
Thank you :)
 
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  • #2
b is asking you why it is that you see a graph as the voltage is being reduced that isn't exactly the same as the graph you saw when it was being increased. This effect is called hysteresis, which is why the question asks why you see hysteresis. Did it occur to you to look up hysteresis on the internet?

As for "c", what do you think of in relation to the term "property of the graph" ?
 
  • #3
I'm still lost on hysteresis. For c) I would suppose the gradient denotes the resistance, because I =V/R
 
  • #4
ninjaduck said:
I'm still lost on hysteresis. For c) I would suppose the gradient denotes the resistance, because I =V/R
You have looked up hysteresis and it doesn't make sense? What DOES it seem to mean to you?
 
  • #5
ninjaduck said:
The question

I've been given this question by a teacher and I'm clueless as to what b) and c) are. Could you please help? I would really appreciate it. The prize is a place to see a lecture on quantum simulation at our local University.
Thank you :)

ninjaduck said:
I'm still lost on hysteresis. For c) I would suppose the gradient denotes the resistance, because I =V/R

Welcome to the PF.

We require that you show some effort on your schoolwork problems here at the PF.

As phinds asks, what have you read about hysteresis? What are some of the physical systems that exhibit it? What kinds of effects can cause hysteresis? What do you think is going on in the light bulb that is showing some hysteresis in the V-I curve? What role do you think the 40Hz frequency plays in all of this?
 
  • #6
You assume I'm one of those people who wants the answer all laid out, well I'm not

I was taken through the question on reddit, and I've been told I have the correct answer, but I don't know why. I'm not looking for the answer but I want to know how I would find it. The deadline is in a week, I'm not in a hurry. My current answer to B would be:

The increase in the filament's temperature also increases the resistance of the filament and a greater change in voltage is required to cause a change in the current. The shallower gradient shows this. When the bulb is decreased, the slope was lower for the turning off than the initial part of turning on. This is because the filament has not had enough time to cool down, and the resistance is still high due to the heat. This causes the smaller gradient for the second line.

But I don't think this answers the question. Hysteresis is the time delay between the effect and the cause of a system, the effect being a value of Ia physical property such as magnetic induction, caused by an electromagnet. I honestly have no idea how this relates to the matter at hand.

EDIT: I don't see how the graph in the question shows a hysteresis curve either, maybe that's the root of the problem
 
  • #7
ninjaduck said:
You assume I'm one of those people who wants the answer all laid out, well I'm not
No, we assume that you are the kind of person who wants to know how to GET answers and learn things. That's the thrust of this forum, to promote learning, not to give answers. I understand that you HAVE the answer and so are not looking for that, but I'm trying to address the sentence that I just quoted.

My current answer to B would be:

The increase in the filament's temperature also increases the resistance of the filament and a greater change in voltage is required to cause a change in the current. The shallower gradient shows this. When the bulb is decreased, the slope was lower for the turning off than the initial part of turning on. This is because the filament has not had enough time to cool down, and the resistance is still high due to the heat. This causes the smaller gradient for the second line.
That's an excellent understanding of hysteresis and is correct for this problem.
But I don't think this answers the question.
But it does.

Hysteresis is the time delay between the effect and the cause of a system, the effect being a value of Ia physical property such as magnetic induction, caused by an electromagnet. I honestly have no idea how this relates to the matter at hand.
Odd since you just did a great job of stating how it DOES relate to the matter at hand.
EDIT: I don't see how the graph in the question shows a hysteresis curve either, maybe that's the root of the problem
Hm ... Do you not see two curves in the hand drawn graph? The lower one is as the voltage went up, heating the filament and the upper one is as the voltage is then lowered but with the filiment already hot, just as you described it above.
 
  • #8
But how does this show hysteresis?
 
  • #9
phinds said:
The lower one is as the voltage went up, heating the filament and the upper one is as the voltage is then lowered but with the filiment already hot, just as you described it above.

Wow I think I've been describing the upper line as the heating curve, and the straight line as the cooling curve.

EDIT: So as the bulb is cooling rapidly, the R in I = V/R is decreasing rapidly but in proportion to V? So there is a straight horizontal line until R reaches a minimum value and V is still decreasing so the sharp downfall happens?

How does 40hz change anything as well?
 
Last edited:

1. What is a lightbulb I/V curve?

A lightbulb I/V curve is a graphical representation of the relationship between the current (I) and voltage (V) of a lightbulb. It shows how the current changes as the voltage is varied, and how the lightbulb responds to different levels of voltage.

2. How is a lightbulb I/V curve measured?

A lightbulb I/V curve is typically measured by connecting the lightbulb to a power source and varying the voltage while recording the corresponding current. This is usually done using a multimeter or other electronic measuring device.

3. What does the shape of a lightbulb I/V curve represent?

The shape of a lightbulb I/V curve represents the electrical characteristics of the lightbulb, including its resistance and how it responds to changes in voltage. It can also indicate the point at which the lightbulb will begin to produce light.

4. How does the temperature of a lightbulb affect its I/V curve?

The temperature of a lightbulb can significantly affect its I/V curve. As the temperature increases, the resistance of the lightbulb decreases, resulting in a steeper I/V curve. This means that the lightbulb will draw more current at a given voltage and produce more light.

5. What can a lightbulb I/V curve tell us about the performance of a lightbulb?

A lightbulb I/V curve can tell us important information about the performance of a lightbulb, including its efficiency and how it responds to changes in voltage. It can also help determine the lifespan of the lightbulb and whether it is functioning properly.

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