In the given circuit, the current (I) through the battery will be

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  • Опубликовано: 6 янв 2025

Комментарии • 14

  • @KRISHNAVENI-zt2dq
    @KRISHNAVENI-zt2dq 11 месяцев назад +9

    Fabulous explanation this is what an student expect 🙏 🙏 🙏 ❤😊

    • @itzshortaze2m656
      @itzshortaze2m656 9 месяцев назад +1

      Noob tu yaha chapter padhne aua hai a solution dekhne

  • @ẞhåm2î_19
    @ẞhåm2î_19 10 месяцев назад +2

    Behad must explanation hai

  • @gingerff1998
    @gingerff1998 10 месяцев назад +2

    fab explantion tq very much sir

  • @roshniWaghmare-ut2qg
    @roshniWaghmare-ut2qg 8 дней назад

    Increadible 🎉

  • @Mr.Zenith-x8h
    @Mr.Zenith-x8h Месяц назад

    Great 🎉 Explanation

  • @anumathai5620
    @anumathai5620 2 месяца назад

    Thankss a lott sirrr..u explainedd VERY VERY VERYY NICELY

  • @boringraj1341
    @boringraj1341 Год назад +1

    king explanation

  • @ManuRai-h5i
    @ManuRai-h5i 10 месяцев назад +1

    Really u explain in a very easy way sir
    Thanku so much sir ♥️
    Pura diode ka concept ik hi ques m clear ho gya sir
    Once again thank you so much sir ♥️😊😊

  • @GamerDude3222
    @GamerDude3222 Год назад +5

    Splendid explanation! I have a doubt, how do we know when to split the current? Like, why doesn't the current split at the point on the left where the two wires with the 10 ohm resistors are joined together?

    • @indiraverma6335
      @indiraverma6335 Год назад +1

      Even if we split the current on the left junction still you would see a wrong orientation of D2 since at left side of battery in this diagram potential would be low.

    • @ananttyagi7372
      @ananttyagi7372 9 месяцев назад +1

      Yes, I had the same doubt, but sometimes, it's better to look at the potential difference across the diodes. If you look at the left part of the diode D2, we can assume the potential to be 0. and on the right part, the potential will definitely be higher. Hence, it would be reverse biased.

  • @Uchihazoro.
    @Uchihazoro. Год назад +2

    Why we will not solve parallel resistor first