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A level physics electricity question 3
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muhammad0112
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#1
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.
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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#2
(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.
![Name: Screenshot 2022-05-24 192419.png
Views: 15
Size: 34.7 KB]()
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.
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.
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.
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ChemEng9081
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#3
(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.
![Name: Screenshot 2022-05-24 192419.png
Views: 15
Size: 34.7 KB]()
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.
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.
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.
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muhammad0112
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#4
(Original post by Mocha Latte)
Hint: express the voltage through P and R before and after Q's filament melts
Hint: express the voltage through P and R before and after Q's filament melts
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cata03
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#5
(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.
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.
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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#6
(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.
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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#7
(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
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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muhammad0112
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#8
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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#9
(Original post by muhammad0112)
Idk maybe i'm just dumb, but I'm still confused on this whole concept 😭.
Idk maybe i'm just dumb, but I'm still confused on this whole concept 😭.
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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