Three bulbs.

  • Thread starter ssj5ankur
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Which of the bulbs shown in the figure will light up first when the switch is closed?
E: cell
K: key
A,B,C: bulbs

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ssj5ankur said:
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Not working.

The Bob (2004 ©)
 
c?

C should light up first?
 
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Highlight:

I'm gonna say they all turn on at the same time. Due to the potential difference between the plates, electrons will go through the circuit in a clockwise fashion. However, electrons are not emitted from the negative plate and shot to the positive one, the electrons come from the actual copper wiring itself. The electrons in the wire at all points are drawn towards the positive plate as soon as the switch is closed. Thus current is established everywhere at the exact same time (but not instantly after the switch is closed).
 
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Explanation

Whozum is right, but can you explain it.
 
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More of a physics problem than a brain teaser. I hid my answer for others to try.
 
whozum said:
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The electrons in the wire at all points are drawn towards the positive plate as soon as the switch is closed.
I don't think this is correct because it would imply that the signal travels faster than the speed of light.
 
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Bulb A will light first because it's closest to the switch, and at the instant the switch closes that's where the potential difference is.

Before the switch closes, there's no potential across the battery (if there were, current would flow through the battery). So it doesn't matter which bulb is closest to which battery terminal.
 
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jimmysnyder said:
I don't think this is correct because it would imply that the signal travels faster than the speed of light.

If you read the line right after it it explains it. Ofcourse its not instant but, as soon as the field is established.
 
whozum said:
Thus current is established everywhere at the exact same time (but not instantly after the switch is closed).
I don't think this is correct either because it would also imply that the signal travels faster than the speed of light.
 
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Gif is still not working.

The Bob (2004 ©)
 
The Bob said:
Gif is still not working.
Click on the link for the image, then click on the "Log Out" link.
 
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jimmysnyder said:
I don't think this is correct either because it would also imply that the signal travels faster than the speed of light.
no it doesnt. re-read please.
 
whozum said:
Thus current is established everywhere at the exact same time (but not instantly after the switch is closed).
What you say, ("current is established everywhere at the exact same time") cannot happen. Not as soon as the switch is closed, and not at any instant after the switch is closed either. The current cannot be established everywhere at the exact same time period because it would imply that a signal travels faster than the speed of light. Try to imagine the circuit to be made of water troughs, water wheels, and a water pump. When the pump is turned on, a pulse of water will propogate along the circuit and the water wheels will not all start to spin at the same time but rather one by one as the initial pulse hits them . Electrical fields aren't that much different, just faster.
 
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chroot said:
To start with, when you connect a battery to a capacitor, the electric field propagates through the wires at nearly the speed of light. It has nothing to do with the movement of any electrons; electrons, in fact, drift only very slowly through a circuit. The electric field which propels the electrons, however, propagates at near light speed.

The current is not like a gush of water coming down a mountain, moving from one place to another -- the current is like a conveyor belt that gets "turned on" everywhere at once in the circuit as the emf (voltage, potential difference) is established through it.
Once the field is established (very soon after the swithc is closed) electrons from all over the wire begin moving, not just at the very beginning of the wire. This is why there is current flowing everywhere once the field is established. This is why the bulbs all turn on at the same time.
 
whozum said:
This is why there is current flowing everywhere once the field is established. This is why the bulbs all turn on at the same time.
What happens in parallel circuits?
 
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What? This isnt a parallel circuit. What's your point?
 
Does the current start to flow at the same instant all along a parallel circuit?
 
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jimmysnyder said:
Does the current start to flow at the same instant all along a parallel circuit?
Sounds to me like you are saying that electrons in the wire that are submersed in an E field will not accelerate behind each other.
 
whozum said:
Sounds to me like you are saying that electrons in the wire that are submersed in an E field will not accelerate behind each other.
No, I was asking a question.
 
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jdavel said:
switch closes that's where the potential difference is.
Before the switch closes, there's no potential across the battery (if there were, current would flow through the battery). So it doesn't matter which bulb is closest to which battery terminal.
Of course there is always a potential accross the battery, where do yo think the potential accross the open switch came from? All closing the switch does is transfer that potential to across all three light bulbs.
 
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whozum said:
Once the field is established (very soon after the swithc is closed) electrons from all over the wire begin moving, not just at the very beginning of the wire. This is why there is current flowing everywhere once the field is established. This is why the bulbs all turn on at the same time.
I like this explaination but it doesn't address "very soon after the switch is closed" who gets the info first to turn on first and what is that info.

As we figure out what info needs to transfer we must also remember to use the Non-Simultaneous effect of SR in a common reference frame over a distance. So what info needs to be transferred, much more than just the switch is closed. Only after the switch is closed can the info on resistance of each light can go from the light to the batter and the available current or “battery impedance” can be communicated to the other members of the circuit. But with added info required from all parts of the circuit at the same time. However SR simultaneous issue says that cannot be true. So I figure the light closest to the observer doing the measurement comes on first. Other observers will get a different answer.

Maybe this does belong in another forum, but it sure teased my brain.
 

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