LR Circuit: Answers to Common Questions

In summary, the bulb will go out eventually. The current through the coil remains the same at the very instant the source of EMF is removed.
  • #1
Anirudhhades
3
1
Homework Statement
Guys in this question I think that once the switch is opened the current doesn’t immediately drop to zero but reduces at some exponential rate but at the very instant the source of EMF is removed the currrent in the circuit must still be equal to the current when switch was closed and hence the brightness of the bulb remains same

The options are:
A) Remains same as before
b) The bulb goes out
C)Glows very brightly

According to answer key the answer is C)
But I think the answer should be A)


Please help me out!
Relevant Equations
I=I (e^-Rt/L)
image.jpg
 
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  • #2
Anirudhhades said:
Homework Statement:: Guys in this question I think that once the switch is opened the current doesn’t immediately drop to zero but reduces at some exponential rate but at the very instant the source of EMF is removed the currrent in the circuit must still be equal to the current when switch was closed and hence the brightness of the bulb remains same

The options are:
A) Remains same as before
b) The bulb goes out
C)Glows very brightly

According to answer key the answer is C)
But I think the answer should be A)Please help me out!
Relevant Equations:: I=I (e^-Rt/L)

View attachment 269810
The branch with the bulb and the switch are conected parallel with the coil. What do you think where is the current higher when the switch is closed? through the bulb or through the coil?
 
  • #3
Aargh! Another poorly written homework problem. Is there any prior information about what the resistance of the solenoid is (or the current through it when the switch is opened)? Maybe from another question or lecture. The question relates to how much solenoid current compared to the lamp current.

Theoretically, inductors (solenoids) have zero resistance. In the real world they don't, and this circuit makes no sense unless that current is limited somehow.
 
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  • #4
A badly written question, but can you answer these hints?:
1) While switched on, does the the coil have a current through it?
2) While switched on, does the the coil have energy stored? If so, what sort?
3) While the switch is (virtually instantly) being opened, what can you say about the time the current (from the battery) takes to drop to zero?
4) While the switch is (virtually instantly) being opened, what can you say about the rate of change of current through the coil? And what can you then say about the induced emf across the coil?
 
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  • #5
Even the answers are written poorly.

I think the correct answer is B) The bulb goes out EVENTUALLY. It's the only thing I'm sure of since the only energy source is removed and I don't believe in perpetual motion machines. But along those lines, your answer of A) It remains the same as before, can be true initially, but it can't stay that way forever.

I'm sorry you have to deal with this, but there is some good learning possible, follow up with the questions from @Steve4Physics
 
  • #6
@DaveE based Off of the options given I think they are asking initially when the source is cut
 
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Likes DaveE
  • #7
Steve4Physics said:
A badly written question, but can you answer these hints?:
1) While switched on, does the the coil have a current through it?
2) While switched on, does the the coil have energy stored? If so, what sort?
3) While the switch is (virtually instantly) being opened, what can you say about the time the current (from the battery) takes to drop to zero?
4) While the switch is (virtually instantly) being opened, what can you say about the rate of change of current through the coil? And what can you then say about the induced emf across the coil?
Thanks a lot
 
  • #8
Anirudhhades said:
Homework Statement:: Guys in this question I think that once the switch is opened the current doesn’t immediately drop to zero but reduces at some exponential rate but at the very instant the source of EMF is removed the currrent in the circuit must still be equal to the current when switch was closed and hence the brightness of the bulb remains same
It's not clear what you mean when you say "the currrent in the circuit." There are three branches in the circuit, and each has a current running through it. Which current stays the same when the switch is opened?
 
  • #9
Anirudhhades said:
Thanks a lot
Hi Anirudhhades. If you attempt to answer my 4 hint-questions (post #4) I'll explain why these lead to choice C.
 

Related to LR Circuit: Answers to Common Questions

1. What is an LR circuit?

An LR circuit is an electrical circuit that contains a resistor (R) and an inductor (L). It is used to study the behavior of electrical current and voltage in a system where inductance is present.

2. What is the purpose of an LR circuit?

The purpose of an LR circuit is to study the relationship between current, voltage, and inductance in a system. It can also be used to filter out certain frequencies in an electrical signal.

3. How does an LR circuit differ from an RC circuit?

An LR circuit differs from an RC circuit in that it contains an inductor instead of a capacitor. This means that the current and voltage in an LR circuit will have a phase difference, while in an RC circuit they will be in phase.

4. How does the inductor affect the behavior of an LR circuit?

The inductor in an LR circuit stores energy in the form of a magnetic field. This causes the current to lag behind the voltage, creating a phase difference. The inductor also resists changes in current, leading to a smoother flow of electricity.

5. What are some real-world applications of LR circuits?

LR circuits have many practical applications, such as in power supplies, motors, and generators. They are also used in electronic filters, radio frequency circuits, and inductive heating systems. Additionally, LR circuits are commonly used in electronic devices to protect them from power surges and spikes.

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