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    (Original post by Jimmy20002012)
    The normal potential divider equation is:

    Voltage output= Voltage in X R2/Total resistance in branch

    So in the January 2013 paper when you are meant to find the voltage output of D-F I thought it would be 6 x 5/15= 2V, but it is the other way round, could you explain?


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    To calculate resistance of a resistor in a potential divider you multiply the cell current by the resistance of the resistor that you are trying to calculate the pd for, divided by the total resistance of the resistors in series.

    So for D-F the calculation would be 12x5000/10000+5000=4V.

    My explanation wasn't very clear but if you're still unsure, just ask!
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    (Original post by PickwickianGeek)
    To calculate resistance of a resistor in a potential divider you multiply the cell current by the resistance of the resistor that you are trying to calculate the pd for, divided by the total resistance of the resistors in series.

    So for D-F the calculation would be 12x5000/10000+5000=4V.

    My explanation wasn't very clear but if you're still unsure, just ask!
    Why do you times by 12, as 6V has been used up already from A-C? Or is it because 12 v goes trough each branch?


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    (Original post by IWantSomeMushu)
    These are terrific notes for electricity!
    yes you are right! Here is particles, quantum phenomena and electricity good notes!!! http://physicsnotes0.tripod.com/id16.html
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    (Original post by Jimmy20002012)
    Why do you times by 12, as 6V has been used up already from A-C? Or is it because 12 v goes trough each branch?


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    BASIC RULE__________ Voltage in series adds up to total. Voltage in parallel is the same. Current in series is the same. Current is parallel adds up to total. How you doing by the way Jimmy? Last day today huh.
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    (Original post by sceezy)
    Can anyone help with question 6b jan 2013 ?????

    PLEASE!!!
    Can you give me a link to where you got the 2013 paper? Cant find it :L
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    (Original post by Jimmy20002012)
    Why do you times by 12, as 6V has been used up already from A-C? Or is it because 12 v goes trough each branch?


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    Jimmy! Do you need any help?:confused:
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    (Original post by StalkeR47)
    BASIC RULE__________ Voltage in series adds up to total. Voltage in parallel is the same. Current in series is the same. Current is parallel adds up to total. How you doing by the way Jimmy? Last day today huh.
    Cool yeah I am okay, how's your last minute revision goi g, your obviously gonna ace this exam. Any prediction for the 6 marker?


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    (Original post by Jimmy20002012)
    Why do you times by 12, as 6V has been used up already from A-C? Or is it because 12 v goes trough each branch?


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    Yep, 12v goes through each branch.
    So for A-C the calculation is 12x20000/40000=6v
    and C-D would be 12-10=2v. This is because the voltage of the circuit must equal the voltage of the battery. Hope this helps!
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    Anyone got any predictions for the paper?
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    (Original post by Jimmy20002012)
    Cool yeah I am okay, how's your last minute revision goi g, your obviously gonna ace this exam. Any prediction for the 6 marker?


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    I am not going to ace the exam but I really want to get an A. To be honest, I am still revising electricity so if you are stuck with anything or any question.. ask me. I do not have any predictions for the 6 marker. However, 6 marker can increase your grade by 1. Just try your best on the 6 marker.
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    (Original post by PickwickianGeek)
    Yep, 12v goes through each branch.
    So for A-C the calculation is 12x20000/40000=6v
    and C-D would be 12-10=2v. This is because the voltage of the circuit must equal the voltage of the battery. Hope this helps!
    That has helped at lot, thanks. The current in each branch won't be the same as they both have different resistances, in the first branch it will be 12/40= 0.3A Nd the second branch will be 12/15= 0.3A which equals 1.2, which is about the goal current as that is 1.1009A


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    If I'm asked to draw a circuit diagram to test whether a component follows ohms law, what symbol do I put for the "component I am testing"?

    hope that made sense lol
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    Do not predict the paper! If you do, you will mostly revise the stuff which you think might come up but it could be totally opposite in the exam and you might end up getting a C. Revise everything!!!!
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    (Original post by sceezy)
    If I'm asked to draw a circuit diagram to test whether a component follows ohms law, what symbol do I put for the "component I am testing"?

    hope that made sense lol
    Remember it should have a constant resistance if a conductor is ohmic. So, you will use a variable resistor to get few measurements, you will use an ammeter and a voltmeter with the circuit to obtain IV graph. You will calculate the resistance. You will see if the graph has a positive gradient (meaning it is a straight line and as current increases, voltage increases). Hope that helps.
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    (Original post by Steroidsman123)
    Can you give me a link to where you got the 2013 paper? Cant find it :L
    I have it downloaded, ill attach it below...
    Attached Images
  1. File Type: pdf AQA-PHYA1-QP-Jan13.pdf (379.1 KB, 77 views)
  2. File Type: pdf AQA-PHYA1-W-MS-JAN13.pdf (126.5 KB, 63 views)
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    (Original post by StalkeR47)
    Remember it should have a constant resistance if a conductor is ohmic. So, you will use a variable resistor to get few measurements, you will use an ammeter and a voltmeter with the circuit to obtain IV graph. You will calculate the resistance. You will see if the graph has a positive gradient (meaning it is a straight line and as current increases, voltage increases). Hope that helps.
    yh i got that but, could u please tell me what the circuit diagram would look like. Also we are testing a component but i dont know what the component so what symbol would i use in the diagram
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    (Original post by sceezy)
    I have it downloaded, ill attach it below...
    what do you need help with?
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    (Original post by StalkeR47)
    what do you need help with?
    the 6 marker and 6b
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    (Original post by sceezy)
    yh i got that but, could u please tell me what the circuit diagram would look like. Also we are testing a component but i dont know what the component so what symbol would i use in the diagram
    You will have to use components. If you use symbols, you will not get any marks!! This is even said in the mark scheme. The circuit diagram would look like this.... Assume the resistor to be a variable resistor.:cool:
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    (Original post by sceezy)
    the 6 marker and 6b
    Ok wait...
 
 
 
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