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A2 Physics | Post your doubts here

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Assalamoalaikum wr wb!

on thursday i think,you asked a question about the graph for diode/rectification one, and u said you didnt et it..sorry cudn't postsome pic that day..
here's it:
hope it helps...
View attachment 9241



http://www.flashscience.com/electricity/smoothing.htm

^check this link to see why there's a flat line in the graph...there'll be a part...when one set is working and the other two are not and vice versa..
walkum aslam
thanks alot..
and the peak value of rectified voltage is less thn the original ac supply???
 
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i think you're still confused... :unsure: don't look at the graph ....look at the circuit and see what happens...although input is changing...polarity at the output is not..right?

waeyyakum..aameen!

May Allah make the exam easy for you too and give you the good what u desire...
The problem with me was that I was looking at the ac wave which is measured by a voltmetre of fixed polarity.
But the polarity of a.c changes, so with respect to the diode, the negative half cycle, that is measured by the voltmeter , is positive.
I already understood all of what you were trying to explain, but I wasnt clear enough to my issue haha.
Im sorry for using up your time, brother, but Allah will InshAllah reward you :)
 
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The problem with me was that I was looking at the ac wave which is measured by a voltmetre of fixed polarity.
But the polarity of a.c changes, so with respect to the diode, the negative half cycle, that is measured by the voltmeter , is positive.
I already understood all of what you were trying to explain, but I wasnt clear enough to my issue haha.
Im sorry for using up your time, brother, but Allah will InshAllah reward you :)

it's ok...so u still confursed? :unsure:
what's exactly confusing you..u don't look at the graph for ac...
look at the circuit..and try to point ur finger and see...how the current flows...
 
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it's ok...so u still confursed? :unsure:
what's exactly confusing you..u don't look at the graph for ac...
look at the circuit..and try to point ur finger and see...how the current flows...
No no i got it, i just said the answer to my problem. That the wave of the ac is measured by a voltmeter of fixed polarity.
But the ac supply polarity changes so the diode detects the negative half cycle as a positive one !
 
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No no i got it, i just said the answer to my problem. That the wave of the ac is measured by a voltmeter of fixed polarity.
But the ac supply polarity changes so the diode detects the negative half cycle as a positive one !
exactly...
 
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From the application booklet:-
"For a satellite having a period of 90 minutes, each orbit crosses the Equator 23 ° to the west of the previous orbit."
Why is that? How was it calculated?
 
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November 2009 Paper 42 Question 2 cii) Examiner report says insufficient data provided, what insufficient data? I calculated it by Work done = change in k.e
30 J = 3/2 Nxkxdelta T
 
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November 2009 Paper 42 Question 2 cii) Examiner report says insufficient data provided, what insufficient data? I calculated it by Work done = change in k.e
30 J = 3/2 Nxkxdelta T
I also thought the same
We can calculate easily
 
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I also thought the same
We can calculate easily
and i didn't even think :D :D

i just saw the ms and wrote a note that question removed :ROFLMAO:
lets think it out.. hmmm
What I did is that since work done by gas is 30 Joules then Internal energy decreased by 30 Joules.
Internal energy = Total Kinetic energies + (Ep=0) [ideal gas ]
-30 Joules = 3/2 N x k x T ? What do you think ? Lets suppose we want to calculate energy change? is it? -30 J = 3/2 N x k x Delta T?
We need to figure this out...
 
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