Wednesday, 15 May 2013

Compute the heat generated while transferring 96000 coulomb of charge



Question 2:
Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.
Solution:
Given Charge (Q) =96000C
Time (t)=1hr=60×60=3600s
Potential difference V=50volts
Now we know that H= VIt
So we have to calculate I first
As I=Q/t
∴ I=96000/3600=80/3 A

16 comments:

  1. This comment has been removed by the author.

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  2. You could have done it directly by H=VQ.

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    Replies
    1. yaa that's too correct but this is also write.... There are many methods to do numericals....

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    2. How can that equation come..

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    3. How did this equation came?

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    4. H=VIT
      I=Q/T
      H=V×Q/T×T
      T AND T CANCEL
      THEN H=VQ

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    5. Answer will be different in each case why ?it should be equal as vq =vit

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    6. Answer is coming different

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  3. Thanks for this!

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  4. why is time multiplyed twice?

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    Replies
    1. its actually ( 60 seconds in a minute x 60minutes in an hour)

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  5. This anwer may be right but in lakhmir singh 4788.4kj is ans

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  6. Replies
    1. =(50×80×3600)/3
      3600/3= 1200
      =50×80×1200j
      =4800000 j
      =4.8×10to the power 6 joule

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  7. Answer is same in both cases.

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  8. Right answer
    But we can do it by using vq formula

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