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A 2 µF capacitor C₁ is first charged to a potential difference of 10V using a battery. Then the battery is removed and the capacitor is connected to an uncharged capacitor C₂ of 8 µF. The charge in C₂ on equilibrium condition is --------- µC. (Round off to the Nearest integer)

 


0 3 Views | Posted a month ago
Asked by Shiksha User

  • 1 Answer

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    Answered by

    alok kumar singh | Contributor-Level 10

    a month ago

    Let the charge on C? be q µC. For the capacitor network:
    q/C? = (C? * 10 - q) / C? ⇒ q/8 = (2 * 10 - q) / 2 ⇒ q = 16

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