Explain How the Lamp in Circuit A Lights When Switch S is closed

In summary, when switch S is closed, the Lamp L, lights. The current through the solenoid causes the magnetic field within the solenoid to cause the two iron reels to attract to one another, thus allowing the current to flow and light the lamp.
  • #1
Richie Smash
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15

Homework Statement


Hello Everyone, a pleasant good morning to all :)
I will attach an image that shows a diagram of a circuit.

The question based on this picture is as follows:
''When switch S is closed, the Lamp L, lights. Explain how this occurs.''

Homework Equations

The Attempt at a Solution


Here is my best attempt at describing what happens.

As the current passes through to the solenoid from circuit B, the magnetic field that is formed within the solenoid causes the two iron reels to attract to one another, thus allowing the current to flow and light the Lamp L.

I'm not 100 percent sure on this, as I see they have a sealed glass tube, of which I don't know the purpose, I do know that magnetic flux does pass through glass, I don't know if it plays some sort of role, as I know glass is also a poor conductor of electricity and heat.

I know that when a solenoid is heated it expands, so I was thinking maybe the heat produced would cause contact between the iron reels due to expansion, but since they are in the glass tube, I have ruled that out.

Would anyone care to share and offer some insight into this problem?
 

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  • #2
Hi Rich,

The thingies are not reels but reeds. The device is a reed switch (aka reed relay) and it works as you describe.
The glass is filled with some inert gas to prevent sparks eroding the contacts.

The current through the solenoid is not so big; it shouldn't heat the coil.
 
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  • #3
Hey thanks BvU, I understand quite well now good article, and I even know the answer to the next question.

It asks if the D.C battery was replaced with an A.C supply would the lamp light?

The answer is no because the AC will constantly change the polarities of the magnetic field of the solenoid causing erratic and unstable contact between the iron reeds.
 
  • #4
Richie Smash said:
Hey thanks BvU, I understand quite well now good article, and I even know the answer to the next question.

It asks if the D.C battery was replaced with an A.C supply would the lamp light?

The answer is no because the AC will constantly change the polarities of the magnetic field of the solenoid causing erratic and unstable contact between the iron reeds.
Depends on the frequency and amplitude of the ac source.
If the frequency is low the behavior is as you describe, assuming adequate amplitude.
But if the frequency is high enough what happens then? Does amplitude enter the picture?
 
  • #5
rude man said:
Depends on the frequency and amplitude of the ac source.
If the frequency is low the behavior is as you describe, assuming adequate amplitude.
But if the frequency is high enough what happens then? Does amplitude enter the picture?

Ok I think that if the frequency is high enough, the amplitude should be low, as I'm picturing... if the polarities are switching fast enough it become sort of constant?
 
  • #6
Frequency and amplitude of the voltage source are both independent variables; you can set either of them the way you like.
If the frequency is high enough then yes the reeds are either open or closed, depending on amplitude of the solenoid field. What is the mechanism to make that happen?
 
  • #7
Hmm, that's a tough question, the mechanism required for that has to be Force produced by a current carrying conductor in a magnetic field?
 
  • #8
Richie Smash said:
Hmm, that's a tough question, the mechanism required for that has to be Force produced by a current carrying conductor in a magnetic field?
That's right, so can you apply F=ma to the reeds?
 
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  • #9
I've been thinking about this hard, the solenoid is producing it's magnetic field, so what is the conductor? the conductor will be the iron reeds perhaps? because once they touch they will create their own magnetic field, however it's the solenoid magnetizing them to make them touch in the place... so the solenoid is interacting with a separate field?

Either way yes F=ma can be applied.
 
  • #10
The conductor current for the lamp is the reed relay. The conductor current for the solenoid is always predicated solely on the ac voltage.
The mag field created by the reed current is ignorable. Even if it's large the direction of that B field is circular, not axial. The reeds react solely to an axial external B field.
So no, the solenoid B field is not influenced by the lamp current as far as interaction with the reeds is concerned.

Let me ask you this: suppose the voltage is a dc pulse of short duration. What would you expect the reeds to do? As you vary the pulse duration? (Ignore the fact that the solenoid current takes time to build up and decay; assume it follows the pulse voltage directly.)
Another question for you: back to the dc voltage case, if you reverse the dc voltage what do the reeds do?
 
  • #11
Well If the Dc voltage is a pulse and ignoring those factors,... I would say that the reeds would make contact for each pulse, and remain in contact for the duration of each pulse also.

My best guess if the dc voltage is reversed, the reeds would not make contact?
 
  • #12
Richie Smash said:
Well If the Dc voltage is a pulse and ignoring those factors,... I would say that the reeds would make contact for each pulse, and remain in contact for the duration of each pulse also.
Even if the pulse is 1 nanosecond?
My best guess if the dc voltage is reversed, the reeds would not make contact?
Why do you say that? What is the basic reason the reeds attract each other? Look carefully at your illustration, note it's the opposite ends of the reeds that are aligned with each other. That's a big hint! What would happen if the reeds were aligned 100% length-to-length?
 
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  • #13
rude man said:
Even if the pulse is 1 nanosecond?Why do you say that? What is the basic reason the reeds attract each other? Look carefully at your illustration, note it's the opposite ends of the reeds that are aligned with each other. That's a big hint! What would happen if the reeds were aligned 100% length-to-length?

Hm... I'm not too sure about the pulse, but I see that I was wrong about the reversing the current, it wouldn't repel the reeds, it would just reverse the poles in the opposite end, but they'd still attract, and if they were length to length, they would repel as there would be like polarities more of them.

Well In my mind I'm just thinking that it requires current for the magnetic field to be generated, so even if it was a pulse it would generate a field for a nanosecond, and perhaps make contact for a nansecond?

Or is the answer somewhere along the lines of, the contact of the reeds is dependent on the duration of the pulse, because even if the magnetic field was produced in the solenoid for one nanosecond, the reeds would need more time to contact, and the instance of the field is faster than the time it would take for that.
 
  • #14
Richie Smash said:
Hm... I'm not too sure about the pulse, but I see that I was wrong about the reversing the current, it wouldn't repel the reeds, it would just reverse the poles in the opposite end, but they'd still attract, and if they were length to length, they would repel as there would be like polarities more of them.
That is 100% correct.
Well In my mind I'm just thinking that it requires current for the magnetic field to be generated, so even if it was a pulse it would generate a field for a nanosecond, and perhaps make contact for a nansecond?
Do the reeds have mass? Can they be instantaneously moved into contact? Remember the 'rest' state for them is OFFf.
Or is the answer somewhere along the lines of, the contact of the reeds is dependent on the duration of the pulse, because even if the magnetic field was produced in the solenoid for one nanosecond, the reeds would need more time to contact, and the instance of the field is faster than the time it would take for that.
That is correct!
 
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1. How does electricity flow through the circuit to light the lamp?

When the switch S is closed, it completes the circuit and allows electricity to flow from the power source through the wires to the lamp. This creates a closed loop for the electricity to travel, activating the lamp and causing it to light up.

2. Why does the lamp turn off when switch S is opened?

When switch S is opened, it breaks the circuit and interrupts the flow of electricity. Without a complete path for the electricity to travel, the lamp is no longer receiving power and therefore turns off.

3. What role does the lamp play in this circuit?

The lamp acts as a load in the circuit, which means it uses up the electrical energy that is flowing through the circuit. This causes the lamp to light up and produce light.

4. Is there any other way to light the lamp without closing switch S?

No, in this specific circuit, the only way to light the lamp is by closing switch S. This allows for the completion of the circuit and the flow of electricity to the lamp.

5. Can the lamp be replaced with a different type of load in this circuit?

Yes, the lamp can be replaced with other types of loads such as a motor or a buzzer. As long as the load allows for the flow of electricity and has a low enough resistance, it can be used in this circuit to create a closed loop and light up when switch S is closed.

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