Power dissipated in bulbs: Solved example (Hindi)

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  • Опубліковано 19 лют 2019
  • Let's tackle questions in which we have circuits with devices (like bulbs) whose power ratings are mentioned.

КОМЕНТАРІ • 50

  • @Yuvrajsingh-3ub
    @Yuvrajsingh-3ub 2 роки тому +7

    I love your videos so much That I forget to comment... But You are Giving Quality Education.. Your Videos are very very Helpful.. No words to thank You..

  • @shagunsingh5505
    @shagunsingh5505 4 роки тому +6

    Sir as u have told power dissipated is heat producing by object then why u re calculating power of bulb in both questions
    Isn't it confusing 🤔

  • @anupama5646
    @anupama5646 3 роки тому +3

    Killer concept👍👍

  • @priya03_94
    @priya03_94 2 роки тому +1

    Ur voice👍

  • @PragnaPatel-kf9qs
    @PragnaPatel-kf9qs 7 місяців тому

    Sir you have explained the concepts in previous videos so well that I could solve these without any clues ..this chapter was honestly so confusing for me but thanks to you that i can do the numericals so easily !! I love the way you explain everything with analogies❤

  • @shadowgamerA
    @shadowgamerA 3 роки тому +4

    Ram Sir please please include an example of bulbs connected in parallel that would be very helpful

  • @naik_shrav
    @naik_shrav 3 роки тому +3

    LOVE YOU SIR..........

  • @garomeoyt2945
    @garomeoyt2945 3 роки тому +3

    Sir ji you are good

  • @asadurrehman7306
    @asadurrehman7306 3 роки тому +2

    ap na 5v battery wala resister ma current find keyu nahi kia?
    please answer

  • @neelamdhanawat840
    @neelamdhanawat840 4 роки тому +3

    Sir u are great

  • @nitinthakur904
    @nitinthakur904 2 роки тому +1

    sir such me maja aa gaya

  • @abhinaamrawat1680
    @abhinaamrawat1680 2 місяці тому

    Great sir 😊

  • @poojachachra120
    @poojachachra120 4 роки тому +4

    Amazing, tysm

  • @SunilKumar-dk5is
    @SunilKumar-dk5is Рік тому

    Sir i have a doubt in the 2nd question, why we can't use the above 2 formula for calculating power. Please sir clear it i have doubt that when we have to use these formula

  • @anmolsingh6158
    @anmolsingh6158 10 місяців тому

    thanks

  • @chandrashekarterdal3794
    @chandrashekarterdal3794 2 роки тому +4

    In question 1 ... Like in general if 10V = 50W then 5 V will be it's half right it should be 25W no sir... Sir plz help.. I am confused.. Even when did first time i got 25 .... First I divided 50/10 to calculate 1V value then I got answer 5 then I multiplied it to 5V to get power of 5V ... Then I got 25 ... Sir plzzzzz help😭😭😭

  • @jeetdattani5535
    @jeetdattani5535 5 років тому +5

    Thank you very much... For this.. I now understood how to solve that kind of problems..

  • @abbasmehdi2923
    @abbasmehdi2923 4 роки тому +4

    At 8 : 59 , see...why same current

  • @abbasmehdi2923
    @abbasmehdi2923 4 роки тому +7

    Sir , in question 1 ;
    We can calculate current and use the direct formula P = VI
    But , when we calculate current I ,
    Then , I = 50/10 = 5ohms
    Now , V = 5 and I = 5 ,
    So , P = VI = 5×5 = 25
    Please help me in this

    • @pratyushsrivastava3405
      @pratyushsrivastava3405 3 роки тому +3

      Always find the resistance bcoz current may or may not be same

    • @chandrashekarterdal3794
      @chandrashekarterdal3794 2 роки тому +1

      I also had the same question🙋

    • @ranjitkumardas5263
      @ranjitkumardas5263 2 роки тому +1

      Look u have taken P= VI
      And then I = P/V = 50/10 = 5
      So in these cases when potential difference is 10 V then current is 5 ampere .. but in 2nd case how can u say that 5 ampere of current will pass through the bulb bcoz potential diff. = 5 V..
      That's why we need to measure resistance because it will remain same in both cases..🙂

    • @chandrashekarterdal3794
      @chandrashekarterdal3794 2 роки тому +1

      @@ranjitkumardas5263 tq for this explenation it helped a lot..... 😊😊😊😊thank you again

    • @chandrashekarterdal3794
      @chandrashekarterdal3794 2 роки тому +1

      @@ranjitkumardas5263 again a question.. Current will not be same in both cases??

  • @shashsankgupta5891
    @shashsankgupta5891 4 роки тому +4

    Sir AAP ekdm mast pdhate ho

  • @sanjeevrajak2627
    @sanjeevrajak2627 5 років тому +5

    Continue ram sir

  • @surjitkaur9236
    @surjitkaur9236 5 років тому +6

    But voltage of each bulb was given that is 20V amd 20V so why couldn't we simply use the formula v^2/R?

  • @sanjeevrajak2627
    @sanjeevrajak2627 5 років тому +4

    Thank you

  • @nalin31081
    @nalin31081 Рік тому

    @ 5:06 Q.2

  • @VG__
    @VG__ 5 років тому +5

    Plzz tell
    Why can't we first find "i" and then apply formula as P=VI..WE are gvn both V and P ..can't we calculate I and then solve it further

    • @proloysrot4114
      @proloysrot4114 4 роки тому +1

      Because current won't be produced from the bulb resistance it will produce from your battery which is 5 voltage.
      so in the pathway of your electricity you have to know how much resistance electricity will face that's why we should know resistance first in this case.
      But when our bulb is been connected with the main line, we first calculate current flow (I) because we have abundant voltage.
      In this case your voltage source has limitation.

    • @proloysrot4114
      @proloysrot4114 4 роки тому +1

      Btw, after knowing resistance, you can calculate how much current will flow, and then you simply put the calculation in P=VI formula. The answer and method will be same

  • @technicalkanhya2221
    @technicalkanhya2221 2 роки тому

    Second question me buld destroyed ho jayega

  • @devanshrastogi805
    @devanshrastogi805 5 років тому +3

    1st like

  • @khushi_sandhu
    @khushi_sandhu 5 років тому +2

    Which app are you using in this video to teach us

  • @mahima5235
    @mahima5235 5 років тому +3

    2 viewer

  • @devanshrastogi805
    @devanshrastogi805 5 років тому +5

    1st view

  • @khushi_sandhu
    @khushi_sandhu 5 років тому +2

    plz tell me

  • @manutv1935
    @manutv1935 4 роки тому +3

    23 volts and 2 amperes find power