# LR circuit - Which bulb gets dim quickly?

• Jahnavi
In summary, when the switch is off, there is a gradual decrease in current through the inductor which delays the bulb that has a resistor in series from dying out. When the switch is opened, current still flows through the inductor and both bulbs get brighter.f

## The Attempt at a Solution

When the switch is off , the current in the inductor doesn't drop to zero instantaneously . The current in the loop consisting of inductor , resistor and two bulbs will decay gradually . Now since there is a complete circuit involving the two bulbs, a gradually decreasing current will flow and both of them will die out simultaneously with some delay . I think answer should be option b) .

Answer given is option d) with the reason that branch having inductor has some inductance which delays B2 to die out promptly .

What is my mistake ?

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You are right.

The main loop can no longer carry any current.

ehild and Jahnavi
If there is current in one part of the loop, will there be current in the complete loop?
If not, where does the current go?

I don't understand your question . There is only one loop after the switch is opened and there is only one current .

Do you disagree with option b) ?

I don't understand your question . There is only one loop after the switch is opened and there is only one current .

Do you disagree with option b) ?
You read my response while I was still editing it.

Option d) is given in at least two books . Also this is a question from past year papers .

I would request you to have a relook at the question again .

Option d) is given in at least two books . Also this is a question from past year papers .

I would request you to have a relook at the question again .
I have no doubt that the answer is B. Answer D, along with the logic provided, would relate to a what happens when the switch is closed. In that case, B1 would light up promptly and B2 would light up with a delay.

The only way something like that could happen when the switch is open would be if there was arcing across the contacts in the switch. But since there is a resistive path across that inductor (B2, B1, R), that won't happen.

Merlin3189 and Jahnavi

## Homework Statement

View attachment 226718

## The Attempt at a Solution

When the switch is off , the current in the inductor doesn't drop to zero instantaneously . The current in the loop consisting of inductor , resistor and two bulbs will decay gradually . Now since there is a complete circuit involving the two bulbs, a gradually decreasing current will flow and both of them will die out simultaneously with some delay . I think answer should be option b) .

Answer given is option d) with the reason that branch having inductor has some inductance which delays B2 to die out promptly .

What is my mistake ?
I think you are right. Before the switch is off, current flows through both bulbs, less through B1 than through B2, as B1 has a resistor connected in series with it, and B2 is connected in series with the inductor which means zero resistance for DC and both branches are connected to the same battery.
When the switch gets opened, current still flows through the inductor, and it does not dies out instantaneously according to Lenz's Law. So there is some current flowing in the loop of two bulbs and R resistor and inductor L, decreasing exponentially with time. This current is the same for both bulbs, but different from that which was flowing when the switch was on. So one bulb gets brighter, (which?) the other gets dimmer at the first instant and both die out simultaneously with delay.

Jahnavi
Thanks a lot ehild

Thanks @.Scott

This current is the same for both bulbs, but different from that which was flowing when the switch was on. So one bulb gets brighter, (which?) the other gets dimmer at the first instant and both die out simultaneously with delay.
When you say brighter, you mean brighter than it was before. Once the switch is open, the current through L will begin to decay starting at the current value that it had settled into with the switch closed. So, "at the first instant", that current will not have changed yet.
Assuming ideal bulbs (respond instantly to current changes), one bulb will become brighter and the other will be at the same brightness. Then they will both decay in brightness together.

Jahnavi
one bulb will become brighter and the other will be at the same brightness.

Why ?

Why ?
That's the question that @ehild is posing to you.
If you read his post, you should be able to pick this up.
If you plotted the brightness of each bulb, both would reach a fairly level value after the switch had been closed for a while.
Then when the switch opens, one would instantly rise, then both would decay towards zero.

Then when the switch opens, one would instantly rise,

B1 would get brighter but B2 would remain same for a moment .Then both get dimmer .Right ?

At the instant when the switch gets open, you can not say anything. No current can stop at once, as all the wires and the resistor and the filaments of the light bulbs have some inductance. You should consider the changing EM field. A bit later, the changes slow down, and you can calculate with a single current in the loop and voltages across the bulbs and resistors and inductor, and apply Kirchhoff's Laws. At this stage, both bulbs are equally bright, and the current changes according to I=I0e-t/τ. You can determine τ but I do not know what I0 should be.

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@Jahnavi It might interest you that the current in Bulb 1 actually changes direction when the switch is opened, but the effect is nearly instantaneous and the current in the reverse direction will be equally effective in keeping the bulb lit for a short time. (Edit: Also, see below).## \\ ## The inductor maintains the current in Bulb 2 in the same direction and develops a voltage to try to maintain the same current by Lenz's law. ## \\ ## Edit: The current in Bulb 2, ## I_2 ##, will be continuous, because of the inductor, even though the switch is opened at ## t=0 ##. It will decay exponentially with the differential equation with initial conditions for the current ## I_2 ## starting at the instant the switch is opened, (at ## t=0 ##). The current in Bulb 1, ## I_1 ##, is discontinuous and actually reverses. The current in Bulb 1 at ## t=0^+ ## will be that of Bulb 2 at and before ## t=0 ##, (in the reverse direction), and if the bulbs are identical, it will result in Bulb 1 getting momentarily brighter, because ## I_1<I_2 ## for ## t<0 ## because of the resistor, and ## |I_1|=|I_2| ## for ## t=0^+ ##. ## \\ ## And you might also ask, "what happens if we also put an inductor in the loop with Bulb 1?". And I think the answer to that is that that is the point where the model of discrete circuit components with the current being the same everywhere throughout the circuit starts to break down.

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Jahnavi
... At this stage, both bulbs are equally bright, ..
Since you say the current in B1 is less than in B2 while the switch is closed, then the filament in B1 will be cooler and have lower resistance.
From the moment the switch opens both bulbs have the same current and the power supplied to B1 is always less than the power to B2. So B1 can never catch up with B2.