http://www.examsolutions.net/mathsr...utorial1a.php
He uses the squaring method here
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 17072016 23:12

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 17072016 23:49
(Original post by PhyM23)
To use S_{∞} = a/(1r) , 1<r<1 , which doesn't hold if 0<x<1 as the series diverges, so the formula can't apply to this set of values.
So we include and the answer is 0<X<root2
But when we say X>0 we are including 1 right and that can't be right because 1 is not a valid solution in our calculations? So our answer could be 0<x<root2 where x is not equal to 1?Last edited by NatoHeadshot; 17072016 at 23:54. 
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 183
 18072016 00:26
(Original post by NatoHeadshot)
So if we combine the condition that X>0 they gave we see that X can be > 0 but less then 1 and be valid
So we include and the answer is 0<X<root2
But when we say X>0 we are including 1 right and that can't be right because 1 is not a valid solution in our calculations? So our answer could be 0<x<root2 where x is not equal to 1? 
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 184
 18072016 20:53
I am having trouble (I always have troubles ) with following problem. The question is interesting. You might want to attempt it before looking at the answer if you haven't already.The problem:Attachment 562508The solution and the trouble: and solution : http://www.physicsandmathstutor.com/pat/solutions2010/ question 11.
Spoiler:Show
How did he know that A=1/6 and B=6/1 but not 2/5?Tagged:Last edited by tangotangopapa2; 18072016 at 21:06. 
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 185
 18072016 20:56
(Original post by tangotangopapa2)
I am having trouble (I always have troubles ) with following problem. The question is interesting. You might want to attempt it before looking at the answer if you haven't already.The problem:Attachment 562508The solution and the trouble:Spoiler:Show
How did he know that A=1/6 and B=6/1 but not 2/5? 
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 186
 18072016 21:01
(Original post by hellomynameisr)
The attachment is invalid it says. Wont let me open it
Solution : http://www.physicsandmathstutor.com/pat/solutions2010/ question 11
How did he know that A and B were 6/1 and 1/6 and not 2/5 and 5/2? 
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 187
 18072016 21:10
(Original post by tangotangopapa2)
I am sorry. I was having trouble with attachments. Edited the post.
Solution : http://www.physicsandmathstutor.com/pat/solutions2010/ question 11
How did he know that A and B were 6/1 and 1/6 and not 2/5 and 5/2?
Perhaps it was just the person that done the answers quickest and easiest way to approach it. 
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 188
 18072016 21:40
(Original post by hellomynameisr)
I think since it says that there is an equal possibility to get all the scores, it wouldnt matter if it was 1/6 or 2/5 as you get the same answer with either.
Perhaps it was just the person that done the answers quickest and easiest way to approach it.
2 d1 + 5d2  7 = S
does not have equal probability for all the scores.
For instant d1 = 1 and d2 = 3 gives S=10 and
d1 = 6 and d2 = 1 gives S =10.
There is no combination that gives S=34. 
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 189
 18072016 23:44
(Original post by tangotangopapa2)
I am having trouble (I always have troubles ) with following problem. The question is interesting. You might want to attempt it before looking at the answer if you haven't already.The problem:Attachment 562508The solution and the trouble: and solution : http://www.physicsandmathstutor.com/pat/solutions2010/ question 11.Spoiler:Show
How did he know that A=1/6 and B=6/1 but not 2/5? 
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 190
 19072016 09:29
(Original post by samb1234)
The important part is the equal probability part. If we look at the list of numbers for 1 and 6 we have 16 and 6,12...36. As you can see there is only one way to get any particular outcome as the only overlap is the 6,6. If we looked at 2 and 5 as a solution then we get some issues with overlap e.g 27 i could have 25 and 2 or 15 and 12 so this condition isnt met
I see that 1/6 is the only way to satisfy all the conditions but how would you figure out that A and B are 1/6 and 6/1 in the first place? They could be any real numbers.
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 19072016 15:45
(Original post by tangotangopapa2)
Thanks for the reply.
I see that 1/6 is the only way to satisfy all the conditions but how would you figure out that A and B are 1/6 and 6/1 in the first place? They could be any real numbers.
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 19072016 17:36
Question 18 here:
https://www2.physics.ox.ac.uk/sites/..._pdf_10581.pdf
Solution:
https://oxfordpat.wordpress.com/oxfo...1/#comment349
When the switch is connected to e, does the capcitor not get charged so should the graph not look like this instead of what was drawn on the first graph
(attached)
or basically should the graphs not be switched around? Because when teh switch is connected to the other end it gets no voltage so it discharges right?
Also, any ideas how he estimated "signifcant time" 
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 19072016 23:43
(Original post by NatoHeadshot)
Question 18 here:
https://www2.physics.ox.ac.uk/sites/..._pdf_10581.pdf
Solution:
https://oxfordpat.wordpress.com/oxfo...1/#comment349
When the switch is connected to e, does the capcitor not get charged so should the graph not look like this instead of what was drawn on the first graph
(attached)
or basically should the graphs not be switched around? Because when teh switch is connected to the other end it gets no voltage so it discharges right?
Also, any ideas how he estimated "signifcant time"
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 194
 20072016 00:04
Also would you mind clearing my graph question? 
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 20072016 00:19
(Original post by NatoHeadshot)
is that a standard thing?
Also would you mind clearing my graph question?
Yours is voltagetime graph and is correct.
Posted from TSR MobileLast edited by tangotangopapa2; 20072016 at 00:26. 
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 196
 20072016 12:41
wait wait, Ive done C1, C2, M1, M2, FP1 and S1.So do I have to self teach C3 and C4 this summer omds fml

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 20072016 14:09
(Original post by CaiusMartius)
wait wait, Ive done C1, C2, M1, M2, FP1 and S1.So do I have to self teach C3 and C4 this summer omds fmlPost rating:1 
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 198
 20072016 17:04
(Original post by tangotangopapa2)
5 RC or 5 times time constant is usually considered to be time to fully charge a capacitor. The first graph of the solution is currenttime graph. Initially, there is no charge so there is high rate of increase of charge I.e. high current. The rate of increase of charge decreases exponentially.
Yours is voltagetime graph and is correct.
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 20072016 18:18
(Original post by NatoHeadshot)
So the Order of the graphs is right in the solution? When the capacitor is connected the battery circuit the current decreases exponentially and when it's one without the battery it increases???
In the second case i.e. while discharging, the current decreases exponentially too. Since, the direction of flow of current is opposite to that when charging, this current has been shown negative.
It is still decreasing nonetheless.
Yes, the order of graph is right.
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 20072016 22:10
Hi, does anyone know what I should learn in prep for the exam? So far I've only learnt AS Physics and Maths, some core 3 and some A2 physics. I'm already planning on learning all of A2 physics and maths this summer. Which F.Maths modules should I learn (I haven't done any of it so far.) And also, which BPhO papers should I do?
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Updated: December 6, 2016
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