AQA A level physics electricity questions help.

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muhammad0112
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#1
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#1
Hi, I don't really understand the answer to this question. The answer is 12v we are already told that lamp X is rated th 12V 36W and lamp Y is rated at 4.5V 2W. I know that the emf is 24 but i don't know why 24 - 12 would give you the pd across R1.
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Answe: 24 - 12 = 12V
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muhammad0112
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#2
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(Original post by muhammad0112)
Hi, I don't really understand the answer to this question. The answer is 12v we are already told that lamp X is rated th 12V 36W and lamp Y is rated at 4.5V 2W. I know that the emf is 24 but i don't know why 24 - 12 would give you the pd across R1.
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Answe: 24 - 12 = 12V
I looked at the other questions, to work out the Pd of R2, I had to do 12v - 4.5v. This shows the the same amount of voltage goes to each junction. But I thought that if a 1 junction has a higher total resistace, more votage will go towards that junction. So we can't assume that the same amount of PD will go to each junction.
Last edited by muhammad0112; 4 weeks ago
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JSJSJSJSJ123
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#3
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(Original post by muhammad0112)
I looked at the other questions, to work out the Pd of R2, I had to do 12v - 4.5v. This shows the the same amount of voltage goes to each junction. But I thought that if a 1 junction has a higher total resistace, more votage will go towards that junction. So we can't assume that the same amount of PD will go to each junction.
can you post the question where you got information to work out R2
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muhammad0112
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#4
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#4
https://www.physicsandmathstutor.com...20Circuits.pdf

It's question 4 of this document.

Here is the markscheme:

https://www.physicsandmathstutor.com...its%2520MS.pdf
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muhammad0112
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#5
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I think question 6ii also uses the same principle, if you don't mind having a look at that. I know that 5.4V will go to the 540 ohm resisistor. and that leaves 9.6V left. but I would think that the thermistor and the 1200 ohm resistor would be given different voltages as they have different resistance.
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BlueChicken
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(Original post by muhammad0112)
https://www.physicsandmathstutor.com...20Circuits.pdf

It's question 4 of this document.

Here is the markscheme:

https://www.physicsandmathstutor.com...its%2520MS.pdf
If you think of the junctions (the black dots) left to right labelled as A to D, then the question tells you that the pd across C-D, i.e. across lamp X, is 12V. As pd from A-D is 24V, the pd across A-B must be 12V, i.e. 24V - 12V.

I'd lookup resistors in series and parallel, including what that means for current and voltage, to convince yourself. Good luck!
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BlueChicken
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(Original post by muhammad0112)
I think question 6ii also uses the same principle, if you don't mind having a look at that. I know that 5.4V will go to the 540 ohm resisistor. and that leaves 9.6V left. but I would think that the thermistor and the 1200 ohm resistor would be given different voltages as they have different resistance.
See my answer above - again, look up pd across components in parallel (and series), i.e. Kirchoff's Laws, and this should become straightforward.
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muhammad0112
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Hi, I've just had a look at voltage across componants in parallel and it makes sense now. Thanks
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BlueChicken
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#9
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(Original post by muhammad0112)
Hi, I've just had a look at voltage across componants in parallel and it makes sense now. Thanks
Excellent! I would’ve explained it, but it’s much easier for you to look it up in notes/text book with diagrams etc. than trying to do it here. Good luck with your exam.
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