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Current in a parallel resistor?

  1. May 14, 2012 #1
    1. The problem statement, all variables and given/known data

    The problem involves a battery of emf 10V [with no internal resistance] and a combination of resistors. Here is the diagram provided:

    A picture of the diagram can be found here: http://gyazo.com/7791710d0645de38cfab59f0ac4740ec

    (Couldn't get the image feature to work)

    The previous questions asked for:
    • Total resistance, which I found to be 5Ω, which is correct
    • Next part says show that each resistor is 3Ω, which I did correctly

    The part that I am stuck on is calculating the current through resistor Y.

    2. Relevant equations
    R=V/I is the only equation I need to use


    3. The attempt at a solution

    The question can be found here (its question 7): http://store.aqa.org.uk/qual/gce/pdf/AQA-PHYA1-W-QP-JUN11.PDF
    The answers can be found here: http://store.aqa.org.uk/qual/gce/pdf/AQA-PHYA1-W-MS-JUN11.PDF

    Since the pd across Y needs to be found before I try to figure out the current, I simply used V = IR, this gave me 6V across Y (2A x 3Ω). So then I simply did I = V/R which gave me an answer of 2A (6V / 3Ω).

    My answer was clearly wrong as current splits in parallel so it cannot be 2A which was stated in the question as the reading of the ammeter, which was in series.

    I couldn't figure it out, so I checked the mark scheme and it said I had to do:

    10V (emf) - (2Ax3Ω) which gives you 4V across Y and then you get 1.3A across Y.

    My question is, why do I have to take away the voltage across Y from the emf? Surely that would just give me the remaining emf in the circuit, not the pd across Y.
     
  2. jcsd
  3. May 14, 2012 #2

    tiny-tim

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    Hi TheGreenMarin! :smile:
    the remaining emf is the pd between those two dots, isn't it? :wink:
     
  4. May 14, 2012 #3
    Yes, I think so, but how would this help me find the current through Y?

    The mark scheme says I have to take the voltage across Y from the emf to give me 4V then use R = V/I to find the current.

    I'm still not sure why I have to take the voltage across Y from the emf, why can't I use the voltage across Y by itself?

    Thanks for reply
     
  5. May 14, 2012 #4

    tiny-tim

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    not following you … what's the difference? :confused:

    from KVL, the sum of the pds must be the emf

    the pd between the dots is the emf minus the pd across Z

    and of course it's the same pd across Y and across W and X

    (and what did you mean by "the voltage across Y by itself"?​
     
  6. May 14, 2012 #5
    Yes, I am confused :confused:

    The question asks for the current going through Y, so I know the resistance of Y is 3Ω, the pd across Y I worked out as 6V, which is wrong. According to the mark scheme I have to do 10-6 to get 4V, which is the pd across Y, but I don't understand, why did I have to do that, why couldn't I just use the 6V I worked out?

    Thanks, electricity isn't my strong point :uhh:
     
  7. May 14, 2012 #6
    That makes more sense :D, so I needed to work out the pd across Z first?
     
  8. May 14, 2012 #7

    tiny-tim

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    let's see where your 6V came from …
    ah, but the current across Y isn't 2A, you'd need to use KCL :wink:
     
  9. May 14, 2012 #8
    I think I know where I went wrong. I was following the circuit from left to right, instead of using conventional current flow, so I never actually passed the Z resistor so I excluded it from my calculations.

    Is that correct?
     
  10. May 14, 2012 #9

    tiny-tim

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    nooo :redface:

    for KVL, you must always go the whole way round any loop
     
  11. May 14, 2012 #10
    I am going to fail this exam. So I have to consider the whole circuit, right I see. I really need help with electricity.
     
  12. May 14, 2012 #11

    tiny-tim

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    in a way, what you've been doing does make sense …

    if you call the potential 0 at one end of the circuit, and V at the other end,

    and if you know what I is,

    then yes you can go from either end (starting at 0 or V as the case may be), subtracting or adding the potential difference of each component until you get to the point you want

    (but that's not easy if you have components in parallel, as with W X and Y)
     
  13. May 14, 2012 #12
    practise makes perfect...
     
  14. May 14, 2012 #13
    You can try these Electricity exam practise questions:
    http://www.online-exam-solutions.co.uk/subjects/physics/physics-as-level/electricity [Broken]

    I got quite low in my first attempt, but then kept going through until i was confident.
    Apparently if you can do them all your on track for top marks.
    I find the 'suggest' and 'Explain' parts the hardest.
     
    Last edited by a moderator: May 6, 2017
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