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What would be the ameter reading?

  1. Jul 25, 2007 #1
    Hi All,

    I have attached a bmp file indicating the diagram. Now, in the diagram shown alongside the cell and the ammeter both have negligible resistence.The resisters are identical with switch K open. the ameter reads 0.6 A . what will be the ammeter reading when the switch is closed?

    Few more questions...

    Q1 . Two resistence when connected in parellal give resultent value of 2 ohm. calculate the value of each resistence .

    Q2 A resister a resistence of 176 ohm. how many of these resisters should be connected in parellal so that their combination draws a current of 5 ampers from a 220 volt supply line.

    hare krishna
    Alok
     

    Attached Files:

  2. jcsd
  3. Jul 25, 2007 #2

    berkeman

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    Staff: Mentor

    Thread moved from General Physics to Homework Help, Intro Physics.

    Welcome to the PF, AlokBehria. Homework and coursework questions should go here in the Homework Help forums, and not in the general technical forums. Also, we require that you show us the relevant equations and show us some of your own work, before we can begin to give you tutorial help and advice.

    So, how would you approach these questions? Also, I think there is something wrong with the questions. When you close switch K, where will the smoke and fire be seen first?
     
  4. Jul 25, 2007 #3
    I will wait for the file to load, but for now I'll help you with your other 2 questions.

    Q1.

    The total resistance of parallel resistors is:

    [tex]R_T = \frac{1}{R_1} + \frac{1}{R_2} + ... + \frac{1}{R_n}[/tex]

    A shortcut to finding the total resistance of 2 parallel resistors is:

    [tex]R_T = \frac{R_1R_2}{R_1 + R_2}[/tex]

    Q2.

    Using the first equation above and Ohm's Law, you have:

    [tex]V = I R_T[/tex]

    [tex]V = I (\frac{1}{R_1} + \frac{1}{R_2} + ... + \frac{1}{R_n})[/tex]

    Now, you know your voltage and your current, so you can find [tex]R_T[/tex]. Once you have [tex]R_T[/tex] and you know what value the resistors need to be, you can find how many resistors you need.
     
    Last edited: Jul 25, 2007
  5. Jul 25, 2007 #4
    Thanks for the help, but I am not able to solve the question as per your suggestion, so would you elaborate more on both the question?

    hare krishna
    Alok
     
  6. Jul 25, 2007 #5
    Which question can't you solve? 1 or 2?
     
  7. Jul 25, 2007 #6
    Q2, please help me.
     
  8. Jul 25, 2007 #7

    berkeman

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    Staff: Mentor

    EugP gave you all the help you need in post #3. Stop trying to have us solve the question for you -- that's not how the PF works. You are *required* to show us your own work in order for us to help you. Please show us some of your work on these questions.
     
  9. Jul 25, 2007 #8
    Alright this should be more than enough information to help you solve the problem.

    Like I said before:

    << Sorry, I have to delete this as too complete of an answer. Wait for more work by the OP. -berkeman- >>
     
    Last edited by a moderator: Jul 25, 2007
  10. Jul 25, 2007 #9
    we used V=IR_t

    [tex]220 =5 (\frac{1}{R_1} + \frac{1}{R_2} + ... + \frac{1}{R_n})[/tex]

    Now if we solve the equestion, we diden't find the answer.
     
    Last edited: Jul 25, 2007
  11. Jul 25, 2007 #10

    berkeman

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    Staff: Mentor

    Remember, that question says that all the resistors are equal and in parallel. What happens to the total resistance when you put two equal-value resistors in parallel? What about 3 in parallel? What about N in parallel?
     
  12. Jul 25, 2007 #11
    No you're right I got a little carried away. Won't happen again.

    AlokBehria, why don't you post any work you've done so far, with as many descriptions as possible of your thought process, so we could GUIDE you to how to solve the problem.
     
  13. Jul 25, 2007 #12
    have you one through the Q2 ? If yes, then where wud you find individual [tex]R_1[/tex] , [tex]R_2[/tex] and Rn
     
  14. Jul 25, 2007 #13
    I'm sorry but I don't understand your question.
     
  15. Jul 25, 2007 #14
    I woluld again repeat my question.

    A resister has a resistence of 176 ohm. how many of these resisters should be connected in parellal so that their combination draws a current of 5 ampers from a 220 volt supply line

    One more request, if you can open up that attached file..

    Now, the question is ..

    In the diagram given in the file, the cell and the ammeter both have negilagable resistence . the resistors are identical with the switch K open,the ammeter reads 0.6 amp. what will be the ammeter reading when the switch is closed. The answer shows the current in ammeter is 0.9 amp.

    hare krishna
    Alok
     
  16. Jul 25, 2007 #15
    Post your work so we can see how far you've gotten.
     
  17. Jul 26, 2007 #16
    ammeter reading

    when three resistors with identical length connected with switch k open,the ameter reads 0.6 amp. reading
    what will be the ammeter reading while the switch is closed.
    it is noted that the cell and the ammiter both have negligiable resistence
     
  18. Jul 26, 2007 #17
    EUGP, if you can direct me a link so that I can go through the manuals to solve my problem.

    hare krishna
    Alok
     
  19. Jul 26, 2007 #18
    What manuals? I don't understand what you want, I gave you all the information you need. I'm not going to give you the answer.
     
  20. Jul 26, 2007 #19

    andrevdh

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    Homework Helper

    When the switch is closed the resistance in the circuit becomes zero, that is the power supply is shorted out. So it is actually not possible to say what the reading will be - each power supply will respond differently by tying to send the maximum possible current throught the wire that is shorting it out.

    As far as the parallel combination of similar resistors goes - note that their total resistance needs to be 220/5 ohm.

    http://www.circuit-magic.com/series_parallel.htm

    http://www.physics.uoguelph.ca/tutorials/ohm/index.html
     
    Last edited: Jul 26, 2007
  21. Jul 26, 2007 #20

    berkeman

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    Staff: Mentor

    That would be the smoke that I was alluding to :eek:
     
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