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OpenStudy (anonymous):

Two light bulbs, B1 and B2, are connected to a battery having appreciable internal resistance as shown in the figure. What happens to the brightness of bulb B1 when we close the switch S? a. The brightness of B1 does not change. b. The brightness of B1 increases permanently. c. The brightness of B1 decreases permanently. d. The brightness of B1 increases temporarily but gradually decreases back to its original brightness. e. The brightness of B1 decreases temporarily but gradually increases back to its original brightness. is it A?

OpenStudy (anonymous):

OpenStudy (roadjester):

I'm thinking A because the battery and bulbs are all in parallel. Therefore the voltage difference across them is all the same.

OpenStudy (roadjester):

Intensity I= dP/dt

OpenStudy (ash2326):

When Switch S is open, there is one path for the current that is through B1. If Switch S is closed, the current drawn from the Battery would be divided, and B1 will have less current flowing through it as compared to the current when switch was open.

OpenStudy (roadjester):

true but the voltage is still the same

OpenStudy (ash2326):

Since power is proportional to the square of current, less power will be drawn by B1.

OpenStudy (roadjester):

At the same time the power is also proportional to the voltage

OpenStudy (ash2326):

That would be true when battery is ideal, in that case voltage across the bulb B1 will be same irrespective of switch being close or open. Here the battery has some internal resistance, when you add another parallel path across the bulb B1,more current will be drawn from the battery and there will be more voltage drop in the internal resistance of the battery and hence less voltage across B1. And yes according to the power being proportional to the voltage, less power for B1 in case when switch is closed.

OpenStudy (ash2326):

@Emineyy do you follow?

OpenStudy (roadjester):

I downloaded a circuit construction kit online when I took E&M; I just used it to get a visual of @Emineyy 's circuit. Opening and closing the circuit does not affect B1. It merely lets current flow through B2

OpenStudy (ash2326):

Did you add a resistor in series with the battery?

OpenStudy (ash2326):

@roadjester

OpenStudy (carmuz):

Nothing happens to the brightness of the light bulbs in the parallel circuit if the battery is capable of supplying the additional current. But you said that the bulbs have appreciable internal resistance, so the original bulb might dim, according to the additional current demand. All of the light bulbs in your house are in parallel across the utility supply to the house. Switching one of them on or off has no effect on the brightness of the others, but what happens if you turn on either a water heating, an iron or the AC?

OpenStudy (roadjester):

With the battery? No, but I did increase the internal resistance of the bulbs. Let me add a resistor to the battery.

OpenStudy (roadjester):

I maxed out the internal resistance of the battery, which happened to be 9 ohms. B1 seems to be permanently dimmer than B2. Odd...

OpenStudy (roadjester):

B1 dims, and B2 is brighter than B1...which I find unususal

OpenStudy (ash2326):

@carmuz Our home supply is an AC supply with a high voltage and has capability of sourcing a good amount of current. Here the battery has limited capability of sourcing current and has some resistance, if adding one bulb won't affect B1, then we can add innumerable no. of bulbs in parallel and still get the same brightness from all and that would be utopia.

OpenStudy (roadjester):

unusual*

OpenStudy (ash2326):

@roadjester decrease the resistance of B1 and increase the resistance of B2

OpenStudy (ash2326):

and you can make B1 brighter than B2 but B1 will be still less bright than when switch S is open.

OpenStudy (roadjester):

I think that part is self-intuitive. It's the part of having different intensities that's bugging me

OpenStudy (roadjester):

when the internal resistance is the same I mean

OpenStudy (carmuz):

Take in count that the questions says: "appreciable internal resistance". If you leave out that phrase, then the answer should be A. But I have doubts because of the "appreciable internal resistance".

OpenStudy (ash2326):

Yes, I agree @carmuz

OpenStudy (roadjester):

well, maximum intensity occurs when the resistance of the source is equivalent to the internal resistance of the load...which I find interesting...not sure if that applies here

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