A level physics electricity question 3

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muhammad0112
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#1
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Hi, I would appreciate it if someone posts and explanation on why lamp P becomes brighter and lamp R becomes dimmer. All 3 lamps are identical. I thnk i need help trying to understand the topic.
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I think that the voltage in lamp P would remain the same, but it would have a faster current. That's as far as I got.
Last edited by muhammad0112; 1 month ago
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Mocha Latte
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(Original post by muhammad0112)
Hi, I would appreciate it if someone posts and explanation on why lamp P becomes brighter and lamp R becomes dimmer. All 3 lamps are identical. I thnk i need help trying to understand the topic.
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I think that the voltage in lamp P would remain the same, but it would have a faster current. That's as far as I got.
Hint: express the voltage through P and R before and after Q's filament melts
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ChemEng9081
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(Original post by muhammad0112)
Hi, I would appreciate it if someone posts and explanation on why lamp P becomes brighter and lamp R becomes dimmer. All 3 lamps are identical. I thnk i need help trying to understand the topic.
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I think that the voltage in lamp P would remain the same, but it would have a faster current. That's as far as I got.
Think about how the resistance changes, and if voltage increases brightness increases
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muhammad0112
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(Original post by Mocha Latte)
Hint: express the voltage through P and R before and after Q's filament melts
I don't know how the voltage would change before and after. That's what i need help understanding. Although, I understand that a higher voltage means a brighter light. But that doesn't make sense because the voltage in lamp P shouldn't change. for example, imagine theres 10v flowing through lamp P, 10v flowing through lamp Q, and 10v through lamp R. If, lamp Q breaks, Each coulomb of charge still carries the same energy (as V =E/Q) so there will still be 10v flowing through lamp P and R. This means that the brightness shouldn't change at all. But that's not correct since the question says that the brightenss changes. this is my confusion.
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cata03
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(Original post by muhammad0112)
I don't know how the voltage would change before and after. That's what i need help understanding. Although, I understand that a higher voltage means a brighter light. But that doesn't make sense because the voltage in lamp P shouldn't change. for example, imagine theres 10v flowing through lamp P, 10v flowing through lamp Q, and 10v through lamp R. If, lamp Q breaks, Each coulomb of charge still carries the same energy (as V =E/Q) so there will still be 10v flowing through lamp P and R. This means that the brightness shouldn't change at all. But that's not correct since the question says that the brightenss changes. this is my confusion.
Edit: said before that voltage doesn't change but it will - resistance of QP pair increases so greater proportion of potential difference used up over QP, therefore the proportion of the potential difference used up over R decreases. Cred to ChemEng9081!!

Without lamp Q, the current through lamp P will increase as it will not be split between Q and P. Meanwhile, the current in lamp R will decrease as the total resistance of the circuit has increased now that P is effectively in series with R.
Last edited by cata03; 1 month ago
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ChemEng9081
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(Original post by muhammad0112)
I don't know how the voltage would change before and after. That's what i need help understanding. Although, I understand that a higher voltage means a brighter light. But that doesn't make sense because the voltage in lamp P shouldn't change. for example, imagine theres 10v flowing through lamp P, 10v flowing through lamp Q, and 10v through lamp R. If, lamp Q breaks, Each coulomb of charge still carries the same energy (as V =E/Q) so there will still be 10v flowing through lamp P and R. This means that the brightness shouldn't change at all. But that's not correct since the question says that the brightenss changes. this is my confusion.
When Q breaks, the resistance of the Q P look increases. That means it’ll get a higher proportion of the voltage
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cata03
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(Original post by ChemEng9081)
When Q breaks, the resistance of the Q P look increases. That means it’ll get a higher proportion of the voltage
Yes you're right, sorry!!
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muhammad0112
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Idk maybe i'm just dumb, but I'm still confused on this whole concept 😭.
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Mocha Latte
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(Original post by muhammad0112)
Idk maybe i'm just dumb, but I'm still confused on this whole concept 😭.
So I'm assuming you've learned about potential dividers?
Let's assume the resistance of each lamp is T and the voltage of the battery is V.

With Q:

The resistance of the QP pair is T/2. The total resistance is 3/2 T.

Hence, the potential difference across P (which is the same as across Q) is (T/2)/ (3T/2) V = V/3. The power of P is thus V^2/9T.

The resistance of R is T. Hence, the potential difference across it is T/(3T/2)V = 2V/3. The power of R is thus 4V^2/9T.

Without Q:
The total resistance is 2T.

The potential difference across P is T/(2T) V= V/2 so its power is V^2/4T. This is the same as R's power.

So, P's power changed from 1/9 (V^2/T) to 1/4 (V^2/T) so it gets brighter while R changed from 4/9 (V^2/T) to 1/4 (V^2/T) so it gets dimmer.
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