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Current electricity

  1. Jul 2, 2009 #1
    1. The problem statement, all variables and given/known data

    A millivoltmeter of resistance 5[tex]\Omega[/tex] gives fullscale deflection,when connected across 75mV.
    Inorder to use this millivoltmeter as an ammeter,a resistance coil with a third terminal C,is connected between terminals A and B ,of the millivoltmeter.
    When a current of 1.5A is flowing between A and C,and a current of 0.15A flows between A and B,the instrument shows full scale deflection.
    Find the resistances of the 2 portions AC and BC

    Answers given are :AC=1/18 ohms
    BC =0.5 ohms
    2. Relevant equations

    Kirchoffs' laws
    [tex]\sum[/tex]E = [tex]\sum[/tex]IR

    3. The attempt at a solution

    I managed to draw a diagram of the circuit(I hope it's correct?),but now, I'm having trouble trying to determine the currents(I and I-i,as labelled in the diagram).
    Using,
    V=IR for the millivoltmeter
    75*10^-3 = i * 5
    I get i= 15mA ,which, I think ,has to be the current flowing through the milivoltmeter inorder to obtain a fullscale deflection,but,
    what is I? Is it 1.5A+0.15A?



    Thank you
     

    Attached Files:

  2. jcsd
  3. Jul 2, 2009 #2

    vk6kro

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    I agree about the meter current.

    Call the shunt resistors R1 and R2. R1 is between A and C. R2 is between B and C.

    Then you take the A to B current and work out the total value of the shunt resistors. All except the 15 mA flows through the shunt resistors and you are given the voltage across them.
    So, you then have a value for R1 + R2.

    When 1.5 amps flows from A to C, 15 mA still flows through the meter (and R2) and the rest through R1, so you have a ratio between R1 and (the total of the meter resistance and R2).

    This gives 2 simultaneous equations. You can then work out R1 and R2.
     
  4. Jul 2, 2009 #3
    Thanks for replying.
    Okay,so the current through the branch ACB = 0.15 - 0.015A = 135mA = I-i
    Now using,
    V = (I-i)(R1+R2)
    75*10-3/135*10-3 = (R1+R2)
    5/9 ohms = R1+R2,

    Now,I don't get this part,
    When 1.5A flows through AC,shouldn't the current through R1 also be 1.5A?
     
  5. Jul 2, 2009 #4

    vk6kro

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    No, because the meter and R2 are across R1 and the total current of 1.5 amps splits between the two paths.
    But you know that one path has 15 mA in it and the other has 1500 - 15 mA in it, so you can work out the ratio between the resistances in each path. It would be 15 / 1485.
    You can then use this to get one of your simultaneous equations.

    If you have a calculator I would get those fractions into actual decimal values if possible.
     
  6. Jul 3, 2009 #5
    Thank you so much,sir
    Sadly,we're not allowed to use calculators,but I've managed to get the answers,without having to convert it to decimal values.
    But, I'm a bit doubtful of the method used.Can you please check it for me?
    5/9 = R1 + R2 -----------(1)

    15*(5+R2) = 1485*R1
    75 + 15R2 = 1485R1
    75 = 1485R1 - 15R2
    5 = 99R1 - R2 ----------(2)

    (1)+(2),
    5/9 + 5 = 100R1
    5 + 45 = 900R1
    R1 = 50/900 = 1/18 ohms

    5/9 = R1 + R2 -----------(1)
    R2 = 5/9 - 1/18 = 1/2= 0.5ohms

    THANK YOU
     
  7. Jul 3, 2009 #6

    vk6kro

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    I think I'll print that out and keep it.

    Most people would find that a difficult problem with a calculator, but you worked it out without one.

    I checked through it and it makes sense so you understand the process.
    I got R1= 0.0505 ohms and R2=0.505 ohms but your fractions may be more accurate.

    Which country are you in? Was this for High School or University?
    How many girls and boys are in your class?
     
  8. Jul 4, 2009 #7
    I'm a A-level student from Sri Lanka.I used to go to an all girls' school,but not anymore...(there were only about 6 students in my class)
    so now I'm doing my ALs privately,i go to classes outside and there are boys and girls(50 max)& even students from popular schools attend these classes.
    I do use calculators when I work sums at home and classes,but we're not allowed to use them at our exams,we're provided with log books instead.since our exams are close,I thought I should stop using cals.

    Anyway thanks for the help
     
  9. Jul 5, 2009 #8

    vk6kro

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    Very interesting but disturbing as well.

    I guess they can't help the class sizes being so big. A class of 50 means each student gets almost no individual attention from the teacher. Have you noticed that those who do get the teacher's attention quickly fire off new questions while they have the chance?

    But they can certainly fix the "no calculators in exams" rule. This is just nonsense.
    If the exam is in arithmetic, sure, you don't allow calculators. But in Physics or Chemistry, the actual calculations are not the object of the exam. It is the logic of assembling the numbers for calculation that matters.

    Some calculators that have just a few numeric memories do not allow any cheating and even if a small range of "permitted" calculators are prescribed, this would be a lot better than using LOG tables. I had forgotten about those. They were horrible.

    Maybe some of the parents could petition for this silly rule to be changed?
    Just tell the teachers they can ask more questions on the exam paper if they allow calculators. They should love that.
    With Scientific calculators now available for $5, class sets could be purchased so that everyone had an equal calculator if $5 might be a problem.
     
  10. Jul 5, 2009 #9
    Well,I don't quite agree with you.
    We don't mind using log books at all.cause our exams(OL or ALs) are sort of designed in a way that we don't really have to deal with decimals,sqrts,etc.(for eg,they say take pi as 3,cos theta,tan theta,sqrt as ...etc)The examiners use alot of approximations & the answers we get using log tables are slightly diff. from using cals,so...they know when you cheat.
    NO WAY!I'd rather work with no LOG books than have more questions added and I'm sure not just the students,even the teachers would agree on this.

    That is a good idea,but $5 would mean 500 Srilankan Rupees,(although my calculator cost me about 900SLR),not everyone could afford it and I don't think the government could pay for all the students(we're talking of 100s n 1000s of students,here).we have more imp. things to worry about.

    well,its true that we don't get as much attention as we should,but it's not that bad.We have very good lecturers and they have alot of experience dealing with classes like these,and we have alot of good schools too(n a few not so gud schools,but i guess you can find them anywhere).& if u live in the city ,& have the money you can go for smaller group classes and even individual lessons.
    but its a bit diff. for my friends from outstation,they have to travel a long way to get to good classes.Althogh some of the best professors conduct classes there too,these tend to get too crowded(abt 300+ students).it's tough for them but I guess that's why their cutoff marks for uni.entrance is relatively low compared to ours.
    Our exams are very competitive scholarship exams,so if u do really...really well,you get free education(be it medicine,dental,engineering etc)but,sadly,only a few get selected based on the Z-score,othere well...once again if you can afford to study abroad,then you're good.
    I know this sounds horrible,but noone complains...afterall we're in a developing country n I'm sure the situation would improve in maybe a couple of yrs(hopefully):smile:

    Anway thanks for your concern,and i really hope i don't get thrown out for going off topic cause i really really like it here
     
  11. Jul 5, 2009 #10

    vk6kro

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    I doubt if anyone would worry about this being off topic since there are just the two of us in this thread. However, delete your reply if you like. I've already got an email of it.

    When you become Prime Minister you can improve the education system. :)
    Sri Lanka's future is looking very bright with recent developments in the North of the country.

    Very interesting to hear your comments, though.
     
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