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    (Original post by Jimmy20002012)
    What about a resistor would voltage go through that if diode is reversed?


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    Here is the basic rule! Forward bias - all current, almost no volts.
    Reverse bias - all volts, almost no current.

    So in reverse bias, the voltage will go through the resistor but there will be negligible current.
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    (Original post by StalkeR47)
    Here is the basic rule! Forward bias - all current, almost no volts.
    Reverse bias - all volts, almost no current.

    So in reverse bias, the voltage will go through the resistor but there will be negligible current.
    So in reverse biased the diode will pretty much not heat up right ?
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    (Original post by StalkeR47)
    Here is the basic rule! Forward bias - all current, almost no volts.
    Reverse bias - all volts, almost no current.

    So in reverse bias, the voltage will go through the resistor but there will be negligible current.
    One last question with diodes:

    In the attachment look at the circuit diagram and the question is what happen to AC and BC if D2 is reversed, the answer for both is 12v, could you explain this?


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    (Original post by posthumus)
    So in reverse biased the diode will pretty much not heat up right ?
    Yes Correct!
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    (Original post by Jimmy20002012)
    One last question with diodes:

    In the attachment look at the circuit diagram and the question is what happen to AC and BC if D2 is reversed, the answer for both is 12v, could you explain this?


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    Where is the attachment?
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    Originally Posted by StalkeR47<br />
    Here is the basic rule! <b>Forward bias - all current, almost no volts.<br />
    Reverse bias - all volts, almost no current.</b> <br />
    So in reverse bias, the voltage will go through the resistor but there will be negligible current.
    <br />
    <br />
    One last question with diodes:<br />
    <br />
    In the attachment look at the circuit diagram and the question is what happen to AC and BC if D2 is reversed, the answer for both is 12v, could you explain this? <br />
    <br />
    <br />
    Posted from TSR Mobile


    Posted from TSR Mobile
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    (Original post by mattj94)
    Hi, I recently set up a revision website. So far AQA Unit 1 is the only module I have for Physics but I hope to add unit 2 notes soon, it would be great if you could check it out! http://www.mattsrevision.com/particl...d-electricity/
    your website and notes are aamazing. I wish you was doing more of the exams I was doing! thanks!
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    (Original post by Jimmy20002012)
    <br />
    <br />
    One last question with diodes:<br />
    <br />
    In the attachment look at the circuit diagram and the question is what happen to AC and BC if D2 is reversed, the answer for both is 12v, could you explain this? <br />
    <br />
    <br />
    Posted from TSR Mobile


    Posted from TSR Mobile
    When D2 is reversed, there will be all voltage and almost no current at that point. So it is 12V at BC. (D2 allows voltage only and D1 allows Current only). As D1 is forward bias, is it allowing only current to flow. However, when D2 is reversed, it is allowing only voltage to flow (no current). So AC is also 12V. Is it clear???
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    (Original post by StalkeR47)
    When D2 is reversed, there will be all voltage and almost no current at that point. So it is 12V at BC. (D2 allows voltage only and D1 allows Current only). As D1 is forward bias, is it allowing only current to flow. However, when D2 is reversed, it is allowing only voltage to flow (no current). So AC is also 12V. Is it clear???
    I still don't get why AC is 12V, surely some current would flow and voltage be used up as D1 is forward biased


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    (Original post by Jimmy20002012)
    I still don't get why AC is 12V, surely some current would flow and voltage be used up as D1 is forward biased


    Posted from TSR Mobile
    The basic rule is that If the diodes were in series (and in opposite direction), no current would flow until the voltage exceeded the breakdown voltage of the diode. However, if the diodes were connected parallel you would see a voltage drop (forward voltage drop of diode) across the diodes. Get it now???
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    (Original post by StalkeR47)
    The basic rule is that If the diodes were in series (and in opposite direction), no current would flow until the voltage exceeded the breakdown voltage of the diode. However, if the diodes were connected parallel you would see a voltage drop (forward voltage drop of diode) across the diodes. Get it now???
    If the pd across AC is 12 V, which is the same as the emf, doesn't that imply that the pd across the 3k Ohm resistor is zero? (which is untrue?)
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    (Original post by StalkeR47)
    The basic rule is that If the diodes were in series (and in opposite direction), no current would flow until the voltage exceeded the breakdown voltage of the diode. However, if the diodes were connected parallel you would see a voltage drop (forward voltage drop of diode) across the diodes. Get it now???
    I think I get it now, so in that example as the diodes and resistors are in series there is no current but voltage which is why there is a 12 V anywhere. So it doesn't matter if the circuit has some forward and reverse diodes, if the circuit consists of one reverse diode there is no current so voltage remains constant everywhere? Could you just explain the parallel diode connection again, sorry for soo many questions, thanks


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    (Original post by Epic Flawless)
    If the pd across AC is 12 V, which is the same as the emf, doesn't that imply that the pd across the 3k Ohm resistor is zero? (which is untrue?)
    Because diodes are in opposite direction to each other, there will be no flow of current, but voltage only. And if there is no current through 3k Ohm resistor, there will be no voltage across 3k Ohm resistor. I.E. v=ir (when I is 0), v=0xr which is also 0. It is clear now?:confused:
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    Can someone help please? JUNE 2012, Q) 5,iii

    stopping d proportional to v^2

    v = 32


    PLEASE HELP!!!!!!!
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    (Original post by Jimmy20002012)
    I think I get it now, so in that example as the diodes and resistors are in series there is no current but voltage which is why there is a 12 V anywhere. So it doesn't matter if the circuit has some forward and reverse diodes, if the circuit consists of one reverse diode there is no current so voltage remains constant everywhere? Could you just explain the parallel diode connection again, sorry for soo many questions, thanks


    Posted from TSR Mobile
    It ok!! In series, no current flow in either direction.
    In parallel, current flow in either direction when the voltage reaches about 0.6V.(for silicon diodes).
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    (Original post by SexyNerd)
    Can someone help please? JUNE 2012, Q) 5,iii

    stopping d proportional to v^2

    v = 32


    PLEASE HELP!!!!!!!
    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.
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    does any1 hav any revision materials for electricity. some of dat ish is a mad ting
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    (Original post by StalkeR47)
    It ok!! In series, no current flow in either direction.
    In parallel, current flow in either direction when the voltage reaches about 0.6V.(for silicon diodes).
    What about the reverse diode position in parallel?


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    (Original post by TaranC)
    does any1 hav any revision materials for electricity. some of dat ish is a mad ting
    Hello!!! I have just asked for that from one of my friend. He has sent me few notes and will send me more later. As I get them, I will surely post them to TSR. Here are few-----http://s1195.photobucket.com/user/po...tml?sort=2&o=0
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    I've attached the answers to all the long questions from june 2009 onwards
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  1. File Type: pdf Physics unit 1 long answers.pdf (175.8 KB, 898 views)
 
 
 
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