AQA Physics Unit 1 PHYA1 20th May 2013

Original post by PickwickianGeek

anyone care to explain 5 b) on this paper? http://filestore.aqa.org.uk/subjects/AQA-PHYA1-W-SQP-07.PDF

You have a total pd of 12V. The resistors are in parallel with a diode in parallel in it's forward bias. (Hint: What is the voltage at 6ohm resistor) ?

Reply 662

Abbseh

Original post by StalkeR47

YEP that is true! But I can help you if you need any. Today is the last day hahahaa :cool:

Oh don't remind me! I want to study physics at uni but damn electricity just annoys me :frown:

I do have one question though, in the context of Q6C Jan 2013 (and others similar) how do you go about talking about proportion?

Up until a week ago I always got it right but all of a sudden I lost it :/

Reply 663

Original post by pushkin_

Hay guys, as I see this thread is pretty active right now. So maaaaybe somebody could explain me 6th question (b) in the 2013 January paper , please ? It's just that in the mark scheme there are just values given, no solutions.. :frown:

p.s. GOOD LUCK TOMORROW EVERYONE, we can do it !!!!!

Hi! I will explain this to you...
Ok.. we have 4 resistors in series. 2 in each series. Total voltage is 12 volts. (Remember voltage in parallel is the same all the way around and in series it adds up to). Since the resistors are in series, they must have voltage adds up to total depending on the amount of resistance. There is an easy way to handle this problem and a harder way to solve. Easy way----- Voltage in parallel----same. So, across ac and ce, they must share same amount of voltage. so it must be 6V for both ac and ce. This is total of 6 Voltage used up in A-E. So you have 6 left. Since the resistance of the resistor is twice the resistance of the thermistor. Resistor BD must get twice the voltage as DF. So, BD must be 4 V and thermistor which is DF or CD must get 2V. so our voltage adds up to total. 6+4+2=12V. Harder way... calculate the current through AE by using V=IR. I=12/4xx10^3 = 3x10^-4A... Use V=IR again to calculate voltage across AC. V=3x10^-4 times 20x10^3 = 6V. The rest can be solved by using V=IR. Hope this helps..

Reply 664

Original post by PickwickianGeek

anyone care to explain 5 b) on this paper? http://filestore.aqa.org.uk/subjects/AQA-PHYA1-W-SQP-07.PDF

As it's a diode, you know that the pd across it is 0.6v, therefore the pd across the resistor is 12-0.6= 11.4v. 11.4/4 = 2.85A. Add this to 12/6 = 2A, and your answer should be 4.85A Is this correct?

Reply 665

Original post by Abbseh

Oh don't remind me! I want to study physics at uni but damn electricity just annoys me :frown:

I do have one question though, in the context of Q6C Jan 2013 (and others similar) how do you go about talking about proportion?

Up until a week ago I always got it right but all of a sudden I lost it :/

Can you give more details on the question you have just asked?

Reply 666

PickwickianGeek

6

Original post by yajman

As it's a diode, you know that the pd across it is 0.6v, therefore the pd across the resistor is 12-0.6= 11.4v. 11.4/4 = 2.85A. Add this to 12/6 = 2A, and your answer should be 4.85A Is this correct?

yeah that's correct, cheers for the explanation. So current is split in a parallel circuit but is the same throughout a series circuit, yes?

Reply 667

Original post by yajman

As it's a diode, you know that the pd across it is 0.6v, therefore the pd across the resistor is 12-0.6= 11.4v. 11.4/4 = 2.85A. Add this to 12/6 = 2A, and your answer should be 4.85A Is this correct?

yajman! you are not allowed to give straight answers! check the guidelines for the student room. that is why I have him a hint.

Reply 668

reemaisthinking

Original post by westarmy

Would you like to balance my pivot by placing weights on either side? :ahee:

That made me giggle and squirm- and I'm not even sure what you meant by that!... I'm still grinning. Like an idiot.

Posted from TSR Mobile

Reply 669

pushkin_

Original post by StalkeR47

Hi! I will explain this to you...
Ok.. we have 4 resistors in series. 2 in each series. Total voltage is 12 volts. (Remember voltage in parallel is the same all the way around and in series it adds up to). Since the resistors are in series, they must have voltage adds up to total depending on the amount of resistance. There is an easy way to handle this problem and a harder way to solve. Easy way----- Voltage in parallel----same. So, across ac and ce, they must share same amount of voltage. so it must be 6V for both ac and ce. This is total of 6 Voltage used up in A-E. So you have 6 left. Since the resistance of the resistor is twice the resistance of the thermistor. Resistor BD must get twice the voltage as DF. So, BD must be 4 V and thermistor which is DF or CD must get 2V. so our voltage adds up to total. 6+4+2=12V. Harder way... calculate the current through AE by using V=IR. I=12/4xx10^3 = 3x10^-4A... Use V=IR again to calculate voltage across AC. V=3x10^-4 times 20x10^3 = 6V. The rest can be solved by using V=IR. Hope this helps..

Ohh, I see now ! I was doing it hard way and messed up with current ratio .. Thanks a lot !!! :redface:

Reply 670

Original post by StalkeR47

yajman! you are not allowed to give straight answers! check the guidelines for the student room. that is why I have him a hint.

Sorry there! As you can see, I'm fairly new to TSR, so I'm extremely sorry for that. :colondollar:

Reply 671

Original post by PickwickianGeek

yeah that's correct, cheers for the explanation. So current is split in a parallel circuit but is the same throughout a series circuit, yes?

That's correct, yes. Apparently, I'm not supposed to give straight answers, so I'll try and be discrete next time. Good luck if you're sitting the exam tomorrow :smile:

Reply 672

Original post by yajman

Sorry there! As you can see, I'm fairly new to TSR, so I'm extremely sorry for that. :colondollar:

It's ok!

Reply 673

Original post by pushkin_

Ohh, I see now ! I was doing it hard way and messed up with current ratio .. Thanks a lot !!! :redface:

No probs! Ratio always confuses me and therefore I tend to use calculations. The reason why I replied all that so quickly is because I just explained all that to someone else before, so I copied my explanations from page 23 of this post. HHAA :cool:

Reply 674

PickwickianGeek

6

Original post by yajman

That's correct, yes. Apparently, I'm not supposed to give straight answers, so I'll try and be discrete next time. Good luck if you're sitting the exam tomorrow :smile:

Ah okay, thanks! Oh, are you not? It's not like it's a practical exam though, surely we are allowed to give away correct calculations on old papers? Thanks, you too if you're sitting it!

Reply 675

Original post by PickwickianGeek

Ah okay, thanks! Oh, are you not? It's not like it's a practical exam though, surely we are allowed to give away correct calculations on old papers? Thanks, you too if you're sitting it!

Yeah, I'm not sure why. I'll be sitting the exam tomorrow too, so thanks for the wishes. Praying that it will be a good paper with not too many difficult questions, particularly the 6-marker. That will make-or-break my chances of getting a good result :s-smilie:

Reply 676

chrishy2012

Does anyone have any idea what the big six marker question could be on?! The one last year was good, hope we have one like that or something on the photoelectric effect

Reply 677

westarmy

Original post by reemaisthinking

That made me giggle and squirm- and I'm not even sure what you meant by that!... I'm still grinning. Like an idiot.

Posted from TSR Mobile

:P well I'm pretty much doing the same! And you know what I meant :colone:

Reply 678