AQA Physics Unit 1 PHYA1 20th May 2014 OFFICIAL Watch

Reecelyons
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I have only ever it ask you to draw the graph for a diode... along with filament lamp and an ohmic conductor,
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ozzyoscy
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(Original post by hltorrance)
You can work out quark structures using the data on the formula structure there is not need to memorise them
Where?
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samlyon
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(Original post by Reecelyons)
I have only ever it ask you to draw the graph for a diode... along with filament lamp and an ohmic conductor,
what is the reversed bias drop off point? I know the forwards bias is 0.6V but I'm sure it was on a question i did and I can't remember the answer
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samlyon
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(Original post by samlyon)
what is the reversed bias drop off point? I know the forwards bias is 0.6V but I'm sure it was on a question i did and I can't remember the answer
Is it -50V?
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Reecelyons
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(Original post by samlyon)
what is the reversed bias drop off point? I know the forwards bias is 0.6V but I'm sure it was on a question i did and I can't remember the answer
Surely because current can only go one way with a diode in place, it just looks like the following... Name:  VI-graph-for-filament-diode.jpg
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MsFahima
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#166
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#166
Just did a past paper.. got 59/70... Good enough?
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MsFahima
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(Original post by Reecelyons)
Surely because current can only go one way with a diode in place, it just looks like the following... Name:  VI-graph-for-filament-diode.jpg
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Don't forget to label the 0.6
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Firuza1029
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(Original post by Reecelyons)
Surely because current can only go one way with a diode in place, it just looks like the following... Name:  VI-graph-for-filament-diode.jpg
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Name:  diode_graph.jpg
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If you extend the reverse bias, it eventually reaches a breakdown voltage where the diode's resistance suddenly drops, so a large current can flow
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simon105
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(Original post by toddmcnugget)
Does anyone have a list of the experiments we're supposed to know such as the resistivity one?
I could do with this too.

Anyone?
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b.kot
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#170
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can anyone help me with potential dividers?
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b.kot
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Also do we need to know Kirchhoff's circuit laws?
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Sumz.96
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Name:  ImageUploadedByStudent Room1400533735.614872.jpg
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I did that question as homework a few months ago. Hope it helps.


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Kingnig
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(Original post by b.kot)
can anyone help me with potential dividers?
yes yes bump this
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imyimy
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(Original post by Kingnig)
yes yes bump this
For potential dividers the formula
Vout=Vin x R1 / (total R)

Sorry not much help aha but that's what I use most of the time


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Dalek1099
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(Original post by MsFahima)
Don't forget to label the 0.6
Are you suppose to know this?What else do you need to know about potential bias because its a topic I have never come across in a diode and in the specimen it asked it and said that current flows past the ammeter when the current in a diode is very high protecting the ammeter.i have done where the current starts flowing in the direction the diode is stopping at a high negative voltage though.The specimen is so hard I got 55/70-be careful if you do it though because some of the answers are wrong.
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Dalek1099
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(Original post by imyimy)
For potential dividers the formula
Vout=Vin x R1 / (total R)

Sorry not much help aha but that's what I use most of the time


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Potential Dividers formulas are unnecessary because the v out basically represents the voltage supplied to the component its as simple as that but AQA try to pretend its a big ew topic but its not really.
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Potatosandwich98
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#177
What do you guys think the big marker will be on?
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Firuza1029
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#178
Would we be expected to know what sensor potential divider circuits look like?
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Wizz1996
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Totally screwed for tomorrow's paper. Should I revise throughout the night?
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Davelittle
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(Original post by JM17)
P=VI, P=I^2/R, P=V^2/R

Swap your P for an E and add on a t.

(Original post by aLeXaNdRa08)
ah, but theres no way in hell that i would have thought of that during an exam
I guess ill just have to memorise it
Thanks

(Original post by JM17)
Well you know Power = Energy / Time, and I'm sure this equation is in the book.

So Energy must equal power times time, so swap E and P and add on t.
P=IV I=Q/t V=W/Q (all given on data sheet)

P=(Q/t)*(W/Q)=W/t=E/t (work is just energy)

Goodluck guys I had this exam last year hope you all do ok!
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