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AQA Physics Unit 1 PHYA1 20th May 2013

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Original post by Jimmy20002012
What about the reverse diode position in parallel?


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Whatever I just said was about diodes in opposite direction. (1 in reverse and the other 1 in forward). If they are both in reverse position in parallel, there would be no flow of current. As I said, if they are in opposite in parallel, the current will flow.
Original post by StalkeR47
Whatever I just said was about diodes in opposite direction. (1 in reverse and the other 1 in forward). If they are both in reverse position in parallel, there would be no flow of current. As I said, if they are in opposite in parallel, the current will flow.


Okay with parallel:

1. If in opposite directions (one forward and one in reverse) voltage flows through and current flows when voltage is 0.6v.

2. If both are reversed no current flows at all, only voltage flows, so voltage if there were resistors would be 0 as V=IR.

In series:

Current does not flow at all at any times even if there were in opposite directions, they have to be in forward in order for current to flow

Am I right :smile:

Last ever question:

If there were combination of diodes in parallel how would current and voltage vary?


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Original post by Jimmy20002012
Okay with parallel:

1. If in opposite directions (one forward and one in reverse) voltage flows through and current flows when voltage is 0.6v.

2. If both are reversed no current flows at all, only voltage flows, so voltage if there were resistors would be 0 as V=IR.

In series:

Current does not flow at all at any times even if there were in opposite directions, they have to be in forward in order for current to flow

Am I right :smile:

Last ever question:

If there were combination of diodes in parallel how would current and voltage vary?


Posted from TSR Mobile


Yes correct! Last Q----It depends if the diodes are forward or reversed.
Original post by StalkeR47
Sure!!! you need to calculate the peak power of the lamp right? Our equation is p=iv------which is P(peak)=I(peak)V(peak)------So, our peak current is 2.8A. (we do not have peak voltage but RMS voltage which is 12V) So, peak voltage would (12 times sqrt2). So, your peak power is---P(peak)=2.8 times (12 times sqrt2) = 48W.---ANSWER.


it appears I have the wrong exam board, i was looking at the date :colondollar: this is a mechanics question im refering too
Original post by StalkeR47
Yes correct! Last Q----It depends if the diodes are forward or reversed.


Say one reverse diode was in parallel with a forward and reverse diode? Thanks soo much for this.


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Original post by SexyNerd
it appears I have the wrong exam board, i was looking at the date :colondollar: this is a mechanics question im refering too


Can you state the question/subject/exam board again please. I will try my best to help you!
Original post by Jimmy20002012
Say one reverse diode was in parallel with a forward and reverse diode? Thanks soo much for this.


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1 reverse and 1 forward would mean that the current will flow. voltage will not.
Original post by ak395
I've attached the answers to all the long questions from june 2009 onwards :smile:


Do you think any of these will be repeated?
I think the 6 marker will be about the photoelectric effect in terms if wave and particles properties and what was observed


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Original post by IWantSomeMushu
Do you think any of these will be repeated?


These are least likely to be repeated since they came up in the exam not long ago.
Original post by StalkeR47
1 reverse and 1 forward would mean that the current will flow. voltage will not.


I though voltage always flows?


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Original post by Jimmy20002012
I though voltage always flows?


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In parallel, if the current flows, voltage will not. Remember the rule?? -----Forward bias - all current, almost no volts.
Reverse bias - all volts, almost no current.
Original post by StalkeR47
In parallel, if the current flows, voltage will not. Remember the rule?? -----Forward bias - all current, almost no volts.
Reverse bias - all volts, almost no current.


And its the complete opposite for series diodes


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Original post by Jimmy20002012
And its the complete opposite for series diodes


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YUP!
Reply 274
Avatar for sceezy
sceezy
HEY, can someone help me with June 2012 question 5(a) (the 6 marker)

I can't even find it in my text book.

PLEASE!!!!!
Original post by sceezy
HEY, can someone help me with June 2012 question 5(a) (the 6 marker)

I can't even find it in my text book.

PLEASE!!!!!


For that kind of question just learn the mark scheme, anyway very unlikely to come up as it came up last year :smile:


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Original post by Jimmy20002012
For that kind of question just learn the mark scheme, anyway very unlikely to come up as it came up last year :smile:


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LOL Jimmy! Are you now clear on diodes? :rolleyes:
Reply 277
Avatar for mattj94
mattj94
Original post by BenChard
your website and notes are aamazing. I wish you was doing more of the exams I was doing! thanks!


Thanks!
What other exams are you doing?

Matt
Reply 278
Avatar for g4adaffi
g4adaffi
8
Please someone can you explain to me what happens to the voltage and current as resistance increases in diffrent components including wires thanks and sorry for the caps.
Original post by g4adaffi
Please someone can you explain to me what happens to the voltage and current as resistance increases in diffrent components including wires thanks and sorry for the caps.


When you increase the resistance, current increases but voltage decreases.

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